Smallest number with given sum of digits and sum of square of digits

Given sum of digits $a$ and sum of square of digits $b$ . Find the smallest number with given sum of digits and sum of the square of digits. The number should not contain more than 100 digits. Print -1 if no such number exists or if the number of digits is more than 100.

Examples:

Input : a = 18, b = 162
Output : 99
Explanation : 99 is the smallest possible number whose sum of digits = 9 + 9 = 18 and sum of squares of digits is 9²+9² = 162.

Input : a = 12, b = 9
Output : -1

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
Since the smallest number can be of 100 digits, it cannot be stored. Hence the first step to solve it will be to find the minimum number of digits which can give us the sum of digits as $a$ and sum of the square of digits as $b$ . To find the minimum number of digits, we can use Dynamic Programming. DP[a][b] signifies the minimum number of digits in a number whose sum of the digits will be $a$ and sum of the square of digits will be $b$ . If there does not exist any such number then DP[a][b] will be -1.

Since the number cannot exceed 100 digits, DP array will be of size 101*8101. Iterate for every digit, and try all possible combination of digits which gives us the sum of digits as $a$ and sum of the square of digits as $b$ . Store the minimum number of digits in DP[a][b] using the below recurrence relation:

DP[a][b] = min( minimumNumberOfDigits(a – i, b – (i * i)) + 1 )
where 1<=i<=9

After getting the minimum number of digits, find the digits. To find the digits, check for all combinations and print those digits which satisfies the condition below:

1 + dp[a – i][b – i * i] == dp[a][b]
where 1<=i<=9

If the condition above is met by any of i, reduce $a$ by i and $b$ by i*i and break. Keep on repeating the above process to find all the digits till $a$ is 0 and $b$ is 0.

Below is the implementation of above approach:

C++

// CPP program to find the Smallest number  
// with given sum of digits and 
// sum of square of digits 
#include <bits/stdc++.h> 
using namespace std; 
  
int dp[901][8101]; 
  
// Top down dp to find minimum number of digits with 
// given sum of dits a and sum of square of digits as b 
int minimumNumberOfDigits(int a, int b) 
{ 
    // Invalid condition  
    if (a > b || a < 0 || b < 0 || a > 900 || b > 8100) 
        return -1; 
      
    // Number of digits satisfied 
    if (a == 0 && b == 0) 
        return 0; 
      
    // Memoization 
    if (dp[a][b] != -1) 
        return dp[a][b]; 
      
    // Initialize ans as maximum as we have to find the   
    // minimum number of digits  
    int ans = 101;  
      
    // Check for all possible combinations of digits 
    for (int i = 9; i >= 1; i--) { 
          
        // recurrence call  
        int k = minimumNumberOfDigits(a - i, b - (i * i));  
          
        // If the combination of digits cannot give sum as a  
        // and sum of square of digits as b  
        if (k != -1) 
            ans = min(ans, k + 1); 
    } 
      
    // Returns the minimum number of digits 
    return dp[a][b] = ans; 
} 
  
// Function to print the digits that gives  
// sum as a and sum of square of digits as b 
void printSmallestNumber(int a,int b) 
{ 
      
    // initialize the dp array as -1 
    memset(dp, -1, sizeof(dp)); 
      
    // base condition  
    dp[0][0] = 0; 
      
    // function call to get the minimum number of digits   
    int k = minimumNumberOfDigits(a, b);  
      
    // When there does not exists any number 
    if (k == -1 || k > 100) 
        cout << "-1"; 
    else { 
        // Printing the digits from the most significant digit 
        while (a > 0 && b > 0) { 
  
            // Trying all combinations  
            for (int i = 1; i <= 9; i++) { 
                // checking conditions for minimum digits 
                if (a >= i && b >= i * i &&  
                    1 + dp[a - i][b - i * i] == dp[a][b]) { 
                    cout << i; 
                    a -= i; 
                    b -= i * i; 
                    break; 
                } 
            } 
        } 
    } 
} 
  
