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# Find number of solutions of a linear equation of n variables

• Difficulty Level : Hard
• Last Updated : 22 Apr, 2021

Given a linear equation of n variables, find number of non-negative integer solutions of it. For example,let the given equation be “x + 2y = 5”, solutions of this equation are “x = 1, y = 2”, “x = 5, y = 0” and “x = 1. It may be assumed that all coefficients in given equation are positive integers.
Example :

```Input:  coeff[] = {1, 2}, rhs = 5
Output: 3
The equation "x + 2y = 5" has 3 solutions.
(x=3,y=1), (x=1,y=2), (x=5,y=0)

Input:  coeff[] = {2, 2, 3}, rhs = 4
Output: 3
The equation "2x + 2y + 3z = 4"  has 3 solutions.
(x=0,y=2,z=0), (x=2,y=0,z=0), (x=1,y=1,z=0)```

We strongly recommend you to minimize your browser and try this yourself first.
We can solve this problem recursively. The idea is to subtract first coefficient from rhs and then recur for remaining value of rhs.

```If rhs = 0
countSol(coeff, 0, rhs, n-1) = 1
Else
countSol(coeff, 0, rhs, n-1) = ∑countSol(coeff, i, rhs-coeff[i], m-1)
where coeff[i]<=rhs and
i varies from 0 to n-1                             ```

Below is recursive implementation of above solution.

## C++

 `// A naive recursive C++ program to``// find number of non-negative solutions``// for a given linear equation``#include``using` `namespace` `std;` `// Recursive function that returns``// count of solutions for given rhs``// value and coefficients coeff[start..end]``int` `countSol(``int` `coeff[], ``int` `start,``             ``int` `end, ``int` `rhs)``{``    ``// Base case``    ``if` `(rhs == 0)``    ``return` `1;` `    ``// Initialize count``    ``// of solutions``    ``int` `result = 0;` `    ``// One by subtract all smaller or``    ``// equal coefficiants and recur``    ``for` `(``int` `i = start; i <= end; i++)``    ``if` `(coeff[i] <= rhs)``        ``result += countSol(coeff, i, end,``                           ``rhs - coeff[i]);` `    ``return` `result;``}` `// Driver Code``int` `main()``{``    ``int` `coeff[] = {2, 2, 5};``    ``int` `rhs = 4;``    ``int` `n = ``sizeof``(coeff) / ``sizeof``(coeff);``    ``cout << countSol(coeff, 0, n - 1, rhs);``    ``return` `0;``}`

## Java

 `// A naive recursive Java program``// to find number of non-negative``// solutions for a given linear equation``import` `java.io.*;` `class` `GFG``{``        ` `    ``// Recursive function that returns``    ``// count of solutions for given``    ``// rhs value and coefficients coeff[start..end]``    ``static` `int` `countSol(``int` `coeff[], ``int` `start,``                        ``int` `end, ``int` `rhs)``    ``{``        ``// Base case``        ``if` `(rhs == ``0``)``        ``return` `1``;``    ` `        ``// Initialize count of solutions``        ``int` `result = ``0``;``    ` `        ``// One by subtract all smaller or``        ``// equal coefficiants and recur``        ``for` `(``int` `i = start; i <= end; i++)``        ``if` `(coeff[i] <= rhs)``            ``result += countSol(coeff, i, end,``                               ``rhs - coeff[i]);``    ` `        ``return` `result;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `coeff[] = {``2``, ``2``, ``5``};``        ``int` `rhs = ``4``;``        ``int` `n = coeff.length;``        ``System.out.println (countSol(coeff, ``0``,``                                     ``n - ``1``, rhs));``            ` `    ``}``}` `// This code is contributed by vt_m.`

## Python3

 `# A naive recursive Python program``# to find number of non-negative``# solutions for a given linear equation` `# Recursive function that returns``# count of solutions for given rhs``# value and coefficients coeff[stat...end]``def` `countSol(coeff, start, end, rhs):` `    ``# Base case``    ``if` `(rhs ``=``=` `0``):``        ``return` `1` `    ``# Initialize count of solutions``    ``result ``=` `0` `    ``# One by one subtract all smaller or``    ``# equal coefficients and recur``    ``for` `i ``in` `range``(start, end``+``1``):``        ``if` `(coeff[i] <``=` `rhs):``            ``result ``+``=` `countSol(coeff, i, end,``                               ``rhs ``-` `coeff[i])` `    ``return` `result` `# Driver Code``coeff ``=` `[``2``, ``2``, ``5``]``rhs ``=` `4``n ``=` `len``(coeff)``print``(countSol(coeff, ``0``, n ``-` `1``, rhs))` `# This code is contributed``# by Soumen Ghosh`

## C#

 `// A naive recursive C# program``// to find number of non-negative``// solutions for a given linear equation``using` `System;` `class` `GFG``{``        ` `    ``// Recursive function that``    ``// returns count of solutions``    ``// for given RHS value and``    ``// coefficients coeff[start..end]``    ``static` `int` `countSol(``int` `[]coeff, ``int` `start,``                        ``int` `end, ``int` `rhs)``    ``{``        ``// Base case``        ``if` `(rhs == 0)``        ``return` `1;``    ` `        ``// Initialize count of solutions``        ``int` `result = 0;``    ` `        ``// One by subtract all smaller or``        ``// equal coefficiants and recur``        ``for` `(``int` `i = start; i <= end; i++)``        ``if` `(coeff[i] <= rhs)``            ``result += countSol(coeff, i, end,``                               ``rhs - coeff[i]);``    ` `        ``return` `result;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]coeff = {2, 2, 5};``        ``int` `rhs = 4;``        ``int` `n = coeff.Length;``        ``Console.Write (countSol(coeff, 0,``                                ``n - 1, rhs));``            ` `    ``}``}` `// This Code is contributed``// by nitin mittal.`

