Smallest N digit number whose sum of square of digits is a Perfect Square - GeeksforGeeks

17 Feb, 2022
Given an integer N, find the smallest N digit number such that the sum of the square of digits (in decimal representation) of the number is also a perfect square. If no such number exists, print -1.Examples: 
Input : N = 2 Output : 34 Explanation: The smallest possible 2 digit number whose sum of square of digits is a perfect square is 34 because 32 + 42 = 52.Input : N = 1 Output : 1 Explanation: The smallest possible 1 digit number is 1 itself.  
Method 1:To solve the problem mentioned above we can use Backtracking. Since we want to find the minimum N digit number satisfying the given condition, the answer will have digits in non-decreasing order. Therefore we generate the possible numbers recursively keeping track of following in each recursive step :
position: the current position of the recursive step i.e. which position digit is being placed.
prev: the previous digit placed because the current digit has to be greater than equal to prev.
sum: the sum of squares of digits placed till now. When digits are placed, this will be used to check whether the sum of squares of all digits placed is a perfect square or not.
A vector which stores what all digits have been placed till this position.
If placing a digit at a position and moving to the next recursive step leads to a possible solution then return 1, else backtrack.Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ implementation to find Smallest N// digit number whose sum of square// of digits is a Perfect Square #include <bits/stdc++.h>using namespace std; // function to check if// number is a perfect squareint isSquare(int n){    int k = sqrt(n);    return (k * k == n);} // function to calculate the// smallest N digit numberint calculate(int pos, int prev,              int sum, vector<int>& v){     if (pos == v.size())        return isSquare(sum);     // place digits greater than equal to prev    for (int i = prev; i <= 9; i++) {        v[pos] = i;        sum += i * i;         // check if placing this digit leads        // to a solution then return it        if (calculate(pos + 1, i, sum, v))            return 1;         // else backtrack        sum -= i * i;    }    return 0;} string minValue(int n){     vector<int> v(n);    if (calculate(0, 1, 0, v)) {         // create a string representing        // the N digit number        string answer = "";        for (int i = 0; i < v.size(); i++)             answer += char(v[i] + '0');         return answer;    }     else        return "-1";} // driver codeint main(){     // initialise N    int N = 2;     cout << minValue(N);     return 0;}
// Java implementation to find Smallest N// digit number whose sum of square// of digits is a Perfect Squareclass GFG{     // function to check if// number is a perfect squarestatic int isSquare(int n){    int k = (int)Math.sqrt(n);    return k * k == n ? 1 : 0;} // Function to calculate the// smallest N digit numberstatic int calculate(int pos, int prev,                     int sum, int[] v){    if (pos == v.length)        return isSquare(sum);     // Place digits greater than equal to prev    for(int i = prev; i <= 9; i++)    {        v[pos] = i;        sum += i * i;         // Check if placing this digit leads        // to a solution then return it        if (calculate(pos + 1, i, sum, v) != 0)            return 1;         // Else backtrack        sum -= i * i;    }    return 0;} static String minValue(int n){    int[] v = new int[n];    if (calculate(0, 1, 0, v) != 0)    {                 // Create a string representing        // the N digit number        String answer = "";                 for(int i = 0; i < v.length; i++)            answer += (char)(v[i] + '0');         return answer;    }    else        return "-1";} // Driver codepublic static void main(String[] args){     // Initialise N    int N = 2;     System.out.println(minValue(N));}} // This code is contributed by jrishabh99
# Python3 implementation to find Smallest N# digit number whose sum of square# of digits is a Perfect Squarefrom math import sqrt # function to check if# number is a perfect squaredef isSquare(n):    k = int(sqrt(n))    return (k * k == n) # function to calculate the# smallest N digit numberdef calculate(pos, prev, sum, v):     if (pos == len(v)):        return isSquare(sum)     # place digits greater than equal to prev    for i in range(prev, 9 + 1):        v[pos] = i        sum += i * i         # check if placing this digit leads        # to a solution then return it        if (calculate(pos + 1, i, sum, v)):            return 1         # else backtrack        sum -= i * i     return 0 def minValue(n):    v = [0]*(n)    if (calculate(0, 1, 0, v)):         # create a representing        # the N digit number        answer = ""        for i in range(len(v)):             answer += chr(v[i] + ord('0'))         return answer     else:        return "-1"  # Driver codeif __name__ == '__main__':     # initialise N    N = 2     print(minValue(N)) # This code is contributed by mohit kumar 29
// C# implementation to find Smallest N// digit number whose sum of square// of digits is a Perfect Squareusing System;class GFG{     // function to check if// number is a perfect squarestatic int isSquare(int n){    int k = (int)Math.Sqrt(n);    return k * k == n ? 