Number of non-negative integral solutions of sum equation

Given a number n (number of variables) and val (sum of the variables), find out how many such non-negative integral solutions are possible.

Examples :

Input : n = 5, val = 1
Output : 5
Explanation:
x1 + x2 + x3 + x4 + x5 = 1
Number of possible solution are : 
(0 0 0 0 1), (0 0 0 1 0), (0 0 1 0 0),
(0 1 0 0 0), (1 0 0 0 0)
Total number of possible solutions are 5
Input : n = 5, val = 4
Output : 70
Explanation:
x1 + x2 + x3 + x4 + x5 = 4
Number of possible solution are: 
(1 1 1 1 0), (1 0 1 1 1), (0 1 1 1 1), 
(2 1 0 0 1), (2 2 0 0 0)........ so on......
Total numbers of possible solutions are 70

Asked in: Microsoft Interview

1. Make a recursive function call to countSolutions(int n, int val)
2. Call this Solution function countSolutions(n-1, val-i) until n = 1 and val >=0 and then return 1.

Below is the implementation of above approach:

C++

// CPP program to find the numbers
// of non negative integral solutions 
#include<iostream>
using namespace std;

// return number of non negative 
// integral solutions
int countSolutions(int n, int val)
{
    // initialize total = 0 
    int total = 0;

    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >=0)
        return 1;

    // iterate the loop till equal the val
    for (int i = 0; i <= val; i++){
        
        // total solution of equations 
        // and again call the recursive 
        // function Solutions(variable,value)
        total += countSolutions(n-1, val-i);
        
    }
    
    // return the total no possible solution 
    return total;
}

// driver code
int main(){
    
    int n = 5;
    int val = 20;
    
    cout<<countSolutions(n, val);
}

//This code is contributed by Smitha Dinesh Semwal

Java

// Java program to find the numbers
// of non negative integral solutions
class GFG {

  // return number of non negative
  // integral solutions
  static int countSolutions(int n, int val) {

    // initialize total = 0
    int total = 0;

    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >= 0)
      return 1;

    // iterate the loop till equal the val
    for (int i = 0; i <= val; i++) {

      // total solution of equations
      // and again call the recursive
      // function Solutions(variable, value)
      total += countSolutions(n - 1, val - i);
    }

    // return the total no possible solution
    return total;
  }

  // Driver code
  public static void main(String[] args) {
    int n = 5;
    int val = 20;

    System.out.print(countSolutions(n, val));
  }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find the numbers
# of non negative integral solutions 

# return number of non negative 
# integral solutions
def countSolutions(n, val):

    # initialize total = 0 
    total = 0

    # Base Case if n = 1 and val >= 0
    # then it should return 1
    if n == 1 and val >=0:
        return 1

    # iterate the loop till equal the val
    for i in range(val+1):
    
        # total solution of equations 
        # and again call the recursive 
        # function Solutions(variable,value)
        total += countSolutions(n-1, val-i)

    # return the total no possible solution 
    return total

# driver code
n = 5
val = 20
print(countSolutions(n, val))

// C# program to find the numbers
// of non negative integral solutions
using System;

class GFG {

    // return number of non negative
    // integral solutions
    static int countSolutions(int n, int val) {
    
        // initialize total = 0
        int total = 0;
    
        // Base Case if n = 1 and val >= 0
        // then it should return 1
        if (n == 1 && val >= 0)
        return 1;
    
        // iterate the loop till equal the val
        for (int i = 0; i <= val; i++) {
        
            // total solution of equations
            // and again call the recursive
            // function Solutions(variable, value)
            total += countSolutions(n - 1, val - i);
        }
    
        // return the total no possible solution
        return total;
    }
    
    // Driver code
    public static void Main()
    {
        int n = 5;
        int val = 20;
    
        Console.WriteLine(countSolutions(n, val));
    }
}

// This code is contributed by Anant vt_m.

