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Ways to sum to N using Natural Numbers up to K with repetitions allowed
  • Last Updated : 04 May, 2021

Given two integers N and K, the task is to find the total number of ways of representing N as the sum of positive integers in the range [1, K], where each integer can be chosen multiple times.

Examples:

Input: N = 8, K = 2
Output: 5
Explanation: All possible ways of representing N as sum of positive integers less than or equal to K are:

  1. {1, 1, 1, 1, 1, 1, 1, 1}, the sum is 8.
  2. {2, 1, 1, 1, 1, 1, 1}, the sum is 8.
  3. {2, 2, 1, 1, 1, 1}, the sum is 8.
  4. 2, 2, 2, 1, 1}, the sum is 8.
  5. {2, 2, 2, 2}}, the sum is 8.

Therefore, the total number of ways is 5.

Input: N = 2, K = 2
Output: 2



Naive Approach: The simplest approach to solve the given problem is to generate all possible combinations of choosing integers over the range [1, K] and count those combinations whose sum is N

Time Complexity: O(KN)
Auxiliary Space: O(1)

Efficient Approach: The above approach has Overlapping Subproblems and Optimal Substructure. Hence, in order to optimize, Dynamic Programming is needed to be performed based on the following observations:

  • Considering dp[i] stores the total number of ways for representing i as the sum of integers lying in the range [1, K], then the transition of states can be defined as:
    • For i in the range [1, K] and for every j in the range [1, N]
    • The value of dp[j] is equal to (dp[j]+ dp[j – i]), for all j ≥ i.

Follow the steps below to solve the problem:

  • Initialize an array, say dp[], with all elements as 0, to store all the recursive states.
  • Initialize dp[0] as 1.
  • Now, iterate over the range [1, K] using a variable i and perform the following steps: 
    • Iterate over the range [1, N], using a variable j, and update the value of dp[j] as dp[j]+ dp[j – i], if j ≥ i.
  • After completing the above steps, print the value of dp[N] as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the total number of
// ways to represent N as the sum of
// integers over the range [1, K]
int NumberOfways(int N, int K)
{
 
    // Initialize a list
    vector<int> dp(N + 1, 0);
   
    // Update dp[0] to 1
    dp[0] = 1;
 
    // Iterate over the range [1, K + 1]
    for (int row = 1; row < K + 1; row++)
    {
 
        // Iterate over the range [1, N + 1]
        for (int col = 1; col < N + 1; col++)
        {
 
            // If col is greater
            // than or equal to row
            if (col >= row)
               
                // Update current
                // dp[col] state
                dp[col] = dp[col] + dp[col - row];
          }
    }
 
    // Return the total number of ways
    return(dp[N]);
}
 
// Driver Code
int main()
{
  int N = 8;
  int K = 2;
 
  cout << (NumberOfways(N, K));
}
 
// This code is contributed by mohit kumar 29.

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to find the total number of
// ways to represent N as the sum of
// integers over the range [1, K]
static int NumberOfways(int N, int K)
{
     
    // Initialize a list
    int[] dp = new int[N + 1];
   
    // Update dp[0] to 1
    dp[0] = 1;
 
    // Iterate over the range [1, K + 1]
    for(int row = 1; row < K + 1; row++)
    {
 
        // Iterate over the range [1, N + 1]
        for(int col = 1; col < N + 1; col++)
        {
             
            // If col is greater
            // than or equal to row
            if (col >= row)
               
                // Update current
                // dp[col] state
                dp[col] = dp[col] + dp[col - row];
          }
    }
 
    // Return the total number of ways
    return(dp[N]);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given inputs
    int N = 8;
    int K = 2;
     
    System.out.println(NumberOfways(N, K));
}
}
 
// This code is contributed by offbeat

Python




# Python program for the above approach
 
# Function to find the total number of
# ways to represent N as the sum of
# integers over the range [1, K]
def NumberOfways(N, K):
   
    # Initialize a list
    dp = [0] * (N + 1)
     
    # Update dp[0] to 1
    dp[0] = 1
     
    # Iterate over the range [1, K + 1]
    for row in range(1, K + 1):
       
        # Iterate over the range [1, N + 1]
        for col in range(1, N + 1):
           
            # If col is greater
            # than or equal to row
            if (col >= row):
               
                # Update current
                # dp[col] state
                dp[col] = dp[col] + dp[col - row]
                 
                 
    # Return the total number of ways
    return(dp[N])
 
# Driver Code
 
N = 8
K = 2
 
print(NumberOfways(N, K))

C#




// C# program for the above approach
using System;
class GFG
{
   
    // Function to find the total number of
    // ways to represent N as the sum of
    // integers over the range [1, K]
    static int NumberOfways(int N, int K)
    {
 
        // Initialize a list
        int[] dp = new int[(N + 1)];
 
        // Update dp[0] to 1
        dp[0] = 1;
 
        // Iterate over the range [1, K + 1]
        for (int row = 1; row < K + 1; row++) {
 
            // Iterate over the range [1, N + 1]
            for (int col = 1; col < N + 1; col++) {
 
                // If col is greater
                // than or equal to row
                if (col >= row)
 
                    // Update current
                    // dp[col] state
                    dp[col] = dp[col] + dp[col - row];
            }
        }
 
        // Return the total number of ways
        return (dp[N]);
    }
 
    // Driver Code
    public static void Main()
    {
        int N = 8;
        int K = 2;
 
        Console.WriteLine(NumberOfways(N, K));
    }
}
 
// This code is contributed by ukasp.
Output: 
5

 

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

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