Count of all possible combinations of K numbers that sums to N

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Given a number N, the task is to count the combinations of K numbers from 1 to N having a sum equal to N, with duplicates allowed.

Example:

Input: N = 7, K = 3
Output:15
Explanation:The combinations which lead to the sum N = 7 are: {1, 1, 5}, {1, 5, 1}, {5, 1, 1}, {2, 1, 4}, {1, 2, 4}, {1, 4, 2}, {2, 4, 1}, {4, 1, 2}, {4, 2, 1}, {3, 1, 3}, {1, 3, 3}, {3, 3, 1}, {2, 2, 3}, {2, 3, 2}, {3, 2, 2}

Input: N = 5, K = 5
Output: 1
Explanation: {1, 1, 1, 1, 1} is the only combination.

Naive Approach: This problem can be solved using recursion and then memoising the result to improve time complexity. To solve this problem, follow the below steps:

Create a function, say countWaysUtil which will accept four parameters that are N, K, sum, and dp. Here N is the sum that K elements are required to have, K is the number of elements consumed, sum is the sum accumulated till now and dp is the matrix to memoise the result. This function will give the number of ways to get the sum in K numbers.
Now initially call countWaysUtil with arguments N, K, sum=0 and dp as a matrix filled with all -1.
In each recursive call:
- Check for the base cases:
  - If the sum is equal to N and K become 0, then return 1.
  - If the sum exceeds N and K is still greater than 0, then return 0.
- Now, run a for loop from 1 to N, to check the result for each outcome.
- Sum all the results in a variable cnt and return cnt after memoising.
Print the answer according to the above observation.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count
// all the possible combinations
// of K numbers having sum equals to N
int countWaysUtil(int N, int K, int sum,
                  vector<vector<int> >& dp)
{
 
    // Base Cases
    if (sum == N and K == 0) {
        return 1;
    }
 
    if (sum >= N and K >= 0) {
        return 0;
    }
 
    if (K < 0) {
        return 0;
    }
 
    // If the result is already memoised
    if (dp[sum][K] != -1) {
        return dp[sum][K];
    }
 
    // Recursive Calls
    int cnt = 0;
    for (int i = 1; i <= N; i++) {
        cnt += countWaysUtil(
            N, K - 1,
            sum + i, dp);
    }
 
    // Returning answer
    return dp[sum][K] = cnt;
}
 
void countWays(int N, int K)
{
    vector<vector<int> > dp(N + 1,
                            vector<int>(
                                K + 1, -1));
    cout << countWaysUtil(N, K, 0, dp);
}
 
// Driver Code
int main()
{
    int N = 7, K = 3;
    countWays(N, K);
}

Java

// Java implementation for the above approach
class GFG {
 
    // Function to count
    // all the possible combinations
    // of K numbers having sum equals to N
    static int countWaysUtil(int N, int K, int sum,
                             int[][] dp)
    {
 
        // Base Cases
        if (sum == N && K == 0) {
            return 1;
        }
 
        if (sum >= N && K >= 0) {
            return 0;
        }
 
        if (K < 0) {
            return 0;
        }
 
        // If the result is already memoised
        if (dp[sum][K] != -1) {
            return dp[sum][K];
        }
 
        // Recursive Calls
        int cnt = 0;
        for (int i = 1; i <= N; i++) {
            cnt += countWaysUtil(N, K - 1, sum + i, dp);
        }
 
        // Returning answer
        return dp[sum][K] = cnt;
    }
 
    static void countWays(int N, int K)
    {
        int[][] dp = new int[N + 1][K + 1];
        for (int i = 0; i < N + 1; i++) {
            for (int j = 0; j < K + 1; j++) {
 
                dp[i][j] = -1;
            }
        }
        System.out.print(countWaysUtil(N, K, 0, dp));
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 7, K = 3;
        countWays(N, K);
    }
}
 
// This code is contributed by ukasp.