// Driver Code 
int main() 
{ 
    int a = 18, b = 162; 
    // Function call to print the smallest number  
    printSmallestNumber(a,b);  
} 

Java

import java.util.Arrays; 
  
// Java program to find the Smallest number  
// with given sum of digits and 
// sum of square of digits 
class GFG { 
  
    static int dp[][] = new int[901][8101]; 
  
// Top down dp to find minimum number of digits with 
// given sum of dits a and sum of square of digits as b 
    static int minimumNumberOfDigits(int a, int b) { 
        // Invalid condition  
        if (a > b || a < 0 || b < 0 || a > 900 || b > 8100) { 
            return -1; 
        } 
  
        // Number of digits satisfied 
        if (a == 0 && b == 0) { 
            return 0; 
        } 
  
        // Memoization 
        if (dp[a][b] != -1) { 
            return dp[a][b]; 
        } 
  
        // Initialize ans as maximum as we have to find the   
        // minimum number of digits  
        int ans = 101; 
  
        // Check for all possible combinations of digits 
        for (int i = 9; i >= 1; i--) { 
  
            // recurrence call  
            int k = minimumNumberOfDigits(a - i, b - (i * i)); 
  
            // If the combination of digits cannot give sum as a  
            // and sum of square of digits as b  
            if (k != -1) { 
                ans = Math.min(ans, k + 1); 
            } 
        } 
  
        // Returns the minimum number of digits 
        return dp[a][b] = ans; 
    } 
  
// Function to print the digits that gives  
// sum as a and sum of square of digits as b 
    static void printSmallestNumber(int a, int b) { 
  
        // initialize the dp array as -1 
        for (int[] row : dp) { 
            Arrays.fill(row, -1); 
        } 
  
        // base condition  
        dp[0][0] = 0; 
  
        // function call to get the minimum number of digits   
        int k = minimumNumberOfDigits(a, b); 
  
        // When there does not exists any number 
        if (k == -1 || k > 100) { 
            System.out.println("-1"); 
        } else { 
            // Printing the digits from the most significant digit 
            while (a > 0 && b > 0) { 
  
                // Trying all combinations  
                for (int i = 1; i <= 9; i++) { 
                    // checking conditions for minimum digits 
                    if (a >= i && b >= i * i 
                            && 1 + dp[a - i][b - i * i] == dp[a][b]) { 
                        System.out.print(i); 
                        a -= i; 
                        b -= i * i; 
                        break; 
                    } 
                } 
            } 
        } 
    } 
  
// Driver Code 
    public static void main(String args[]) { 
        int a = 18, b = 162; 
        // Function call to print the smallest number  
        printSmallestNumber(a, b); 
    } 
} 
  
// This code is contributed by PrinciRaj19992 

Python3

# Python3 program to find the Smallest number 
# with given sum of digits and 
# sum of square of digits 
  
dp=[[-1 for i in range(8101)]for i in range(901)] 
  
# Top down dp to find minimum number of digits with 
# given sum of dits a and sum of square of digits as b 
def minimumNumberOfDigits(a,b): 
    # Invalid condition  
    if (a > b or a < 0 or b < 0 or a > 900 or b > 8100): 
        return -1
          
    # Number of digits satisfied 
    if (a == 0 and b == 0): 
        return 0
          
    # Memoization 
    if (dp[a][b] != -1): 
        return dp[a][b] 
          
    # Initialize ans as maximum as we have to find the 
    # minimum number of digits 
    ans = 101
      
    #Check for all possible combinations of digits 
    for i in range(9,0,-1): 
          
        # recurrence call 
        k = minimumNumberOfDigits(a - i, b - (i * i)) 
          
        # If the combination of digits cannot give sum as a 
        # and sum of square of digits as b  
        if (k != -1): 
            ans = min(ans, k + 1) 
              
