Mobile Numeric Keypad Problem - GeeksforGeeks

24 May, 2021
Given the mobile numeric keypad. You can only press buttons that are up, left, right or down to the current button. You are not allowed to press bottom row corner buttons (i.e. * and # ). 
Given a number N, find out the number of possible numbers of given length. Examples: For N=1, number of possible numbers would be 10 (0, 1, 2, 3, ...., 9) For N=2, number of possible numbers would be 36 Possible numbers: 00,08 11,12,14 22,21,23,25 and so on. If we start with 0, valid numbers will be 00, 08 (count: 2) If we start with 1, valid numbers will be 11, 12, 14 (count: 3) If we start with 2, valid numbers will be 22, 21, 23,25 (count: 4) If we start with 3, valid numbers will be 33, 32, 36 (count: 3) If we start with 4, valid numbers will be 44,41,45,47 (count: 4) If we start with 5, valid numbers will be 55,54,52,56,58 (count: 5) .................................... ....................................We need to print the count of possible numbers.
N = 1 is trivial case, number of possible numbers would be 10 (0, 1, 2, 3, ...., 9) For N > 1, we need to start from some button, then move to any of the four direction (up, left, right or down) which takes to a valid button (should not go to *, #). Keep doing this until N length number is obtained (depth first traversal).
Recursive Solution: Mobile Keypad is a rectangular grid of 4X3 (4 rows and 3 columns) Lets say Count(i, j, N) represents the count of N length numbers starting from position (i, j) 
If N = 1
  Count(i, j, N) = 10  
Else
  Count(i, j, N) = Sum of all Count(r, c, N-1) where (r, c) is new 
                   position after valid move of length 1 from current 
                   position (i, j)
Following is the implementation of above recursive formula. 
C++
Java
Python3
C#
PHP
Javascript
// A Naive Recursive C program to count number of possible numbers// of given length#include <stdio.h> // left, up, right, down move from current locationint row[] = {0, 0, -1, 0, 1};int col[] = {0, -1, 0, 1, 0}; // Returns count of numbers of length n starting from key position// (i, j) in a numeric keyboard.int getCountUtil(char keypad[][3], int i, int j, int n){    if (keypad == NULL || n <= 0)        return 0;     // From a given key, only one number is possible of length 1    if (n == 1)        return 1;     int k=0, move=0, ro=0, co=0, totalCount = 0;     // move left, up, right, down from current location and if    // new location is valid, then get number count of length    // (n-1) from that new position and add in count obtained so far    for (move=0; move<5; move++)    {        ro = i + row[move];        co = j + col[move];        if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&           keypad[ro][co] != '*' && keypad[ro][co] != '#')        {            totalCount += getCountUtil(keypad, ro, co, n-1);        }    }     return totalCount;} // Return count of all possible numbers of length n// in a given numeric keyboardint getCount(char keypad[][3], int n){    // Base cases    if (keypad == NULL || n <= 0)        return 0;    if (n == 1)        return 10;     int i=0, j=0, totalCount = 0;    for (i=0; i<4; i++)  // Loop on keypad row    {        for (j=0; j<3; j++)   // Loop on keypad column        {            // Process for 0 to 9 digits            if (keypad[i][j] != '*' && keypad[i][j] != '#')            {                // Get count when number is starting from key                // position (i, j) and add in count obtained so far                totalCount += getCountUtil(keypad, i, j, n);            }        }    }    return totalCount;} // Driver program to test above functionint main(int argc, char *argv[]){   char keypad[4][3] = {{'1','2','3'},                        {'4','5','6'},                        {'7','8','9'},                        {'*','0','#'}};   printf("Count for numbers of length %d: %dn", 1, getCount(keypad, 1));   printf("Count for numbers of length %d: %dn", 2, getCount(keypad, 2));   printf("Count for numbers of length %d: %dn", 3, getCount(keypad, 3));   printf("Count for numbers of length %d: %dn", 4, getCount(keypad, 4));   printf("Count for numbers of length %d: %dn", 5, getCount(keypad, 5));    return 0;}
// A Naive Recursive Java program// to count number of possible// numbers of given lengthclass GfG{ // left, up, right, down// move from current locationstatic int row[] = {0, 0, -1, 0, 1};static int col[] = {0, -1, 0, 1, 0}; // Returns count of numbers of length// n starting from key position// (i, j) in a numeric keyboard.static int getCountUtil(char keypad[][],                        int i, int j, int n){    if (keypad == null || n <= 0)        return 0;     // From a given key, only one    // number is possible of length 1    if (n == 1)        return 1;     int k = 0, move = 0, ro = 0, co = 0, totalCount = 0;     // move left, up, right, down    // from current location and if    // new location is valid, then    // get number count of length    // (n-1) from that new position    // and add in count obtained so far    for (move=0; move<5; move++)    {        ro = i + row[move];        co = j + col[move];        if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&        keypad[ro][co] != '*' && keypad[ro][co] != '#')        {            totalCount += getCountUtil(keypad, ro, co, n - 1);        }    }    return totalCount;} // Return count of all possible numbers of length n// in a given numeric keyboardstatic int getCount(char keypad[][], int n){    // Base cases    if (keypad == null || n <= 0)        return 0;    if (n == 1)        return 10;     int i = 0, j = 0, totalCount = 0;    for (i = 0; i < 4; i++) // Loop on keypad row    {        for (j=0; j<3; j++) // Loop on keypad column        {            // Process for 0 to 9 digits            if (keypad[i][j] != '*' && keypad[i][j] != '#')            {                // Get count when number is starting from key                // position (i, j) and add in count obtained so far                totalCount += getCountUtil(keypad, i, j, n);            }        }    }    return totalCount;} // Driver codepublic static void main(String[] args){    char keypad[][] = {{'1','2','3'},                        {'4','5','6'},                        {'7','8','9'},                        {'*','0','#'}};    System.out.printf("Count for numbers of"+                    " length %d: %d", 1, getCount(keypad, 1));    System.out.printf("\nCount for numbers of" +                    "length %d: %d", 2, getCount(keypad, 2));    System.out.printf("\nCount for numbers of" +                    "length %d: %d", 3, getCount(keypad, 3));    System.out.printf("\nCount for numbers of" +                    "length %d: %d", 4, getCount(keypad, 4));    System.out.printf("\nCount for numbers of" +                    "length %d: %d", 5, getCount(keypad, 5));}} // This code has been contributed by 29AjayKumar