1 : 0;} // Function to calculate the// smallest N digit numberstatic int calculate(int pos, int prev,                     int sum, int[] v){    if (pos == v.Length)        return isSquare(sum);     // Place digits greater than equal to prev    for(int i = prev; i <= 9; i++)    {        v[pos] = i;        sum += i * i;         // Check if placing this digit leads        // to a solution then return it        if (calculate(pos + 1, i, sum, v) != 0)            return 1;         // Else backtrack        sum -= i * i;    }    return 0;} static string minValue(int n){    int[] v = new int[n];    if (calculate(0, 1, 0, v) != 0)    {                 // Create a string representing        // the N digit number        string answer = "";                 for(int i = 0; i < v.Length; i++)            answer += (char)(v[i] + '0');         return answer;    }    else        return "-1";} // Driver codepublic static void Main(){     // Initialise N    int N = 2;     Console.Write(minValue(N));}}
<script> // Javascript implementation to find Smallest N// digit number whose sum of square// of digits is a Perfect Square // function to check if// number is a perfect squarefunction isSquare(n){    let k = Math.floor(Math.sqrt(n));    return k * k == n ? 1 : 0;}  // Function to calculate the// smallest N digit numberfunction calculate(pos, prev, sum, v){    if (pos == v.length)        return isSquare(sum);      // Place digits greater than equal to prev    for(let i = prev; i <= 9; i++)    {        v[pos] = i;        sum += i * i;          // Check if placing this digit leads        // to a solution then return it        if (calculate(pos + 1, i, sum, v) != 0)            return 1;          // Else backtrack        sum -= (i * i);    }    return 0;}  function minValue(n){    let v = Array.from({length: n}, (_, i) => 0);    if (calculate(0, 1, 0, v) != 0)    {                  // Create a string representing        // the N digit number        let answer = "";                  for(let i = 0; i < v.length; i++)            answer += (v[i]  + 0);          return answer;    }    else        return "-1";}  // Driver Code         // Initialise N    let N = 2;    document.write(minValue(N));   // This code is contributed by sanjoy_62.</script>
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Time Complexity: O(sqrt(n))
Auxiliary Space: O(n)
Method 2:The above-mentioned problem can also be solved using Dynamic Programming. If we observe the question carefully we see that it can be converted to the standard Coin Change problem. Given N as the number of digits, the base answer will be N 1’s, the sum of the square of whose digits will be N.  
If N itself is a perfect square then the N times 1 will be the final answer.
Otherwise, we will have to replace some 1’s in the answer with other digits from 2-9. Each replacement in the digit will increase the sum of the square by a certain amount and since 1 can be changed to only 8 other possible digits there are only 8 such possible increments. For example, if 1 is changed to 2, then increment will be 22 – 12 = 3. Similarly, all possible changes are : {3, 8, 15, 24, 35, 48, 63, 80}.
So the problem now can be interpreted as having 8 kinds of coins of the aforementioned values and we can use any coin any number of times to create the required sum. The sum of squares will lie in the range of N (all digits are 1) to 81 * N (all digits are 9). We just have to consider perfect square sums in the range and use the idea of coin change to find the N digits that will be in the answer. One important point we need to take into account is that we have to find the smallest N digit number not the number with the smallest square sum of digits.Below is the implementation of the above-mentioned approach: 
C++
Java
// C++ implementation to find the Smallest// N digit number whose sum of square// of digits is a Perfect Square#include <bits/stdc++.h>using namespace std;long long value[8100006];int first[8100006];// array for all possible changesint coins[8] = { 3, 8, 15, 24, 35, 48, 63, 80 }; void coinChange(){    const long long inf = INT_MAX;     // iterating till 81 * N    // since N is at max 10^5    for (int x = 1; x <= 8100005; x++) {         value[x] = inf;         for (auto c : coins) {            if (x - c >= 0 && value[x - c] + 1 < value[x]) {                value[x] = min(value[x], value[x - c] + 1);                 // least value of coin                first[x] = c;            }        }    }} // function to find the// minimum possible valuestring minValue(int n){     // applying coin change for all the numbers    coinChange();     string answer = "";     // check if number is    // perfect square or not    if ((sqrt(n) * sqrt(n)) == n) {        for (int i = 0; i < n; i++)             answer += "1";         return answer;    }     long long hi = 81 * n;    long long lo = sqrt(n);     // keeps a check whether    // number is found or not    bool found = false;     long long upper = 81 * n;    long long lower = n;     // sorting suffix strings    string suffix;    bool suf_init = false;     while ((lo * lo) <= hi) {        lo++;         long long curr = lo * lo;         long long change = curr - n;         if (value[change] <= lower) {             // build a suffix string            found = true;             if (lower > value[change]) {                // number to be used