    # Returns the minimum number of digits 
    dp[a][b] = ans 
    return ans 
  
# Function to print the digits that gives 
# sum as a and sum of square of digits as b 
def printSmallestNumber(a,b): 
    # initialize the dp array as 
    for i in range(901): 
        for j in range(8101): 
            dp[i][j]=-1
              
    # base condition 
    dp[0][0] = 0
      
    # function call to get the minimum number of digits 
    k = minimumNumberOfDigits(a, b) 
      
    # When there does not exists any number 
    if (k == -1 or k > 100): 
        print(-1,end='') 
    else: 
        # Printing the digits from the most significant digit 
          
        while (a > 0 and b > 0): 
              
            # Trying all combinations 
            for i in range(1,10): 
                  
                #checking conditions for minimum digits 
                if (a >= i and b >= i * i and
                    1 + dp[a - i][b - i * i] == dp[a][b]): 
                    print(i,end='') 
                    a -= i 
                    b -= i * i 
                    break
# Driver Code 
if __name__=='__main__': 
    a = 18
    b = 162
# Function call to print the smallest number 
    printSmallestNumber(a,b) 
      
# This code is contributed by sahilshelangia 

C#

// C# program to find the Smallest number  
// with given sum of digits and 
// sum of square of digits 
using System; 
public class GFG { 
   
    static int [,]dp = new int[901,8101]; 
   
// Top down dp to find minimum number of digits with 
// given sum of dits a and sum of square of digits as b 
    static int minimumNumberOfDigits(int a, int b) { 
        // Invalid condition  
        if (a > b || a < 0 || b < 0 || a > 900 || b > 8100) { 
            return -1; 
        } 
   
        // Number of digits satisfied 
        if (a == 0 && b == 0) { 
            return 0; 
        } 
   
        // Memoization 
        if (dp[a,b] != -1) { 
            return dp[a,b]; 
        } 
   
        // Initialize ans as maximum as we have to find the   
        // minimum number of digits  
        int ans = 101; 
   
        // Check for all possible combinations of digits 
        for (int i = 9; i >= 1; i--) { 
   
            // recurrence call  
            int k = minimumNumberOfDigits(a - i, b - (i * i)); 
   
            // If the combination of digits cannot give sum as a  
            // and sum of square of digits as b  
            if (k != -1) { 
                ans = Math.Min(ans, k + 1); 
            } 
        } 
   
        // Returns the minimum number of digits 
        return dp[a,b] = ans; 
    } 
   
// Function to print the digits that gives  
// sum as a and sum of square of digits as b 
    static void printSmallestNumber(int a, int b) { 
   
        // initialize the dp array as -1 
        for (int i = 0; i < dp.GetLength(0); i++) 
            for (int j = 0; j < dp.GetLength(1); j++) 
                   dp[i, j] = -1; 
  
   
        // base condition  
        dp[0,0] = 0; 
   
        // function call to get the minimum number of digits   
        int k = minimumNumberOfDigits(a, b); 
   
        // When there does not exists any number 
        if (k == -1 || k > 100) { 
            Console.WriteLine("-1"); 
        } else { 
            // Printing the digits from the most significant digit 
            while (a > 0 && b > 0) { 
   
                // Trying all combinations  
                for (int i = 1; i <= 9; i++) { 
                    // checking conditions for minimum digits 
                    if (a >= i && b >= i * i 
                            && 1 + dp[a - i,b - i * i] == dp[a,b]) { 
                        Console.Write(i); 
                        a -= i; 
                        b -= i * i; 
                        break; 
                    } 
                } 
            } 
        } 
    } 
   
// Driver Code 
    public static void Main() { 
        int a = 18, b = 162; 
        // Function call to print the smallest number  
        printSmallestNumber(a, b); 
    } 
} 
   
// This code is contributed by PrinciRaj19992 

Output:

Time Complexity : O(900*8100*9)
Auxiliary Space : O(900*8100)

Note: Time complexity is in terms of numbers as we are trying all possible combinations of digits.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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