# A Naive Recursive Python program to count number of possible numbers# of given length # left, up, right, down move from current locationrow = [0, 0, -1, 0, 1]col = [0, -1, 0, 1, 0] # Returns count of numbers of length n starting from key position# (i, j) in a numeric keyboard.def getCountUtil(keypad, i, j, n):    if (keypad == None or n <= 0):        return 0     # From a given key, only one number is possible of length 1    if (n == 1):        return 1    k = 0    move = 0    ro = 0    co = 0    totalCount = 0     # move left, up, right, down from current location and if    # new location is valid, then get number count of length    # (n-1) from that new position and add in count obtained so far    for move in range(5):        ro = i + row[move]        co = j + col[move]        if (ro >= 0 and ro <= 3 and co >= 0 and co <= 2 and                keypad[ro][co] != '*' and keypad[ro][co] != '#'):            totalCount += getCountUtil(keypad, ro, co, n - 1)    return totalCount # Return count of all possible numbers of length n# in a given numeric keyboarddef getCount(keypad, n):         # Base cases    if (keypad == None or n <= 0):        return 0    if (n == 1):        return 10    i = 0    j = 0    totalCount = 0    for i in range(4):  # Loop on keypad row        for j in range(3):   # Loop on keypad column                         # Process for 0 to 9 digits            if (keypad[i][j] != '*' and keypad[i][j] != '#'):               # Get count when number is starting from key                # position (i, j) and add in count obtained so far                totalCount += getCountUtil(keypad, i, j, n)    return totalCount # Driver codekeypad = [['1', '2', '3'],          ['4', '5', '6'],          ['7', '8', '9'],          ['*', '0', '#']]print("Count for numbers of length 1:", getCount(keypad, 1))print("Count for numbers of length 2:", getCount(keypad, 2))print("Count for numbers of length 3:", getCount(keypad, 3))print("Count for numbers of length 4:", getCount(keypad, 4))print("Count for numbers of length 5:", getCount(keypad, 5)) # This code is contributed by subhammahato348
// A Naive Recursive C# program// to count number of possible// numbers of given lengthusing System; class GfG{ // left, up, right, down// move from current locationstatic int []row = {0, 0, -1, 0, 1};static int []col = {0, -1, 0, 1, 0}; // Returns count of numbers of length// n starting from key position// (i, j) in a numeric keyboard.static int getCountUtil(char [,]keypad,                        int i, int j, int n){    if (keypad == null || n <= 0)        return 0;     // From a given key, only one    // number is possible of length 1    if (n == 1)        return 1;     int k = 0, move = 0, ro = 0, co = 0, totalCount = 0;     // move left, up, right, down    // from current location and if    // new location is valid, then    // get number count of length    // (n-1) from that new position    // and add in count obtained so far    for (move=0; move<5; move++)    {        ro = i + row[move];        co = j + col[move];        if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&        keypad[ro,co] != '*' && keypad[ro,co] != '#')        {            totalCount += getCountUtil(keypad, ro, co, n - 1);        }    }    return totalCount;} // Return count of all possible numbers of length n// in a given numeric keyboardstatic int getCount(char [,]keypad, int n){    // Base cases    if (keypad == null || n <= 0)        return 0;    if (n == 1)        return 10;     int i = 0, j = 0, totalCount = 0;    for (i = 0; i < 4; i++) // Loop on keypad row    {        for (j = 0; j < 3; j++) // Loop on keypad column        {            // Process for 0 to 9 digits            if (keypad[i, j] != '*' && keypad[i, j] != '#')            {                // Get count when number is starting from key                // position (i, j) and add in count obtained so far                totalCount += getCountUtil(keypad, i, j, n);            }        }    }    return totalCount;} // Driver codepublic static void Main(){    char [,]keypad = {{'1','2','3'},                        {'4','5','6'},                        {'7','8','9'},                        {'*','0','#'}};    Console.Write("Count for numbers of"+                    " length {0}: {1}", 1, getCount(keypad, 1));    Console.Write("\nCount for numbers of" +                    "length {0}: {1}", 2, getCount(keypad, 2));    Console.Write("\nCount for numbers of" +                    "length {0}: {1}", 3, getCount(keypad, 3));    Console.Write("\nCount for numbers of" +                    "length {0}: {1}", 4, getCount(keypad, 4));    Console.Write("\nCount for numbers of" +                    "length {0}: {1}", 5, getCount(keypad, 5));}} /* This code contributed by PrinciRaj1992 */
<?php// A Naive Recursive PHP program// to count number of possible// numbers of given length// left, up, right, down// move from current location// static $row = array(0, 0, -1, 0, 1);//static $col = array(0, -1, 0, 1, 0); // Returns count of numbers of length// n starting from key position// (i, j) in a numeric keyboard.function getCountUtil($keypad,                        $i, $j, $n){    static $row= array(0,0,-1,0,1);    static $col= array(0,-1,0,1,0);    if ($keypad == null || $n <= 0)        return 0;     // From a given key, only one    // number is possible of length 1    if ($n == 1)        return 1;     $k = 0; $move = 0; $ro = 0; $co = 0; $totalCount = 0;     // move left, up, right, down    // from current location and if    // new location is valid, then    // get number count of length    // (n-1) from that new position    // and add in count obtained so far    for ($move = 0; $move < 5; $move++)    {        $ro = $i + $row[$move];        $co = $j + $col[$move];        if ($ro >= 0 && $ro <= 3 && $co >=0 && $co <= 2 &&        $keypad[$ro][$co] != '*' && $keypad[$ro][$co] != '#')        {            $totalCount += getCountUtil($keypad, $ro, $co, $n - 1);        }    }    return $totalCount;} // Return count of all possible numbers of length n// in a given numeric keyboardfunction getCount($keypad, $n){    // Base cases    if ($keypad == null || $n <= 0)        return 0;    if ($n == 1)        return 10;     $i = 0; $j = 0; $totalCount = 0;    for ($i = 0; $i < 4; $i++) // Loop on keypad row    {        for ($j = 0; $j < 3; $j++) // Loop on keypad column        {            // Process for 0 to 9 digits            if ($keypad[$i][$j] != '*' && $keypad[$i][$j] != '#')            {                // Get count when number is starting from key                // position (i, j) and add in count obtained so far                $totalCount += getCountUtil($keypad, $i, $j, $n);            }        }    }    return $totalCount;} // Driver code{    $keypad = array(array('1','2','3'),                        array('4','5','6'),                        array('7','8','9'),                        array('*','0','#'));    echo("Count for numbers of"." length". getCount($keypad, 1));    echo("\nCount for numbers of" .                    " length ". getCount($keypad, 2));    echo("\nCount for numbers of" .                    " length ".getCount($keypad, 3));    echo("\nCount for numbers of" .                    " length ".getCount($keypad, 4));    echo("\nCount for numbers of" .                    " length ".getCount($keypad, 5));} // This code has been contributed by Code_Mech.