JavaScript

<script>

// JavaScript program to find the numbers
// of non negative integral solutions

  // return number of non negative
  // integral solutions
  function countSolutions(n, val) {
  
    // initialize total = 0
    let total = 0;
  
    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >= 0)
      return 1;
  
    // iterate the loop till equal the val
    for (let i = 0; i <= val; i++) {
  
      // total solution of equations
      // and again call the recursive
      // function Solutions(variable, value)
      total += countSolutions(n - 1, val - i);
    }
  
    // return the total no possible solution
    return total;
  }

// Driver code
        
        let n = 5;
        let val = 20;
  
    document.write(countSolutions(n, val));
        
</script>

PHP

<?php
// PHP program to find the numbers
// of non negative integral solutions 

// return number of non negative 
// integral solutions
function countSolutions($n, $val)
{
    // initialize total = 0 
    $total = 0;

    // Base Case if n = 1 and 
    // val >= 0 then it should 
    // return 1
    if ($n == 1 && $val >=0)
        return 1;

    // iterate the loop 
    // till equal the val
    for ($i = 0; $i <= $val; $i++)
    {
        
        // total solution of equations 
        // and again call the recursive 
        // function Solutions(variable,value)
        $total += countSolutions($n - 1, 
                                 $val - $i);
        
    }
    
    // return the total 
    // no possible solution 
    return $total;
}

// Driver Code
$n = 5;
$val = 20;

echo countSolutions($n, $val);

// This code is contributed by nitin mittal. 
?>

Output :

Dynamic Programming Approach:

(Using Memoization)

C++

// CPP program to find the numbers
// of non negative integral solutions
#include<bits/stdc++.h>
using namespace std;

int dp[1001][1001];

// return number of non negative
// integral solutions
int countSolutions(int n, int val)
{
    // initialize total = 0
    int total = 0;

    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >=0) {
        return 1;
    }
  
    // If a value already present in dp,
    // return it
    if(dp[n][val] != -1) {
        return dp[n][val];
    }
    
    // iterate the loop till equal the val
    for (int i = 0; i <= val; i++){
        
        // total solution of equations
        // and again call the recursive
        // function Solutions(variable,value)
        total += countSolutions(n-1, val-i);
        
    }
    
    // Store the value in dp
    dp[n][val] = total;
    
    // Return dp
    return dp[n][val];
}

// driver code
int main(){
    
    int n = 5;
    int val = 20;
    
    memset(dp, -1, sizeof(dp));
    
    cout << countSolutions(n, val);
}

Java

// Java program to find the numbers
// of non negative integral solutions
import java.util.*;
public class GFG
{
  static int dp[][] = new int[1001][1001];

  // return number of non negative
  // integral solutions
  static int countSolutions(int n, int val)
  {
    // initialize total = 0
    int total = 0;

    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >=0) {
      return 1;
    }

    // If a value already present in dp,
    // return it
    if(dp[n][val] != -1) {
      return dp[n][val];
    }

    // iterate the loop till equal the val
    for (int i = 0; i <= val; i++){

      // total solution of equations
      // and again call the recursive
      // function Solutions(variable,value)
      total += countSolutions(n-1, val-i);

    }

    // Store the value in dp
    dp[n][val] = total;

    // Return dp
    return dp[n][val];
  }

  // driver code
  public static void main(String args[]){

    int n = 5;
    int val = 20;

    for(int i = 0; i < 1001; i++) {
      for(int j = 0; j < 1001; j++) {
        dp[i][j]=-1;
      }
    }

    System.out.println(countSolutions(n, val));
  }
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python3 program to find the numbers
# of non negative integral solutions 

# Taking the matrix as globally
dp = [[-1 for i in range(1001)] 
          for j in range(1001)]

# Return number of non negative 
# integral solutions
def countSolutions(n, val):

    # Initialize total = 0 
    total = 0

    # Base Case if n = 1 and val >= 0
    # then it should return 1
    if n == 1 and val >= 0:
        return 1
    
    # If a value is already present
    # in dp
    if (dp[n][val] != -1):
        return dp[n][val]
    
    # Iterate the loop till equal the val
    for i in range(val + 1):
    
        # total solution of equations 
        # and again call the recursive 
        # function Solutions(variable,value)
        total += countSolutions(n - 1, val - i)

    # Return the total no possible solution 
    dp[n][val] = total
    return dp[n][val]

# Driver code
n = 5
val = 20

print(countSolutions(n, val))

# This code is contributed by Samim Hossain Mondal.

// C# program to find the numbers
// of non negative integral solutions
using System;
class GFG
{
  static int [,]dp = new int[1001, 1001];

  // return number of non negative
  // integral solutions
  static int countSolutions(int n, int val)
  {
    // initialize total = 0
    int total = 0;