Python3

# Python3 program for the above approach
 
# Function to count all the possible 
# combinations of K numbers having 
# sum equals to N
def countWaysUtil(N, K, sum, dp):
 
    # Base Cases
    if (sum == N and K == 0):
        return 1
 
    if (sum >= N and K >= 0):
        return 0
 
    if (K < 0):
        return 0
 
    # If the result is already memoised
    if (dp[sum][K] != -1):
        return dp[sum][K]
 
    # Recursive Calls
    cnt = 0
    for i in range(1, N+1):
        cnt += countWaysUtil(N, K - 1, sum + i, dp)
 
    # Returning answer
    dp[sum][K] = cnt
    return dp[sum][K]
 
def countWays(N, K):
     
    dp = [[-1 for _ in range(K + 1)] 
              for _ in range(N + 1)]
    print(countWaysUtil(N, K, 0, dp))
 
# Driver Code
if __name__ == "__main__":
 
    N = 7
    K = 3
     
    countWays(N, K)
 
# This code is contributed by rakeshsahni

C#

// C# implementation for the above approach
using System;
class GFG
{
     
// Function to count
// all the possible combinations
// of K numbers having sum equals to N
static int countWaysUtil(int N, int K, int sum,
                  int [,]dp)
{
 
    // Base Cases
    if (sum == N && K == 0) {
        return 1;
    }
 
    if (sum >= N && K >= 0) {
        return 0;
    }
 
    if (K < 0) {
        return 0;
    }
 
    // If the result is already memoised
    if (dp[sum, K] != -1) {
        return dp[sum, K];
    }
 
    // Recursive Calls
    int cnt = 0;
    for (int i = 1; i <= N; i++) {
        cnt += countWaysUtil(
            N, K - 1,
            sum + i, dp);
    }
 
    // Returning answer
    return dp[sum, K] = cnt;
}
 
static void countWays(int N, int K)
{
    int [,]dp = new int[N + 1, K + 1];
    for(int i = 0; i < N + 1; i++) {
        for(int j = 0; j < K + 1; j++) {
             
            dp[i, j] = -1;
        }
    }
    Console.Write(countWaysUtil(N, K, 0, dp));
}
 
// Driver Code
public static void Main()
{
    int N = 7, K = 3;
    countWays(N, K);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
// Javascript program for the above approach
 
// Function to count
// all the possible combinations
// of K numbers having sum equals to N
function countWaysUtil(N, K, sum, dp) {
 
    // Base Cases
    if (sum == N && K == 0) {
        return 1;
    }
 
    if (sum >= N && K >= 0) {
        return 0;
    }
 
    if (K < 0) {
        return 0;
    }
 
    // If the result is already memoised
    if (dp[sum][K] != -1) {
        return dp[sum][K];
    }
 
    // Recursive Calls
    let cnt = 0;
    for (let i = 1; i <= N; i++) {
        cnt += countWaysUtil(
            N, K - 1,
            sum + i, dp);
    }
 
    // Returning answer
    return dp[sum][K] = cnt;
}
 
function countWays(N, K) {
    let dp = new Array(N + 1).fill(0).map(() => new Array(K + 1).fill(-1))
    document.write(countWaysUtil(N, K, 0, dp));
}
 
// Driver Code
let N = 7, K = 3;
countWays(N, K);
 
// This code is contributed by saurabh_jaiswal.
</script>

Output

Time Complexity: O(N*K)
Space Complexity: O(N*K)

Efficient Approach: This problem can also be solved using the binomial theorem. As the required sum is N with K elements, so suppose the K numbers are:

a1 + a2 + a3 + a4 + …….. + aK = N
According to the standard principle of partitioning in the binomial theorem, the above equation has a solution which is ^N+K-1C_K-1, where K>=0.
But in our case, K>=1.
So, therefore N should be substituted with N-K and the equation becomes ^N-1C_K-1

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Method to find factorial of given number
int factorial(int n)
{
    if (n == 0)
        return 1;
         
    return n * factorial(n - 1);
}
 
// Function to count all the possible 
// combinations of K numbers having
// sum equals to N
int totalWays(int N, int K)
{
     
    // If N<K
    if (N < K)
        return 0;
     
    // Storing numerator
    int  n1 = factorial(N - 1);
     
    // Storing denominator
    int n2 = factorial(K - 1) * factorial(N - K);
    int ans = (n1 / n2);
     