<script> // A Naive Recursive Javascript program// to count number of possible// numbers of given length          // left, up, right, down    // move from current location    let row=[0, 0, -1, 0, 1];    let col=[0, -1, 0, 1, 0];         // Returns count of numbers of length    // n starting from key position    // (i, j) in a numeric keyboard.    function getCountUtil(keypad,i,j,n)    {        if (keypad == null || n <= 0)        {return 0;}                 // From a given key, only one        // number is possible of length 1        if (n == 1)            return 1;                 let k = 0, move = 0, ro = 0, co = 0, totalCount = 0;                 // move left, up, right, down        // from current location and if        // new location is valid, then        // get number count of length        // (n-1) from that new position        // and add in count obtained so far        for (move=0; move<5; move++)        {            ro = i + row[move];            co = j + col[move];            if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&            keypad[ro][co] != '*' && keypad[ro][co] != '#')            {                totalCount += getCountUtil(keypad, ro, co, n - 1);            }        }        return totalCount;    }         // Return count of all possible numbers of length n    // in a given numeric keyboard     function getCount(keypad,n)    {        // Base cases        if (keypad == null || n <= 0)            return 0;        if (n == 1)            return 10;              let i = 0, j = 0, totalCount = 0;        for (i = 0; i < 4; i++) // Loop on keypad row        {            for (j=0; j<3; j++) // Loop on keypad column            {                // Process for 0 to 9 digits                if (keypad[i][j] != '*' && keypad[i][j] != '#')                {                    // Get count when number is starting from key                    // position (i, j) and add in count obtained so far                    totalCount += getCountUtil(keypad, i, j, n);                }            }        }        return totalCount;    }         // Driver code    let keypad=[['1','2','3'],['4','5','6'],['7','8','9'],['*','0','#']];    document.write("Count for numbers of"+                    " length ", 1,": ", getCount(keypad, 1));    document.write("<br>Count for numbers of" +                    "length ", 2,": ", getCount(keypad, 2));    document.write("<br>Count for numbers of" +                    "length ", 3,": ", getCount(keypad, 3));    document.write("<br>Count for numbers of" +                    "length ", 4,": ", getCount(keypad, 4));    document.write("<br>Count for numbers of" +                    "length ", 5,": ", getCount(keypad, 5));                         // This code is contributed by avanitrachhadiya2155     </script>
Output: 
Count for numbers of length 1: 10
Count for numbers of length 2: 36
Count for numbers of length 3: 138
Count for numbers of length 4: 532
Count for numbers of length 5: 2062
Dynamic Programming There are many repeated traversal on smaller paths (traversal for smaller N) to find all possible longer paths (traversal for bigger N). See following two diagrams for example. In this traversal, for N = 4 from two starting positions (buttons ‘4’ and ‘8’), we can see there are few repeated traversals for N = 2 (e.g. 4 -> 1, 6 -> 3, 8 -> 9, 8 -> 7 etc).  
 
Since the problem has both properties: Optimal Substructure and Overlapping Subproblems, it can be efficiently solved using dynamic programming. 
Following is the program for dynamic programming implementation. 
C++
Java
Python3
C#
Javascript
// A Dynamic Programming based C program to count number of// possible numbers of given length#include <stdio.h> // Return count of all possible numbers of length n// in a given numeric keyboardint getCount(char keypad[][3], int n){    if(keypad == NULL || n <= 0)        return 0;    if(n == 1)        return 10;     // left, up, right, down move from current location    int row[] = {0, 0, -1, 0, 1};    int col[] = {0, -1, 0, 1, 0};     // taking n+1 for simplicity - count[i][j] will store    // number count starting with digit i and length j    int count[10][n+1];    int i=0, j=0, k=0, move=0, ro=0, co=0, num = 0;    int nextNum=0, totalCount = 0;     // count numbers starting with digit i and of lengths 0 and 1    for (i=0; i<=9; i++)    {        count[i][0] = 0;        count[i][1] = 1;    }     // Bottom up - Get number count of length 2, 3, 4, ... , n    for (k=2; k<=n; k++)    {        for (i=0; i<4; i++)  // Loop on keypad row        {            for (j=0; j<3; j++)   // Loop on keypad column            {                // Process for 0 to 9 digits                if (keypad[i][j] != '*' && keypad[i][j] != '#')                {                    // Here we are counting the numbers starting with                    // digit keypad[i][j] and of length k keypad[i][j]                    // will become 1st digit, and we need to look for                    // (k-1) more digits                    num = keypad[i][j] - '0';                    count[num][k] = 0;                     // move left, up, right, down from current location                    // and if new location is valid, then get number                    // count of length (k-1) from that new digit and                    // add in count we found so far                    for (move=0; move<5; move++)                    {                        ro = i + row[move];                        co = j + col[move];                        if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&                           keypad[ro][co] != '*' && keypad[ro][co] != '#')                        {                            nextNum = keypad[ro][co] - '0';                            count[num][k] += count[nextNum][k-1];                        }                    }                }            }        }    }     // Get count of all possible numbers of length "n" starting    // with digit 0, 1, 2, ..., 9    totalCount = 0;    for (i=0; i<=9; i++)        totalCount += count[i][n];    return totalCount;} // Driver program to test above functionint main(int argc, char *argv[]){   char keypad[4][3] = {{'1','2','3'},                        {'4','5','6'},                        {'7','8','9'},                        {'*','0','#'}};   printf("Count for numbers of length %d: %dn", 1, getCount(keypad, 1));   printf("Count for numbers of length %d: %dn", 2, getCount(keypad, 2));   printf("Count for numbers of length %d: %dn", 3, getCount(keypad, 3));   printf("Count for numbers of length %d: %dn", 4, getCount(keypad, 4));   printf("Count for numbers of length %d: %dn", 5, getCount(keypad, 5));    return 0;}