## PHP

 ``

## Javascript

 ``

Output :

`3`

The time complexity of above solution is exponential. We can solve this problem in Pseudo Polynomial Time (time complexity is dependent on numeric value of input) using Dynamic Programming. The idea is similar to Dynamic Programming solution Subset Sum problem. Below is Dynamic Programming based implementation.

## C++

 `// A Dynamic programming based C++``// program to find number of non-negative``// solutions for a given linear equation``#include``using` `namespace` `std;` `// Returns count of solutions for``// given rhs and coefficients coeff[0..n-1]``int` `countSol(``int` `coeff[], ``int` `n, ``int` `rhs)``{``    ``// Create and initialize a table``    ``// to store results of subproblems``    ``int` `dp[rhs + 1];``    ``memset``(dp, 0, ``sizeof``(dp));``    ``dp = 1;` `    ``// Fill table in bottom up manner``    ``for` `(``int` `i = 0; i < n; i++)``    ``for` `(``int` `j = coeff[i]; j <= rhs; j++)``        ``dp[j] += dp[j - coeff[i]];` `    ``return` `dp[rhs];``}` `// Driver Code``int` `main()``{``    ``int` `coeff[] = {2, 2, 5};``    ``int` `rhs = 4;``    ``int` `n = ``sizeof``(coeff) / ``sizeof``(coeff);``    ``cout << countSol(coeff, n, rhs);``    ``return` `0;``}`

## Java

 `// A Dynamic programming based Java program``// to find number of non-negative solutions``// for a given linear equation``import` `java.util.Arrays;` `class` `GFG``{``    ``// Returns counr of solutions for given``    ``// rhs and coefficients coeff[0..n-1]``    ``static` `int` `countSol(``int` `coeff[],``                        ``int` `n, ``int` `rhs)``    ``{``        ` `        ``// Create and initialize a table to``        ``// store results of subproblems``        ``int` `dp[] = ``new` `int``[rhs + ``1``];``        ``Arrays.fill(dp, ``0``);``        ``dp[``0``] = ``1``;``    ` `        ``// Fill table in bottom up manner``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``for` `(``int` `j = coeff[i]; j <= rhs; j++)``            ``dp[j] += dp[j - coeff[i]];``    ` `        ``return` `dp[rhs];``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `coeff[] = {``2``, ``2``, ``5``};``        ``int` `rhs = ``4``;``        ``int` `n = coeff.length;``        ``System.out.print(countSol(coeff, n, rhs));``    ``}``}` `// This code is contributed by Anant Agarwal`

## Python3

 `# A Dynamic Programming based``# Python program to find number``# of non-negative solutions for``# a given linear equation` `# Returns count of solutions for given``# rhs and coefficients coeff[0...n-1]``def` `countSol(coeff, n, rhs):` `    ``# Create and initialize a table``    ``# to store results of subproblems``    ``dp ``=` `[``0` `for` `i ``in` `range``(rhs ``+` `1``)]``    ``dp[``0``] ``=` `1` `    ``# Fill table in bottom up manner``    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(coeff[i], rhs ``+` `1``):``            ``dp[j] ``+``=` `dp[j ``-` `coeff[i]]` `    ``return` `dp[rhs]` `# Driver Code``coeff ``=` `[``2``, ``2``, ``5``]``rhs ``=` `4``n ``=` `len``(coeff)``print``(countSol(coeff, n, rhs))` `# This code is contributed``# by Soumen Ghosh`

## C#

 `// A Dynamic programming based``// C# program to find number of``// non-negative solutions for a``// given linear equation``using` `System;` `class` `GFG``{``    ``// Returns counr of solutions``    ``// for given rhs and coefficients``    ``// coeff[0..n-1]``    ``static` `int` `countSol(``int` `[]coeff,``                        ``int` `n, ``int` `rhs)``    ``{``        ` `        ``// Create and initialize a``        ``// table to store results``        ``// of subproblems``        ``int` `[]dp = ``new` `int``[rhs + 1];``        ` `        ``// Arrays.fill(dp, 0);``        ``dp = 1;``    ` `        ``// Fill table in``        ``// bottom up manner``        ``for` `(``int` `i = 0; i < n; i++)``        ``for` `(``int` `j = coeff[i]; j <= rhs; j++)``            ``dp[j] += dp[j - coeff[i]];``    ` `        ``return` `dp[rhs];``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]coeff = {2, 2, 5};``        ``int` `rhs = 4;``        ``int` `n = coeff.Length;``        ``Console.Write(countSol(coeff,``                               ``n, rhs));``    ``}``}` `// This code is contributed``// by shiv_bhakt.`

## PHP

 ``

## Javascript

 ``

Output :

`3`

Time Complexity of above solution is O(n * rhs)