    // Returning answer
    return ans;
}
 
// Driver code
int main() 
{
    int N = 7;
    int K = 3;
    int ans = totalWays(N, K);
     
    cout << ans;
     
    return 0;
}
 
// This code is contributed by Shubham Singh

C

// C Program for the above approach
#include <stdio.h>
 
 // method to find factorial of given number
 int factorial(int n)
 {
    if (n == 0)
      return 1;
 
    return n*factorial(n - 1);
 }
 
 // Function to count
 // all the possible combinations
 // of K numbers having sum equals to N
 int totalWays(int N, int K) {
 
    // If N<K
    if (N < K)
      return 0;
 
    // Storing numerator
    int  n1 = factorial(N - 1);
 
    // Storing denominator
    int n2 = factorial(K - 1)*factorial(N - K);
    int ans = (n1/n2);
 
    // Returning answer
    return ans;
 }
 
  // Driver method
  int main() 
  {
 
    int N = 7;
    int K = 3;
    int ans = totalWays(N, K);
    printf("%d",ans);
    return 0;
  }
   
// This code is contributed by Shubham Singh

Java

// Java Program for the above approach
class Solution{
 
  // method to find factorial of given number
  static int factorial(int n)
  {
    if (n == 0)
      return 1;
 
    return n*factorial(n - 1);
  }
 
  // Function to count
  // all the possible combinations
  // of K numbers having sum equals to N
  static  int totalWays(int N, int K) {
 
    // If N<K
    if (N < K)
      return 0;
 
    // Storing numerator
    int  n1 = factorial(N - 1);
 
    // Storing denominator
    int n2 = factorial(K - 1)*factorial(N - K);
    int ans = (n1/n2);
 
    // Returning answer
    return ans;
  }
 
  // Driver method
  public static void main(String[] args) 
  {
 
    int N = 7;
    int K = 3;
    int ans = totalWays(N, K);
    System.out.println(ans);
  }
}
 
// This code is contributed by umadevi9616

Python3

# Python Program for the above approach
 
from math import factorial
 
 
class Solution:
 
    # Function to count
    # all the possible combinations
    # of K numbers having sum equals to N
    def totalWays(self, N, K):
 
        # If N<K
        if (N < K):
            return 0
 
        # Storing numerator
        n1 = factorial(N-1)
 
        # Storing denominator
        n2 = factorial(K-1)*factorial(N-K)
 
        ans = (n1//n2)
 
        # Returning answer
        return ans
 
 
# Driver Code
if __name__ == '__main__':
    N = 7
    K = 3
    ob = Solution()
    ans = ob.totalWays(N, K)
    print(ans)

C#

// C# Program for the above approach
using System;
public class Solution {
 
  // method to find factorial of given number
  static int factorial(int n) {
    if (n == 0)
      return 1;
 
    return n * factorial(n - 1);
  }
 
  // Function to count
  // all the possible combinations
  // of K numbers having sum equals to N
  static int totalWays(int N, int K) {
 
    // If N<K
    if (N < K)
      return 0;
 
    // Storing numerator
    int n1 = factorial(N - 1);
 
    // Storing denominator
    int n2 = factorial(K - 1) * factorial(N - K);
    int ans = (n1 / n2);
 
    // Returning answer
    return ans;
  }
 
  // Driver method
  public static void Main(String[] args) {
 
    int N = 7;
    int K = 3;
    int ans = totalWays(N, K);
    Console.WriteLine(ans);
  }
}
 
// This code is contributed by umadevi9616

Javascript

<script>
// Javascript program for the above approach
 
// Method to find factorial of given number
function factorial(n)
{
    if (n == 0)
        return 1;
         
    return n * factorial(n - 1);
}
 
// Function to count all the possible 
// combinations of K numbers having
// sum equals to N
function totalWays( N, K)
{
     
    // If N<K
    if (N < K)
        return 0;
     
    // Storing numerator
    let  n1 = factorial(N - 1);
     
    // Storing denominator
    let n2 = factorial(K - 1) * factorial(N - K);
    let ans = (n1 / n2);
     
    // Returning answer
    return ans;
}
 
// Driver code
let N = 7;
let K = 3;
let ans = totalWays(N, K);
 
document.write(ans);
 
// This code is contributed by Shubham Singh
</script>

Output

Time complexity: O(N)
Auxiliary Space: O(1)

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