// A Dynamic Programming based Java program to// count number of possible numbers of given lengthclass GFG{     // Return count of all possible numbers of length n// in a given numeric keyboardstatic int getCount(char keypad[][], int n){    if(keypad == null || n <= 0)        return 0;    if(n == 1)        return 10;     // left, up, right, down move from current location    int row[] = {0, 0, -1, 0, 1};    int col[] = {0, -1, 0, 1, 0};     // taking n+1 for simplicity - count[i][j] will store    // number count starting with digit i and length j    int [][]count = new int[10][n + 1];    int i = 0, j = 0, k = 0, move = 0,             ro = 0, co = 0, num = 0;    int nextNum = 0, totalCount = 0;     // count numbers starting with digit i    // and of lengths 0 and 1    for (i = 0; i <= 9; i++)    {        count[i][0] = 0;        count[i][1] = 1;    }     // Bottom up - Get number count of length 2, 3, 4, ... , n    for (k = 2; k <= n; k++)    {        for (i = 0; i < 4; i++) // Loop on keypad row        {            for (j = 0; j < 3; j++) // Loop on keypad column            {                // Process for 0 to 9 digits                if (keypad[i][j] != '*' &&                    keypad[i][j] != '#')                {                    // Here we are counting the numbers starting with                    // digit keypad[i][j] and of length k keypad[i][j]                    // will become 1st digit, and we need to look for                    // (k-1) more digits                    num = keypad[i][j] - '0';                    count[num][k] = 0;                     // move left, up, right, down from current location                    // and if new location is valid, then get number                    // count of length (k-1) from that new digit and                    // add in count we found so far                    for (move = 0; move < 5; move++)                    {                        ro = i + row[move];                        co = j + col[move];                        if (ro >= 0 && ro <= 3 && co >= 0 &&                            co <= 2 && keypad[ro][co] != '*' &&                                       keypad[ro][co] != '#')                        {                            nextNum = keypad[ro][co] - '0';                            count[num][k] += count[nextNum][k - 1];                        }                    }                }            }        }    }     // Get count of all possible numbers of length "n"    // starting with digit 0, 1, 2, ..., 9    totalCount = 0;    for (i = 0; i <= 9; i++)        totalCount += count[i][n];    return totalCount;} // Driver Codepublic static void main(String[] args){    char keypad[][] = {{'1','2','3'},                       {'4','5','6'},                       {'7','8','9'},                       {'*','0','#'}};    System.out.printf("Count for numbers of length %d: %d\n", 1,                                           getCount(keypad, 1));    System.out.printf("Count for numbers of length %d: %d\n", 2,                                           getCount(keypad, 2));    System.out.printf("Count for numbers of length %d: %d\n", 3,                                           getCount(keypad, 3));    System.out.printf("Count for numbers of length %d: %d\n", 4,                                           getCount(keypad, 4));    System.out.printf("Count for numbers of length %d: %d\n", 5,                                           getCount(keypad, 5));}} // This code is contributed by Rajput-Ji
# A Dynamic Programming based C program to count number of# possible numbers of given length # Return count of all possible numbers of length n# in a given numeric keyboarddef getCount(keypad, n):    if (keypad == None or n <= 0):        return 0    if (n == 1):        return 10     # left, up, right, down move from current location    row = [0, 0, -1, 0, 1]    col = [0, -1, 0, 1, 0]     # taking n+1 for simplicity - count[i][j] will store    # number count starting with digit i and length j    # count[10][n+1]    count = [[0]*(n + 1)]*10    i = 0    j = 0    k = 0    move = 0    ro = 0    co = 0    num = 0    nextNum = 0    totalCount = 0     # count numbers starting with    # digit i and of lengths 0 and 1    for i in range(10):        count[i][0] = 0        count[i][1] = 1     # Bottom up - Get number    # count of length 2, 3, 4, ... , n    for k in range(2, n + 1):        for i in range(4):  # Loop on keypad row            for j in range(3):   # Loop on keypad column                               # Process for 0 to 9 digits                if (keypad[i][j] != '*' and keypad[i][j] != '#'):                                       # Here we are counting the numbers starting with                    # digit keypad[i][j] and of length k keypad[i][j]                    # will become 1st digit, and we need to look for                    # (k-1) more digits                    num = ord(keypad[i][j]) - 48                    count[num][k] = 0                     # move left, up, right, down from current location                    # and if new location is valid, then get number                    # count of length (k-1) from that new digit and                    # add in count we found so far                    for move in range(5):                        ro = i + row[move]                        co = j + col[move]                        if (ro >= 0 and ro <= 3 and co >= 0 and co <= 2 and                                keypad[ro][co] != '*' and keypad[ro][co] != '#'):                            nextNum = ord(keypad[ro][co]) - 48                            count[num][k] += count[nextNum][k - 1]     # Get count of all possible numbers of length "n" starting    # with digit 0, 1, 2, ..., 9    totalCount = 0    for i in range(10):        totalCount += count[i][n]    return totalCount # Driver codeif __name__ == "__main__":    keypad = [['1','2','3'],            ['4','5','6'],            ['7','8','9'],            ['*','0','#']]         print("Count for numbers of length", 1, ":", getCount(keypad, 1))    print("Count for numbers of length", 2, ":", getCount(keypad, 2))    print("Count for numbers of length", 3, ":", getCount(keypad, 3))    print("Count for numbers of length", 4, ":", getCount(keypad, 4))    print("Count for numbers of length", 5, ":", getCount(keypad, 5))     # This code is contributed by subhammahato348
// A Dynamic Programming based C# program to// count number of possible numbers of given Lengthusing System; class GFG{     // Return count of all possible numbers of Length n// in a given numeric keyboardstatic int getCount(char [,]keypad, int n){    if(keypad == null || n <= 0)        return 0;    if(n == 1)        return 10;     // left, up, right, down move    // from current location    int []row = {0, 0, -1, 0, 1};    int []col = {0, -1, 0, 1, 0};     // taking n+1 for simplicity - count[i,j]    // will store number count starting with    // digit i and.Length j    int [,]count = new int[10,n + 1];    int i = 0, j = 0, k = 0, move = 0,              ro = 0, co = 0, num = 0;    int nextNum = 0, totalCount = 0;     // count numbers starting with digit i    // and of.Lengths 0 and 1    for (i = 0; i <= 9; i++)    {        count[i, 0] = 0;        count[i, 1] = 1;    }     // Bottom up - Get number count of    // Length 2, 3, 4, ... , n    for (k = 2; k <= n; k++)    {        for (i = 0; i < 4; i++) // Loop on keypad row        {            for (j = 0; j < 3; j++) // Loop on keypad column            {                // Process for 0 to 9 digits                if (keypad[i, j] != '*' &&                    keypad[i, j] != '#')                {                    // Here we are counting the numbers starting with                    // digit keypad[i,j] and of.Length k keypad[i,j]                    // will become 1st digit, and we need to look for                    // (k-1) more digits                    num = keypad[i, j] - '0';                    count[num, k] = 0;                     // move left, up, right, down from current location                    // and if new location is valid, then get number                    // count of.Length (k-1) from that new digit and                    //.Add in count we found so far                    for (move = 0; move < 5; move++)                    {                        ro = i + row[move];                        co = j + col[move];                        if (ro >= 0 && ro <= 3 && co >= 0 &&                            co <= 2 && keypad[ro, co] != '*' &&                                       keypad[ro, co] != '#')                        {                            nextNum = keypad[ro, co] - '0';                            count[num, k] += count[nextNum, k - 1];                        }                    }                }            }        }    }     // Get count of all possible numbers of.Length "n"    // starting with digit 0, 1, 2, ..., 9    totalCount = 0;    for (i = 0; i <= 9; i++)        totalCount += count[i, n];    return totalCount;} // Driver Codepublic static void Main(String[] args){    char [,]keypad = {{'1', '2', '3'},                      {'4', '5', '6'},                      {'7', '8', '9'},                      {'*', '0', '#'}};    Console.Write("Count for numbers of.Length {0}: {1}\n", 1,                                        getCount(keypad, 1));    Console.Write("Count for numbers of.Length {0}: {1}\n", 2,                                        getCount(keypad, 2));    Console.Write("Count for numbers of.Length {0}: {1}\n", 3,                                        getCount(keypad, 3));    Console.Write("Count for numbers of.Length {0}: {1}\n", 4,                                        getCount(keypad, 4));    Console.Write("Count for numbers of.Length {0}: {1}\n", 5,                                        getCount(keypad, 5));}} // This code is contributed by Rajput-Ji
<script> // A Dynamic Programming based Javascript program to// count number of possible numbers of given length // Return count of all possible numbers of length n// in a given numeric keyboardfunction getCount(keypad, n){    if (keypad == null || n <= 0)        return 0;             if (n == 1)        return 10;         // left, up, right, down move    // from current location       let row = [ 0, 0, -1, 0, 1 ];    let col = [ 0, -1, 0, 1, 0 ];         // Taking n+1 for simplicity - count[i][j]    // will store number count starting with    // digit i and length j    let count = new Array(10);    for(let i = 0; i < 10; i++)    {        count[i] = new Array(n + 1);        for(let j = 0; j < n + 1; j++)        {            count[i][j] = 0;        }    }         let i = 0, j = 0, k = 0, move = 0,         ro = 0, co = 0, num = 0;    let nextNum = 0, totalCount = 0;         // count numbers starting with digit i    // and of lengths 0 and 1    for(i = 0; i <= 9; i++)    {        count[i][0] = 0;        count[i][1] = 1;    }      // Bottom up - Get number count    // of length 2, 3, 4, ... , n    for(k = 2; k <= n; k++)    {                 // Loop on keypad row        for(i = 0; i < 4; i++)        {                         // Loop on keypad column            for(j = 0; j < 3; j++)            {                                 // Process for 0 to 9 digits                if (keypad[i][j] != '*' &&                    keypad[i][j] != '#')                {                                         // Here we are counting the numbers starting with                    // digit keypad[i][j] and of length k keypad[i][j]                    // will become 1st digit, and we need to look for                    // (k-1) more digits                    num = keypad[i][j].charCodeAt(0) -                                   '0'.charCodeAt(0);                    count[num][k] = 0;                      // Move left, up, right, down from current location                    // and if new location is valid, then get number                    // count of length (k-1) from that new digit and                    // add in count we found so far                    for(move = 0; move < 5; move++)                    {                        ro = i + row[move];                        co = j + col[move];                        if (ro >= 0 && ro <= 3 && co >= 0 &&                            co <= 2 && keypad[ro][co] != '*' &&                                       keypad[ro][co] != '#')                        {                            nextNum = keypad[ro][co].charCodeAt(0) -                                                 '0'.charCodeAt(0);                            count[num][k] += count[nextNum][k - 1];                        }                    }                }            }        }    }      // Get count of all possible numbers of length "n"    // starting with digit 0, 1, 2, ..., 9    totalCount = 0;    for(i = 0; i <= 9; i++)        totalCount += count[i][n];             return totalCount;     } // Driver Codelet keypad = [ [ '1','2','3' ],               [ '4','5','6' ],               [ '7','8','9' ],               [ '*','0','#' ] ];                document.write("Count for numbers of length " +         1 + " : " + getCount(keypad, 1) + "<br>")document.write("Count for numbers of length " +         2 + " : " + getCount(keypad, 2) + "<br>")document.write("Count for numbers of length " +         3 + " : " + getCount(keypad, 3) + "<br>")document.write("Count for numbers of length " +         4 + " : " + getCount(keypad, 4) + "<br>")document.write("Count for numbers of length " +         5 + " : " + getCount(keypad, 5) + "<br>") // This code is contributed by rag2127 </script>
Output: 
Count for numbers of length 1: 10
Count for numbers of length 2: 36
Count for numbers of length 3: 138
Count for numbers of length 4: 532
Count for numbers of length 5: 2062
A Space Optimized Solution: The above dynamic programming approach also runs in O(n) time and requires O(n) auxiliary space, as only one for loop runs n times, other for loops runs for constant time. We can see that nth iteration needs data from (n-1)th iteration only, so we need not keep the data from older iterations. We can have a space efficient dynamic programming approach with just two arrays of size 10. Thanks to Nik for suggesting this solution.