    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >=0) {
      return 1;
    }

    // If a value already present in dp,
    // return it
    if(dp[n, val] != -1) {
      return dp[n, val];
    }

    // iterate the loop till equal the val
    for (int i = 0; i <= val; i++){

      // total solution of equations
      // and again call the recursive
      // function Solutions(variable,value)
      total += countSolutions(n-1, val-i);

    }

    // Store the value in dp
    dp[n, val] = total;

    // Return dp
    return dp[n, val];
  }

  // driver code
  public static void Main(){

    int n = 5;
    int val = 20;

    for(int i = 0; i < 1001; i++) {
      for(int j = 0; j < 1001; j++) {
        dp[i, j] = -1;
      }
    }

    Console.Write(countSolutions(n, val));
  }
}

// This code is contributed by Samim Hossain Mondal.

JavaScript

<script>
// JavaScript program to find the numbers
// of non negative integral solutions
  var dp = new Array(1001);
  
  // Loop to create 2D array using 1D array
  for (var i = 0; i < dp.length; i++) {
      dp[i] = new Array(1001);
  }
  
  // return number of non negative
  // integral solutions
  function countSolutions(n, val) {
  
    // initialize total = 0
    let total = 0;
  
    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >= 0)
      return 1;
    
    // if a value is already 
    // present in dp
    if(dp[n][val] != -1) 
        return dp[n][val];
    
    // iterate the loop till equal the val
    for (let i = 0; i <= val; i++) {
  
      // total solution of equations
      // and again call the recursive
      // function Solutions(variable, value)
      total += countSolutions(n - 1, val - i);
    }
  
    // return the total no possible solution
    return dp[n][val] = total;
  }

// Driver code
let n = 5;
let val = 20;

for(let i = 0; i < 1001; i++) {
    for(let j = 0; j < 1001; j++) {
        dp[i][j] = -1;
    }
}
document.write(countSolutions(n, val));

// This code is contributed by Samim Hossain Mondal.   
</script>

Output

Time Complexity: O(n * val)

Auxiliary Space: O(n * val)

Dynamic Programming Approach: (Bottom-Up)

The approach used here is a bottom-up dynamic programming approach. We start with the simplest case (one variable) and build up the solution to the n variables.

We create a two-dimensional array called dp with dimensions n+1 x val+1, where dp[i][j] represents the number of solutions to the equation with i variables and sum j.
Since there is only one variable in the first row (i=1), the number of solutions to the equation is always 1. Thus, initialize the first row of dp with all 1's.
For each subsequent row (i > 1) and each column j, we calculate dp[i][j] as the sum of dp[i-1][j-k] for all possible values of k between 0 and j. Because, for each solution to the equation with i variables and sum j, we can obtain a solution to the equation with i-1 variables and sum j-k by subtracting k from one of the variables.
After computing all values of dp[n][j], we return dp[n][val], which represents the number of non-negative integral solutions to the equation with n variables and sum val.

Here is the Java Implementation of above algorithm

C++

#include <iostream>
using namespace std;

// Function to count the number of 
// non-negative integral solutions
int Solutions(int n, int val)
{
    int dp[n + 1][val + 1];
    // Initialize the first row of the dp array dp[1][j] will have 1 solution for every j
    for (int i = 0; i <= val; i++) {
        dp[1][i] = 1;
    }
    // Compute the number of solutions for each i and j using previous solutions
    for (int i = 2; i <= n; i++) {
        for (int j = 0; j <= val; j++) {
            dp[i][j] = 0;
            for (int k = 0; k <= j; k++) {
                dp[i][j] += dp[i - 1][j - k];
            }
        }
    }
    // Return the solution for n variables and sum val
    return dp[n][val];
}
// Driver code
int main()
{
    int n = 5;
    int val = 20;
    // Function call
    cout << Solutions(n, val) << endl;
    return 0;
}

Java

// Java program to find the numbers
// of non negative integral solutions using bottom up
// approach
import java.io.*;

class GFG {
    // return number of non negative
    // integral solutions
    public static int countSolutions(int n, int val)
    {
        // Create a 2D array to store the number of
        // solutions dp[i][j] will store the number of
        // solutions for i variables and sum j
        int dp[][] = new int[n + 1][val + 1];

