Number of n-digits non-decreasing integers

10 Sep, 2021
Given an integer n > 0, which denotes the number of digits, the task to find the total number of n-digit positive integers which are non-decreasing in nature. A non-decreasing integer is one in which all the digits from left to right are in non-decreasing form. ex: 1234, 1135, ..etc. Note: Leading zeros also count in non-decreasing integers such as 0000, 0001, 0023, etc are also non-decreasing integers of 4-digits. Examples :  
Input : n = 1
Output : 10
Numbers are 0, 1, 2, ...9.

Input : n = 2
Output : 55

Input : n = 4
Output : 715
 
Naive Approach: We generate all possible n-digit numbers and then for each number we check whether it is non-decreasing or not. Time Complexity : (n*10^n), where 10^n is for generating all possible n-digits numbers and n is for checking whether a particular number is non-decreasing or not.Dynamic Programming: If we fill digits one by one from left to right, the following conditions hold.  
If current last digit is 9, we can fill only 9s in remaining places. So only one solution is possible if current last digit is 9.If current last digit is less than 9, then we can recursively compute count using following formula.  
If current last digit is 9, we can fill only 9s in remaining places. So only one solution is possible if current last digit is 9.
If current last digit is less than 9, then we can recursively compute count using following formula.  
a[i][j] = a[i-1][j] + a[i][j + 1] 
         For every digit j smaller than 9.

We consider previous length count and count
to be increased by all greater digits.
We build a matrix a[][] where a[i][j] = count of all valid i-digit non-decreasing integers with j or greater than j as the leading digit. The solution is based on below observations. We fill this matrix column-wise, first calculating a[1][9] then using this value to compute a[2][8] and so on. At any instant if we wish to calculate a[i][j] means number of i-digits non-decreasing integers with leading digit as j or digit greater than j, we should add up a[i-1][j] (number of i-1 digit integers which should start from j or greater digit, because in this case if we place j as its left most digit then our number will be i-digit non-decreasing number) and a[i][j+1] (number of i-digit integers which should start with digit equals to greater than j+1). So, a[i][j] = a[i-1][j] + a[i][j+1].  
 