C++
C
Java
Python3
C#
Javascript
// A Space Optimized C++ program to count number of possible numbers// of given length#include <bits/stdc++.h>using namespace std; // Return count of all possible numbers of length n// in a given numeric keyboardint getCount(char keypad[][3], int n){    if (keypad == NULL || n <= 0)        return 0;    if (n == 1)        return 10;     // odd[i], even[i] arrays represent count of numbers starting    // with digit i for any length j    int odd[10], even[10];    int i = 0, j = 0, useOdd = 0, totalCount = 0;     for (i = 0; i <= 9; i++)        odd[i] = 1; // for j = 1     for (j = 2; j <= n; j++) // Bottom Up calculation from j = 2 to n    {        useOdd = 1 - useOdd;         // Here we are explicitly writing lines for each number 0        // to 9. But it can always be written as DFS on 4X3 grid        // using row, column array valid moves        if (useOdd == 1)        {            even[0] = odd[0] + odd[8];            even[1] = odd[1] + odd[2] + odd[4];            even[2] = odd[2] + odd[1] + odd[3] + odd[5];            even[3] = odd[3] + odd[2] + odd[6];            even[4] = odd[4] + odd[1] + odd[5] + odd[7];            even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6];            even[6] = odd[6] + odd[3] + odd[5] + odd[9];            even[7] = odd[7] + odd[4] + odd[8];            even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9];            even[9] = odd[9] + odd[6] + odd[8];        }        else        {            odd[0] = even[0] + even[8];            odd[1] = even[1] + even[2] + even[4];            odd[2] = even[2] + even[1] + even[3] + even[5];            odd[3] = even[3] + even[2] + even[6];            odd[4] = even[4] + even[1] + even[5] + even[7];            odd[5] = even[5] + even[2] + even[4] + even[8] + even[6];            odd[6] = even[6] + even[3] + even[5] + even[9];            odd[7] = even[7] + even[4] + even[8];            odd[8] = even[8] + even[0] + even[5] + even[7] + even[9];            odd[9] = even[9] + even[6] + even[8];        }    }     // Get count of all possible numbers of length "n" starting    // with digit 0, 1, 2, ..., 9    totalCount = 0;    if (useOdd == 1)    {        for (i = 0; i <= 9; i++)            totalCount += even[i];    }    else    {        for (i = 0; i <= 9; i++)            totalCount += odd[i];    }    return totalCount;} // Driver program to test above functionint main(){    char keypad[4][3] = {{'1', '2', '3'},                         {'4', '5', '6'},                         {'7', '8', '9'},                         {'*', '0', '#'}};    cout << "Count for numbers of length 1: " << getCount(keypad, 1) << endl;    cout << "Count for numbers of length 2: " << getCount(keypad, 2) << endl;    cout << "Count for numbers of length 3: " << getCount(keypad, 3) << endl;    cout << "Count for numbers of length 4: " << getCount(keypad, 4) << endl;    cout << "Count for numbers of length 5: " << getCount(keypad, 5) << endl;     return 0;} //This code is contributed by Mayank Tyagi
// A Space Optimized C program to count number of possible numbers// of given length#include <stdio.h> // Return count of all possible numbers of length n// in a given numeric keyboardint getCount(char keypad[][3], int n){    if(keypad == NULL || n <= 0)        return 0;    if(n == 1)        return 10;     // odd[i], even[i] arrays represent count of numbers starting    // with digit i for any length j    int odd[10], even[10];    int i = 0, j = 0, useOdd = 0, totalCount = 0;     for (i=0; i<=9; i++)        odd[i] = 1;  // for j = 1     for (j=2; j<=n; j++) // Bottom Up calculation from j = 2 to n    {        useOdd = 1 - useOdd;         // Here we are explicitly writing lines for each number 0        // to 9. But it can always be written as DFS on 4X3 grid        // using row, column array valid moves        if(useOdd == 1)        {            even[0] = odd[0] + odd[8];            even[1] = odd[1] + odd[2] + odd[4];            even[2] = odd[2] + odd[1] + odd[3] + odd[5];            even[3] = odd[3] + odd[2] + odd[6];            even[4] = odd[4] + odd[1] + odd[5] + odd[7];            even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6];            even[6] = odd[6] + odd[3] + odd[5] + odd[9];            even[7] = odd[7] + odd[4] + odd[8];            even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9];            even[9] = odd[9] + odd[6] + odd[8];        }        else        {            odd[0] = even[0] + even[8];            odd[1] = even[1] + even[2] + even[4];            odd[2] = even[2] + even[1] + even[3] + even[5];            odd[3] = even[3] + even[2] + even[6];            odd[4] = even[4] + even[1] + even[5] + even[7];            odd[5] = even[5] + even[2] + even[4] + even[8] + even[6];            odd[6] = even[6] + even[3] + even[5] + even[9];            odd[7] = even[7] + even[4] + even[8];            odd[8] = even[8] + even[0] + even[5] + even[7] + even[9];            odd[9] = even[9] + even[6] + even[8];        }    }     // Get count of all possible numbers of length "n" starting    // with digit 0, 1, 2, ..., 9    totalCount = 0;    if(useOdd == 1)    {        for (i=0; i<=9; i++)            totalCount += even[i];    }    else    {        for (i=0; i<=9; i++)            totalCount += odd[i];    }    return totalCount;} // Driver program to test above functionint main(){    char keypad[4][3] = {{'1','2','3'},        {'4','5','6'},        {'7','8','9'},        {'*','0','#'}    };    printf("Count for numbers of length %d: %dn", 1, getCount(keypad, 1));    printf("Count for numbers of length %d: %dn", 2, getCount(keypad, 2));    printf("Count for numbers of length %d: %dn", 3, getCount(keypad, 3));    printf("Count for numbers of length %d: %dn", 4, getCount(keypad, 4));    printf("Count for numbers of length %d: %dn", 5, getCount(keypad, 5));     return 0;}