        // Initialize the first row of the dp array
        // dp[1][j] will have 1 solution for every j
        for (int i = 0; i <= val; i++) {
            dp[1][i] = 1;
        }

        // Compute the number of solutions for each i and j
        // using previous solutions
        for (int i = 2; i <= n; i++) {
            for (int j = 0; j <= val; j++) {
                dp[i][j] = 0;
                // For each k from 0 to j, add the number of
                // solutions from the previous row
                // dp[i][j] = dp[i-1][j] + dp[i-1][j-1] +
                // ... + dp[i-1][j-k]
                for (int k = 0; k <= j; k++) {
                    dp[i][j] += dp[i - 1][j - k];
                }
            }
        }
        // Return the solution for n variables and sum val
        return dp[n][val];
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 5;
        int val = 20;

        // function call
        System.out.println(countSolutions(n, val));
    }
}

// This code is contributed by Tapesh(tapeshdua420)

Python3

def solutions(n, val):
    dp = [[0 for _ in range(val + 1)] for _ in range(n + 1)]

    # Initialize the first row of the dp array
    for i in range(val + 1):
        dp[1][i] = 1

    # Compute the number of solutions for each i and j using previous solutions
    for i in range(2, n + 1):
        for j in range(val + 1):
            dp[i][j] = 0
            for k in range(j + 1):
                dp[i][j] += dp[i - 1][j - k]

    # Return the solution for n variables and sum val
    return dp[n][val]

# Driver code
if __name__ == "__main__":
    n = 5
    val = 20
    # Function call
    print(solutions(n, val))

using System;

class GFG
{
    // Function to return the number of non-negative integral solutions
    public static int CountSolutions(int n, int val)
    {
        // Create a 2D array to store the number of solutions
        // dp[i][j] will store the number of solutions for i variables and sum j
        int[,] dp = new int[n + 1, val + 1];

        // Initialize the first row of the dp array
        // dp[1][j] will have 1 solution for every j
        for (int i = 0; i <= val; i++)
        {
            dp[1, i] = 1;
        }

        // Compute the number of solutions for each i and j
        // using previous solutions
        for (int i = 2; i <= n; i++)
        {
            for (int j = 0; j <= val; j++)
            {
                dp[i, j] = 0;
                // For each k from 0 to j, add the number of solutions from the previous row
                // dp[i][j] = dp[i-1][j] + dp[i-1][j-1] + ... + dp[i-1][j-k]
                for (int k = 0; k <= j; k++)
                {
                    dp[i, j] += dp[i - 1, j - k];
                }
            }
        }
        // Return the solution for n variables and sum val
        return dp[n, val];
    }

    // Main method
    public static void Main(string[] args)
    {
        int n = 5;
        int val = 20;

        // Function call
        Console.WriteLine(CountSolutions(n, val));
    }
}

JavaScript

// JavaScript program to find the numbers
// of non negative integral solutions using bottom up
// approach

// return number of non negative
// integral solutions
function countSolutions(n, val) 
{
    // Create a 2D array to store the number of
    // solutions dp[i][j] will store the number of
    // solutions for i variables and sum j
    const dp = new Array(n + 1).fill(null).map(() => new Array(val + 1).fill(0));
    
    // Initialize the first row of the dp array
    // dp[1][j] will have 1 solution for every j
    for (let i = 0; i <= val; i++) {
        dp[1][i] = 1;
    }
    
    // Compute the number of solutions for each i and j
    // using previous solutions
    for (let i = 2; i <= n; i++) {
        for (let j = 0; j <= val; j++) {
            dp[i][j] = 0;
            // For each k from 0 to j, add the number of
            // solutions from the previous row
            // dp[i][j] = dp[i-1][j] + dp[i-1][j-1] +
            // ... + dp[i-1][j-k]
            
            for (let k = 0; k <= j; k++) {
                dp[i][j] += dp[i - 1][j - k];
            }
        }
    }
    
    // Return the solution for n variables and sum val
    return dp[n][val];
}

// Driver code
const n = 5;
const val = 20;
// function call
console.log(countSolutions(n, val));

Output

Complexity Analysis

Time Complexity : O(n * val^2), where n is the number of variables and val is the given sum.

Auxiliary Space : O( n * val), creating a 2D array of size (n + 1) * (val + 1).

Number of non-negative integral solutions of sum equation

Complexity Analysis

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