C++
Java
Python3
C#
PHP
Javascript
// C++ program for counting n digit numbers with// non decreasing digits#include <bits/stdc++.h>using namespace std; // Returns count of non- decreasing numbers with// n digits.int nonDecNums(int n){    /* a[i][j] = count of all possible number    with i digits having leading digit as j */    int a[n + 1][10];     // Initialization of all 0-digit number    for (int i = 0; i <= 9; i++)        a[0][i] = 1;     /* Initialization of all i-digit    non-decreasing number leading with 9*/    for (int i = 1; i <= n; i++)        a[i][9] = 1;     /* for all digits we should calculate    number of ways depending upon leading    digits*/    for (int i = 1; i <= n; i++)        for (int j = 8; j >= 0; j--)            a[i][j] = a[i - 1][j] + a[i][j + 1];     return a[n][0];} // driver programint main(){    int n = 2;    cout << "Non-decreasing digits = "        << nonDecNums(n) << endl;    return 0;}
// Java program for counting n digit numbers with// non decreasing digitsimport java.io.*; class GFG {     // Function that returns count of non- decreasing numbers    // with n digits    static int nonDecNums(int n)    {        // a[i][j] = count of all possible number        // with i digits having leading digit as j        int[][] a = new int[n + 1][10];         // Initialization of all 0-digit number        for (int i = 0; i <= 9; i++)            a[0][i] = 1;         // Initialization of all i-digit        // non-decreasing number leading with 9        for (int i = 1; i <= n; i++)            a[i][9] = 1;         // for all digits we should calculate        // number of ways depending upon leading        // digits        for (int i = 1; i <= n; i++)            for (int j = 8; j >= 0; j--)                a[i][j] = a[i - 1][j] + a[i][j + 1];         return a[n][0];    }     // driver program    public static void main(String[] args)    {        int n = 2;        System.out.println("Non-decreasing digits = " + nonDecNums(n));    }} // Contributed by Pramod Kumar
# Python3 program for counting n digit# numbers with non decreasing digitsimport numpy as np # Returns count of non- decreasing# numbers with n digits.def nonDecNums(n) :             # a[i][j] = count of all possible number    # with i digits having leading digit as j    a = np.zeros((n + 1, 10))     # Initialization of all 0-digit number    for i in range(10) :        a[0][i] = 1     # Initialization of all i-digit    # non-decreasing number leading with 9    for i in range(1, n + 1) :        a[i][9] = 1     # for all digits we should calculate    # number of ways depending upon    # leading digits    for i in range(1, n + 1) :        for j in range(8, -1, -1) :            a[i][j] = a[i - 1][j] + a[i][j + 1]     return int(a[n][0]) # Driver Codeif __name__ == "__main__" :     n = 2    print("Non-decreasing digits = ",                       nonDecNums(n)) # This code is contributed by Ryuga
// C# function to find number of diagonals// in n sided convex polygonusing System; class GFG {         // Function that returns count of non-    // decreasing numbers with n digits    static int nonDecNums(int n)    {        // a[i][j] = count of all possible number        // with i digits having leading digit as j        int[, ] a = new int[n + 1, 10];         // Initialization of all 0-digit number        for (int i = 0; i <= 9; i++)            a[0, i] = 1;         // Initialization of all i-digit        // non-decreasing number leading with 9        for (int i = 1; i <= n; i++)            a[i, 9] = 1;         // for all digits we should calculate        // number of ways depending upon leading        // digits        for (int i = 1; i <= n; i++)            for (int j = 8; j >= 0; j--)                a[i, j] = a[i - 1, j] + a[i, j + 1];         return a[n, 0];    }     // driver program    public static void Main()    {        int n = 2;        Console.WriteLine("Non-decreasing digits = " +                                       nonDecNums(n));    }} // This code is contributed by Sam007
<?php// PHP program for counting// n digit numbers with// non decreasing digits // Returns count of non-// decreasing numbers with// n digits. function nonDecNums($n){    /* a[i][j] = count of    all possible number    with i digits having    leading digit as j */     // Initialization of    // all 0-digit number    for ($i = 0; $i <= 9; $i++)        $a[0][$i] = 1;     /* Initialization of all    i-digit non-decreasing    number leading with 9*/    for ($i = 1; $i <= $n; $i++)        $a[$i][9] = 1;     /* for all digits we should    calculate number of ways    depending upon leading digits*/    for ($i = 1; $i <= $n; $i++)        for ($j = 8; $j >= 0; $j--)            $a[$i][$j] = $a[$i - 1][$j] +                         $a[$i][$j + 1];     return $a[$n][0];} // Driver Code$n = 2;echo "Non-decreasing digits = ",            nonDecNums($n),"\n"; // This code is contributed by m_kit?>
<script>     // Javascript program for counting n digit    // numbers with non decreasing digits         // Function that returns count    // of non- decreasing numbers    // with n digits    function nonDecNums(n)    {        // a[i][j] = count of all possible number        // with i digits having leading digit as j        let a = new Array(n + 1)        for (let i = 0; i < n + 1; i++)        {            a[i] = new Array(10);        }           // Initialization of all 0-digit number        for (let i = 0; i <= 9; i++)            a[0][i] = 1;           // Initialization of all i-digit        // non-decreasing number leading with 9        for (let i = 1; i <= n; i++)            a[i][9] = 1;           // for all digits we should calculate        // number of ways depending upon leading        // digits        for (let i = 1; i <= n; i++)            for (let j = 8; j >= 0; j--)                a[i][j] = a[i - 1][j] + a[i][j + 1];           return a[n][0];    }         let n = 2;      document.write(    "Non-decreasing digits = " + nonDecNums(n)    );     </script>
Non-decreasing digits = 55
Output :  
Non-decreasing digits = 55
Time Complexity : O(10*n) equivalent to O(n).
Another Approach: 
If we observe, we can see that 0 has to be placed before 1-9, 1 has to be placed before 2-9 and so on. As we are asked to find non-decreasing integers, 111223 is a valid non-decreasing integer which means same digit can occur conscuetively.
Example 1: When N=2, we have 11C9, which is equal to 55.
Example 2: When N=5, we have 14C9, which is equal to 2002.
C++
Java
Python3
C#
Javascript
// CPP program To calculate Number of n-digits non-decreasing integers//Contributed by Parishrut Kushwaha//#include <bits/stdc++.h>using namespace std; // Returns factorial of nlong long int fact(int n){    long long int res = 1;    for (int i = 2; i <= n; i++)        res = res * i;    return res;}  // returns nCrlong long int nCr(int n, int r){    return fact(n) / (fact(r) * fact(n - r));} // Driver codeint main(){    int n = 2;    cout <<"Number of Non-Decreasing digits: "<< nCr(n+9,9);    return 0;}
// Java program To calculate Number// of n-digits non-decreasing integersclass GFG {     // Returns factorial of n    static long fact(int n)    {        long res = 1;        for (int i = 2; i <= n; i++)            res = res * i;        return res;    }     // returns nCr    static long nCr(int n, int r)    {        return fact(n) / (fact(r) * fact(n - r));    }     // Driver code    public static void main(String[] args)    {        int n = 2;        System.out.println(            "Number of Non-Decreasing digits: "            + nCr(n + 9, 9));    }} // This code is contributed by rajsanghavi9.
# Python program To calculate Number of n-digits non-decreasing integers#Contributed by Parishrut Kushwaha## Returns factorial of ndef fact(n):         res = 1    for i in range (2,n+1):        res = res * i    return res     # returns nCrdef nCr(n, r):    return fact(n) // ((fact(r) * fact(n - r))) # Driver coden = 2print("Number of Non-Decreasing digits: " , nCr(n+9,9))     # This code is contributed by shivanisinghss2110   
// C# program To calculate Number// of n-digits non-decreasing integersusing System; class GFG {     // Returns factorial of n    static long fact(int n)    {        long res = 1;        for (int i = 2; i <= n; i++)            res = res * i;        return res;    }     // returns nCr    static long nCr(int n, int r)    {        return fact(n) / (fact(r) * fact(n - r));    }     // Driver code    public static void Main(String[] args)    {        int n = 2;        Console.Write("Number of Non-Decreasing digits: " + nCr(n + 9, 9));    }} // This code is contributed by shivanisinghss2110
<script>// JavaScript program To calculate Number// of n-digits non-decreasing integers     // Returns factorial of n    function fact( n)    {        var res = 1;        for (var i = 2; i <= n; i++)            res = res * i;        return res;    }     // returns nCr    function nCr(n, r)    {        return fact(n) / (fact(r) * fact(n - r));    }     // Driver code        var n = 2;        document.write("Number of Non-Decreasing digits: " + nCr(n + 9, 9)); // This code is contributed by shivanisinghss2110. </script>
Number of Non-Decreasing digits: 55
Time Complexity :  O(n).
Space Complexity: O(n) .This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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