// A Space Optimized Java program to// count number of possible numbers// of given lengthclass GFG{ // Return count of all possible numbers of// length n in a given numeric keyboardstatic int getCount(char keypad[][], int n){    if(keypad == null || n <= 0)        return 0;    if(n == 1)        return 10;     // odd[i], even[i] arrays represent count of    // numbers starting with digit i for any length j    int []odd = new int[10];    int []even = new int[10];    int i = 0, j = 0, useOdd = 0, totalCount = 0;     for (i = 0; i <= 9; i++)        odd[i] = 1; // for j = 1         // Bottom Up calculation from j = 2 to n    for (j = 2; j <= n; j++)    {        useOdd = 1 - useOdd;         // Here we are explicitly writing lines        // for each number 0 to 9. But it can always be        // written as DFS on 4X3 grid using row,        // column array valid moves        if(useOdd == 1)        {            even[0] = odd[0] + odd[8];            even[1] = odd[1] + odd[2] + odd[4];            even[2] = odd[2] + odd[1] +                      odd[3] + odd[5];            even[3] = odd[3] + odd[2] + odd[6];            even[4] = odd[4] + odd[1] +                      odd[5] + odd[7];            even[5] = odd[5] + odd[2] + odd[4] +                               odd[8] + odd[6];            even[6] = odd[6] + odd[3] +                      odd[5] + odd[9];            even[7] = odd[7] + odd[4] + odd[8];            even[8] = odd[8] + odd[0] + odd[5] +                               odd[7] + odd[9];            even[9] = odd[9] + odd[6] + odd[8];        }        else        {            odd[0] = even[0] + even[8];            odd[1] = even[1] + even[2] + even[4];            odd[2] = even[2] + even[1] +                     even[3] + even[5];            odd[3] = even[3] + even[2] + even[6];            odd[4] = even[4] + even[1] +                     even[5] + even[7];            odd[5] = even[5] + even[2] + even[4] +                               even[8] + even[6];            odd[6] = even[6] + even[3] +                     even[5] + even[9];            odd[7] = even[7] + even[4] + even[8];            odd[8] = even[8] + even[0] + even[5] +                               even[7] + even[9];            odd[9] = even[9] + even[6] + even[8];        }    }     // Get count of all possible numbers of    // length "n" starting with digit 0, 1, 2, ..., 9    totalCount = 0;    if(useOdd == 1)    {        for (i = 0; i <= 9; i++)            totalCount += even[i];    }    else    {        for (i = 0; i <= 9; i++)            totalCount += odd[i];    }    return totalCount;} // Driver Codepublic static void main(String[] args){    char keypad[][] = {{'1','2','3'},                       {'4','5','6'},                       {'7','8','9'},                       {'*','0','#'}};    System.out.printf("Count for numbers of length %d: %d\n", 1,                                           getCount(keypad, 1));    System.out.printf("Count for numbers of length %d: %d\n", 2,                                           getCount(keypad, 2));    System.out.printf("Count for numbers of length %d: %d\n", 3,                                           getCount(keypad, 3));    System.out.printf("Count for numbers of length %d: %d\n", 4,                                           getCount(keypad, 4));    System.out.printf("Count for numbers of length %d: %d\n", 5,                                           getCount(keypad, 5));}} // This code is contributed by PrinciRaj1992
# A Space Optimized Python program to count# number of possible numbers# of given length # Return count of all possible numbers# of length n# in a given numeric keyboarddef getCount(keypad, n):     if(not keypad or n <= 0):        return 0    if(n == 1):        return 10     # odd[i], even[i] arrays represent    # count of numbers starting    # with digit i for any length j    odd = [0]*10    even = [0]*10    i = 0    j = 0    useOdd = 0    totalCount = 0     for i in range(10):        odd[i] = 1 # for j = 1     for j in range(2,n+1): # Bottom Up calculation from j = 2 to n             useOdd = 1 - useOdd         # Here we are explicitly writing lines for each number 0        # to 9. But it can always be written as DFS on 4X3 grid        # using row, column array valid moves        if(useOdd == 1):                     even[0] = odd[0] + odd[8]            even[1] = odd[1] + odd[2] + odd[4]            even[2] = odd[2] + odd[1] + odd[3] + odd[5]            even[3] = odd[3] + odd[2] + odd[6]            even[4] = odd[4] + odd[1] + odd[5] + odd[7]            even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6]            even[6] = odd[6] + odd[3] + odd[5] + odd[9]            even[7] = odd[7] + odd[4] + odd[8]            even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9]            even[9] = odd[9] + odd[6] + odd[8]                 else:                     odd[0] = even[0] + even[8]            odd[1] = even[1] + even[2] + even[4]            odd[2] = even[2] + even[1] + even[3] + even[5]            odd[3] = even[3] + even[2] + even[6]            odd[4] = even[4] + even[1] + even[5] + even[7]            odd[5] = even[5] + even[2] + even[4] + even[8] + even[6]            odd[6] = even[6] + even[3] + even[5] + even[9]            odd[7] = even[7] + even[4] + even[8]            odd[8] = even[8] + even[0] + even[5] + even[7] + even[9]            odd[9] = even[9] + even[6] + even[8]     # Get count of all possible numbers of length "n" starting    # with digit 0, 1, 2, ..., 9    totalCount = 0    if(useOdd == 1):        for i in range(10):            totalCount += even[i]         else:        for i in range(10):            totalCount += odd[i]     return totalCount # Driver program to test above functionif __name__ == "__main__":    keypad = [['1','2','3'],            ['4','5','6'],            ['7','8','9'],            ['*','0','#']]         print("Count for numbers of length ",1,": ", getCount(keypad, 1))    print("Count for numbers of length ",2,": ", getCount(keypad, 2))    print("Count for numbers of length ",3,": ", getCount(keypad, 3))    print("Count for numbers of length ",4,": ", getCount(keypad, 4))    print("Count for numbers of length ",5,": ", getCount(keypad, 5))     # This code is contributed by# ChitraNayal
// A Space Optimized C# program to// count number of possible numbers// of given lengthusing System;     class GFG{ // Return count of all possible numbers of// length n in a given numeric keyboardstatic int getCount(char [,]keypad, int n){    if(keypad == null || n <= 0)        return 0;    if(n == 1)        return 10;     // odd[i], even[i] arrays represent count of    // numbers starting with digit i for any length j    int []odd = new int[10];    int []even = new int[10];    int i = 0, j = 0, useOdd = 0, totalCount = 0;     for (i = 0; i <= 9; i++)        odd[i] = 1; // for j = 1         // Bottom Up calculation from j = 2 to n    for (j = 2; j <= n; j++)    {        useOdd = 1 - useOdd;         // Here we are explicitly writing lines        // for each number 