Count of n digit numbers whose sum of digits equals to given sum

3.5

Given two integers ‘n’ and ‘sum’, find count of all n digit numbers with sum of digits as ‘sum’. Leading 0’s are not counted as digits.
1 <= n <= 100 and 1 <= sum <= 500

Example:

Input:  n = 2, sum = 2
Output: 2
Explanation: Numbers are 11 and 20

Input:  n = 2, sum = 5
Output: 5
Explanation: Numbers are 14, 23, 32, 41 and 50

Input:  n = 3, sum = 6
Output: 21

The idea is simple, we subtract all values from 0 to 9 from given sum and recur for sum minus that digit. Below is recursive formula.

    countRec(n, sum) = ∑countRec(n-1, sum-x)
                            where 0 =< x <= 9 and
                                 sum-x >= 0

    One important observation is, leading 0's must be
    handled explicitly as they are not counted as digits.
    So our final count can be written as below.
    finalCount(n, sum) = ∑countRec(n-1, sum-x)
                           where 1 =< x <= 9 and
                                 sum-x >= 0

Below is a simple recursive solution based on above recursive formula.

C++

// A recursive program to count numbers with sum
// of digits as given 'sum'
#include<bits/stdc++.h>
using namespace std;

// Recursive function to count 'n' digit numbers
// with sum of digits as 'sum'. This function
// considers leading 0's also as digits, that is
// why not directly called
unsigned long long int countRec(int n, int sum)
{
    // Base case
    if (n == 0)
       return sum == 0;

    // Initialize answer
    unsigned long long int ans = 0;

    // Traverse through every digit and count
    // numbers beginning with it using recursion
    for (int i=0; i<=9; i++)
       if (sum-i >= 0)
          ans += countRec(n-1, sum-i);

    return ans;
}

// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining digits.
unsigned long long int finalCount(int n, int sum)
{
    // Initialize final answer
    unsigned long long int ans = 0;

    // Traverse through every digit from 1 to
    // 9 and count numbers beginning with it
    for (int i = 1; i <= 9; i++)
      if (sum-i >= 0)
         ans += countRec(n-1, sum-i);

    return ans;
}

// Driver program
int main()
{
    int n = 2, sum = 5;
    cout << finalCount(n, sum);
    return 0;
}

Java

class sum_dig
{
	// Recursive function to count 'n' digit numbers
	// with sum of digits as 'sum'. This function
	// considers leading 0's also as digits, that is
	// why not directly called
	static int countRec(int n, int sum)
	{
	    // Base case
	    if (n == 0)
	       return sum == 0 ?1:0;
	 
	    // Initialize answer
	    int ans = 0;
	 
	    // Traverse through every digit and count
	    // numbers beginning with it using recursion
	    for (int i=0; i<=9; i++)
	       if (sum-i >= 0)
	          ans += countRec(n-1, sum-i);
	 
	    return ans;
	}
	 
	// This is mainly a wrapper over countRec. It
	// explicitly handles leading digit and calls
	// countRec() for remaining digits.
	static int finalCount(int n, int sum)
	{
	    // Initialize final answer
	    int ans = 0;
	 
	    // Traverse through every digit from 1 to
	    // 9 and count numbers beginning with it
	    for (int i = 1; i <= 9; i++)
	      if (sum-i >= 0)
	         ans += countRec(n-1, sum-i);
	 
	    return ans;
	}

	/* Driver program to test above function */ 
	public static void main (String args[])
	{
		  int n = 2, sum = 5;
    	  System.out.println(finalCount(n, sum));
	}
}/* This code is contributed by Rajat Mishra */


Output:
5

The time complexity of above solution is exponential. If we draw the complete recursion tree, we can observer that many subproblems are solved again and again. For example, if we start with n = 3 and sum = 10, we can reach n = 1, sum = 8, by considering digit sequences 1,1 or 2, 0.
Since same suproblems are called again, this problem has Overlapping Subprolems property. So min square sum problem has both properties (see this and this) of a dynamic programming problem.

Below is Memoization based the implementation.

C++

// A memoization based recursive program to count 
// numbers with sum of n as given 'sum'
#include<bits/stdc++.h>
using namespace std;

// A lookup table used for memoization
unsigned long long int lookup[101][501];

// Memoizatiob based implementation of recursive
// function
unsigned long long int countRec(int n, int sum)
{
    // Base case
    if (n == 0)
       return sum == 0;

    // If this subproblem is already evaluated,
    // return the evaluated value
    if (lookup[n][sum] != -1)
       return lookup[n][sum];

    // Initialize answer
    unsigned long long int ans = 0;

    // Traverse through every digit and
    // recursively count numbers beginning
    // with it
    for (int i=0; i<10; i++)
       if (sum-i >= 0)
          ans += countRec(n-1, sum-i);

    return lookup[n][sum] = ans;
}

// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining n.
unsigned long long int finalCount(int n, int sum)
{
    // Initialize all entries of lookup table
    memset(lookup, -1, sizeof lookup);

    // Initialize final answer
    unsigned long long int ans = 0;

    // Traverse through every digit from 1 to
    // 9 and count numbers beginning with it
    for (int i = 1; i <= 9; i++)
      if (sum-i >= 0)
         ans += countRec(n-1, sum-i);
    return ans;
}

// Driver program
int main()
{
    int n = 3, sum = 5;
    cout << finalCount(n, sum);
    return 0;
}

Java

class sum_dig
{
	// A lookup table used for memoization
	static int lookup[][] = new int[101][501];
	 
	// Memoizatiob based implementation of recursive
	// function
	static int countRec(int n, int sum)
	{
	    // Base case
	    if (n == 0)
	       return sum == 0 ? 1 : 0;
	 
	    // If this subproblem is already evaluated,
	    // return the evaluated value
	    if (lookup[n][sum] != -1)
	       return lookup[n][sum];
	 
	    // Initialize answer
	    int ans = 0;
	 
	    // Traverse through every digit and
	    // recursively count numbers beginning
	    // with it
	    for (int i=0; i<10; i++)
	       if (sum-i >= 0)
	          ans += countRec(n-1, sum-i);
	 
	    return lookup[n][sum] = ans;
	}
	 
	// This is mainly a wrapper over countRec. It
	// explicitly handles leading digit and calls
	// countRec() for remaining n.
	static int finalCount(int n, int sum)
	{
	    // Initialize all entries of lookup table
	    for(int i = 0;i<=100;++i){
	    	for(int j=0;j<=50000;++j){
	    		lookup[i][j] = -1;
	    	}
	    }
	 
	    // Initialize final answer
	    int ans = 0;
	 
	    // Traverse through every digit from 1 to
	    // 9 and count numbers beginning with it
	    for (int i = 1; i <= 9; i++)
	      if (sum-i >= 0)
	         ans += countRec(n-1, sum-i);
	    return ans;
	}

	/* Driver program to test above function */ 
	public static void main (String args[])
	{
		  int n = 2, sum = 5;
    	  System.out.println(finalCount(n, sum));
	}
}/* This code is contributed by Rajat Mishra */

Output:

15

Asked in: Drishti-Soft

Thanks to Gaurav Ahirwar for suggesting above solution.

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