for updation of lower,                // first values that will be used                // to construct the final number later                lower = value[change];                upper = change;                suffix = "";                suf_init = true;                int len = change;                 while (len > 0) {                    int k = sqrt(first[len] + 1);                    suffix = suffix + char(k + 48);                    len = len - first[len];                }            }             else if (lower == value[change]) {                string tempsuf = "";                int len = change;                while (len > 0) {                    int k = sqrt(first[len] + 1);                    tempsuf = tempsuf + char(k + 48);                    len = len - first[len];                }                 if (tempsuf < suffix or suf_init == false) {                    lower = value[change];                    upper = change;                    suffix = tempsuf;                    suf_init = true;                }            }        }    }    // check if number is found    if (found) {        // construct the number from first values        long long x = lower;        for (int i = 0; i < (n - x); i++)            answer += "1";         long long temp = upper;         // fill in rest of the digits        while (temp > 0) {            int dig = sqrt(first[temp] + 1);            temp = temp - first[temp];            answer += char(dig + '0');        }        return answer;    }    else        return "-1";} // driver codeint main(){     // initialise N    int N = 2;     cout << minValue(N);     return 0;}
// Java implementation to find the Smallest// N digit number whose sum of square// of digits is a Perfect Squareimport java.util.*; class GFG {  static long[] value = new long[(int)8100006];  static int[] first = new int[8100006];     // array for all possible changes  static int coins[] = { 3, 8, 15, 24, 35, 48, 63, 80 };   public static void coinChange()  {    final long inf = Integer.MAX_VALUE;     // iterating till 81 * N    // since N is at max 10^5    for (int x = 1; x <= 8100005; x++) {       value[x] = inf;       for (int c : coins) {        if (x - c >= 0            && value[x - c] + 1 < value[x]) {          value[x] = Math.min(value[x],                              value[x - c] + 1);           // least value of coin          first[x] = c;        }      }    }  }   // function to find the  // minimum possible value  public static String minValue(int n)  {     // applying coin change for all the numbers    coinChange();    String answer = "";     // check if number is    // perfect square or not    if ((Math.sqrt(n) * Math.sqrt(n)) == n) {      for (int i = 0; i < n; i++)         answer += "1";       return answer;    }     long hi = 81 * n;    long lo = (long)Math.sqrt(n);     // keeps a check whether    // number is found or not    boolean found = false;     long upper = 81 * n;    long lower = n;     // sorting suffix strings    String suffix = "";    boolean suf_init = false;     while ((lo * lo) <= hi) {      lo++;       long curr = lo * lo;      long change = curr - n;       if (value[(int)change] <= lower) {         // build a suffix string        found = true;         if (lower > value[(int)change])        {                     // number to be used for updation of          // lower, first values that will be used          // to construct the final number later          lower = value[(int)change];          upper = change;          suffix = "";          suf_init = true;          int len = (int)change;           while (len > 0) {            int k = (int)Math.sqrt(first[len]                                   + 1);            suffix = suffix + (char)(k + 48);            len = len - first[len];          }        }         else if (lower == value[(int)change]) {          String tempsuf = "";          int len = (int)change;          while (len > 0) {            int k = (int)Math.sqrt(first[len]                                   + 1);            tempsuf = tempsuf + (char)(k + 48);            len = len - first[len];          }           if ((tempsuf.compareTo(suffix) < 0)              || (suf_init == false)) {            lower = value[(int)change];            upper = change;            suffix = tempsuf;            suf_init = true;          }        }      }    }         // check if number is found    if (found)    {             // construct the number from first values      long x = lower;      for (int i = 0; i < (n - x); i++)        answer += "1";       long temp = upper;       // fill in rest of the digits      while (temp > 0) {        int dig          = (int)Math.sqrt(first[(int)temp] + 1);        temp = temp - first[(int)temp];        answer += (char)(dig + '0');      }      return answer;    }    else      return "-1";  }   // Driver code  public static void main(String[] args)  {    // initialise N    int N = 2;     System.out.println(minValue(N));  }} // This code is contributed by Palak Gupta
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Time Complexity : O(81 * N)
Auxiliary Space: O(81 * 105) 
mohit kumar 29
jrishabh99
ukasp
sanjoy_62
sumitgumber28
as5853535
gulshankumarar231
rishavmahato348
ninja_hattori
Backtracking
Dynamic Programming
Mathematical
Dynamic Programming
Mathematical
Backtracking
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