0 to 9. But it can always be        // written as DFS on 4X3 grid using row,        // column array valid moves        if(useOdd == 1)        {            even[0] = odd[0] + odd[8];            even[1] = odd[1] + odd[2] + odd[4];            even[2] = odd[2] + odd[1] +                      odd[3] + odd[5];            even[3] = odd[3] + odd[2] + odd[6];            even[4] = odd[4] + odd[1] +                      odd[5] + odd[7];            even[5] = odd[5] + odd[2] + odd[4] +                               odd[8] + odd[6];            even[6] = odd[6] + odd[3] +                      odd[5] + odd[9];            even[7] = odd[7] + odd[4] + odd[8];            even[8] = odd[8] + odd[0] + odd[5] +                               odd[7] + odd[9];            even[9] = odd[9] + odd[6] + odd[8];        }        else        {            odd[0] = even[0] + even[8];            odd[1] = even[1] + even[2] + even[4];            odd[2] = even[2] + even[1] +                     even[3] + even[5];            odd[3] = even[3] + even[2] + even[6];            odd[4] = even[4] + even[1] +                     even[5] + even[7];            odd[5] = even[5] + even[2] + even[4] +                               even[8] + even[6];            odd[6] = even[6] + even[3] +                     even[5] + even[9];            odd[7] = even[7] + even[4] + even[8];            odd[8] = even[8] + even[0] + even[5] +                               even[7] + even[9];            odd[9] = even[9] + even[6] + even[8];        }    }     // Get count of all possible numbers of    // length "n" starting with digit 0, 1, 2, ..., 9    totalCount = 0;    if(useOdd == 1)    {        for (i = 0; i <= 9; i++)            totalCount += even[i];    }    else    {        for (i = 0; i <= 9; i++)            totalCount += odd[i];    }    return totalCount;} // Driver Codepublic static void Main(String[] args){    char [,]keypad = {{'1','2','3'},                      {'4','5','6'},                      {'7','8','9'},                      {'*','0','#'}};    Console.Write("Count for numbers of length {0}: {1}\n", 1,                                        getCount(keypad, 1));    Console.Write("Count for numbers of length {0}: {1}\n", 2,                                        getCount(keypad, 2));    Console.Write("Count for numbers of length {0}: {1}\n", 3,                                        getCount(keypad, 3));    Console.Write("Count for numbers of length {0}: {1}\n", 4,                                        getCount(keypad, 4));    Console.Write("Count for numbers of length {0}: {1}\n", 5,                                        getCount(keypad, 5));}} // This code is contributed by 29AjayKumar
<script> // A Space Optimized javascript program to// count number of possible numbers// of given length // Return count of all possible numbers of// length n in a given numeric keyboardfunction getCount(keypad , n){    if(keypad == null || n <= 0)        return 0;    if(n == 1)        return 10;     // odd[i], even[i] arrays represent count of    // numbers starting with digit i for any length j    var odd = Array.from({length: 10}, (_, i) => 0);    var even = Array.from({length: 10}, (_, i) => 0);    var i = 0, j = 0, useOdd = 0, totalCount = 0;     for (i = 0; i <= 9; i++)        odd[i] = 1; // for j = 1         // Bottom Up calculation from j = 2 to n    for (j = 2; j <= n; j++)    {        useOdd = 1 - useOdd;         // Here we are explicitly writing lines        // for each number 0 to 9. But it can always be        // written as DFS on 4X3 grid using row,        // column array valid moves        if(useOdd == 1)        {            even[0] = odd[0] + odd[8];            even[1] = odd[1] + odd[2] + odd[4];            even[2] = odd[2] + odd[1] +                      odd[3] + odd[5];            even[3] = odd[3] + odd[2] + odd[6];            even[4] = odd[4] + odd[1] +                      odd[5] + odd[7];            even[5] = odd[5] + odd[2] + odd[4] +                               odd[8] + odd[6];            even[6] = odd[6] + odd[3] +                      odd[5] + odd[9];            even[7] = odd[7] + odd[4] + odd[8];            even[8] = odd[8] + odd[0] + odd[5] +                               odd[7] + odd[9];            even[9] = odd[9] + odd[6] + odd[8];        }        else        {            odd[0] = even[0] + even[8];            odd[1] = even[1] + even[2] + even[4];            odd[2] = even[2] + even[1] +                     even[3] + even[5];            odd[3] = even[3] + even[2] + even[6];            odd[4] = even[4] + even[1] +                     even[5] + even[7];            odd[5] = even[5] + even[2] + even[4] +                               even[8] + even[6];            odd[6] = even[6] + even[3] +                     even[5] + even[9];            odd[7] = even[7] + even[4] + even[8];            odd[8] = even[8] + even[0] + even[5] +                               even[7] + even[9];            odd[9] = even[9] + even[6] + even[8];        }    }     // Get count of all possible numbers of    // length "n" starting with digit 0, 1, 2, ..., 9    totalCount = 0;    if(useOdd == 1)    {        for (i = 0; i <= 9; i++)            totalCount += even[i];    }    else    {        for (i = 0; i <= 9; i++)            totalCount += odd[i];    }    return totalCount;} // Driver Code var keypad = [['1','2','3'],                   ['4','5','6'],                   ['7','8','9'],                   ['*','0','#']];                    document.write("Count for numbers of length "+ 1+": "+                                       getCount(keypad, 1));document.write("<br>Count for numbers of length  "+ 2+": "+                                       getCount(keypad, 2));document.write("<br>Count for numbers of length "+ 3+": "+                                       getCount(keypad, 3));document.write("<br>Count for numbers of length "+ 4+": "+                                       getCount(keypad, 4));document.write("<br>Count for numbers of length "+ 5+": "+                                       getCount(keypad, 5));  // This code is contributed by 29AjayKumar </script>
Output: 
Count for numbers of length 1: 10
Count for numbers of length 2: 36
Count for numbers of length 3: 138
Count for numbers of length 4: 532
Count for numbers of length 5: 2062
This article is contributed by Anurag Singh. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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