# Ways to sum to N using array elements with repetition allowed

Given a set of m distinct positive integers and a value ‘N’. The problem is to count the total number of ways we can form ‘N’ by doing sum of the array elements. Repetitions and different arrangements are allowed.**Examples :**

Input : arr = {1, 5, 6}, N = 7 Output : 6 Explanation:- The different ways are:1+1+1+1+1+1+11+1+51+5+15+1+11+66+1Input : arr = {12, 3, 1, 9}, N = 14 Output : 150

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**Approach:** The approach is based on the concept of dynamic programming.

countWays(arr, m, N)Declare and initializecount[N + 1]= {0} count[0] = 1 fori= 1 to N forj= 0 to m - 1 if i >= arr[j] count[i] += count[i - arr[j]] return count[N]

Below is the implementation of above approach.

## C++

`// C++ implementation to count ways` `// to sum up to a given value N` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// function to count the total` `// number of ways to sum up to 'N'` `int` `countWays(` `int` `arr[], ` `int` `m, ` `int` `N)` `{` ` ` `int` `count[N + 1];` ` ` `memset` `(count, 0, ` `sizeof` `(count));` ` ` ` ` `// base case` ` ` `count[0] = 1;` ` ` ` ` `// count ways for all values up` ` ` `// to 'N' and store the result` ` ` `for` `(` `int` `i = 1; i <= N; i++)` ` ` `for` `(` `int` `j = 0; j < m; j++)` ` ` `// if i >= arr[j] then` ` ` `// accumulate count for value 'i' as` ` ` `// ways to form value 'i-arr[j]'` ` ` `if` `(i >= arr[j])` ` ` `count[i] += count[i - arr[j]];` ` ` ` ` `// required number of ways` ` ` `return` `count[N];` ` ` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = {1, 5, 6};` ` ` `int` `m = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `int` `N = 7;` ` ` `cout << ` `"Total number of ways = "` ` ` `<< countWays(arr, m, N);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to count ways ` `// to sum up to a given value N` `class` `Gfg` `{` ` ` `static` `int` `arr[] = {` `1` `, ` `5` `, ` `6` `};` ` ` ` ` `// method to count the total number` ` ` `// of ways to sum up to 'N'` ` ` `static` `int` `countWays(` `int` `N)` ` ` `{` ` ` `int` `count[] = ` `new` `int` `[N + ` `1` `];` ` ` ` ` `// base case` ` ` `count[` `0` `] = ` `1` `;` ` ` ` ` `// count ways for all values up` ` ` `// to 'N' and store the result` ` ` `for` `(` `int` `i = ` `1` `; i <= N; i++)` ` ` `for` `(` `int` `j = ` `0` `; j < arr.length; j++)` ` ` ` ` `// if i >= arr[j] then` ` ` `// accumulate count for value 'i' as` ` ` `// ways to form value 'i-arr[j]'` ` ` `if` `(i >= arr[j])` ` ` `count[i] += count[i - arr[j]];` ` ` ` ` `// required number of ways` ` ` `return` `count[N];` ` ` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `N = ` `7` `;` ` ` `System.out.println(` `"Total number of ways = "` ` ` `+ countWays(N));` ` ` `}` `}` |

## Python3

`# Python3 implementation to count` `# ways to sum up to a given value N` `# Function to count the total` `# number of ways to sum up to 'N'` `def` `countWays(arr, m, N):` ` ` `count ` `=` `[` `0` `for` `i ` `in` `range` `(N ` `+` `1` `)]` ` ` ` ` `# base case` ` ` `count[` `0` `] ` `=` `1` ` ` ` ` `# Count ways for all values up` ` ` `# to 'N' and store the result` ` ` `for` `i ` `in` `range` `(` `1` `, N ` `+` `1` `):` ` ` `for` `j ` `in` `range` `(m):` ` ` `# if i >= arr[j] then` ` ` `# accumulate count for value 'i' as` ` ` `# ways to form value 'i-arr[j]'` ` ` `if` `(i >` `=` `arr[j]):` ` ` `count[i] ` `+` `=` `count[i ` `-` `arr[j]]` ` ` ` ` `# required number of ways` ` ` `return` `count[N]` ` ` `# Driver Code` `arr ` `=` `[` `1` `, ` `5` `, ` `6` `]` `m ` `=` `len` `(arr)` `N ` `=` `7` `print` `(` `"Total number of ways = "` `,` ` ` `countWays(arr, m, N))` ` ` `# This code is contributed by Anant Agarwal.` |

## C#

`// C# implementation to count ways ` `// to sum up to a given value N` `using` `System;` `class` `Gfg` `{` ` ` `static` `int` `[]arr = {1, 5, 6};` ` ` ` ` `// method to count the total number` ` ` `// of ways to sum up to 'N'` ` ` `static` `int` `countWays(` `int` `N)` ` ` `{` ` ` `int` `[]count = ` `new` `int` `[N+1];` ` ` ` ` `// base case` ` ` `count[0] = 1;` ` ` ` ` `// count ways for all values up` ` ` `// to 'N' and store the result` ` ` `for` `(` `int` `i = 1; i <= N; i++)` ` ` `for` `(` `int` `j = 0; j < arr.Length; j++)` ` ` ` ` `// if i >= arr[j] then` ` ` `// accumulate count for value 'i' as` ` ` `// ways to form value 'i-arr[j]'` ` ` `if` `(i >= arr[j])` ` ` `count[i] += count[i - arr[j]];` ` ` ` ` `// required number of ways` ` ` `return` `count[N];` ` ` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `N = 7;` ` ` `Console.Write(` `"Total number of ways = "` ` ` `+ countWays(N));` ` ` `}` `}` `//This code is contributed by nitin mittal.` |

## PHP

`<?php` `// PHP implementation to count ways` `// to sum up to a given value N` `// function to count the total` `// number of ways to sum up to 'N'` `function` `countWays(` `$arr` `, ` `$m` `, ` `$N` `)` `{` ` ` `$count` `= ` `array_fill` `(0,` `$N` `+ 1,0);` ` ` ` ` `// base case` ` ` `$count` `[0] = 1;` ` ` ` ` `// count ways for all values up` ` ` `// to 'N' and store the result` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$N` `; ` `$i` `++)` ` ` `for` `(` `$j` `= 0; ` `$j` `< ` `$m` `; ` `$j` `++)` ` ` `// if i >= arr[j] then` ` ` `// accumulate count for value 'i' as` ` ` `// ways to form value 'i-arr[j]'` ` ` `if` `(` `$i` `>= ` `$arr` `[` `$j` `])` ` ` `$count` `[` `$i` `] += ` `$count` `[` `$i` `- ` `$arr` `[` `$j` `]];` ` ` ` ` `// required number of ways` ` ` `return` `$count` `[` `$N` `];` ` ` `}` `// Driver code` `$arr` `= ` `array` `(1, 5, 6);` `$m` `= ` `count` `(` `$arr` `);` `$N` `= 7;` `echo` `"Total number of ways = "` `,countWays(` `$arr` `, ` `$m` `, ` `$N` `);` `// This code is contributed by Ryuga` `?>` |

## Javascript

`<script>` ` ` `// JavaScript implementation to count ways ` ` ` `// to sum up to a given value N` ` ` ` ` `let arr = [1, 5, 6];` ` ` ` ` `// method to count the total number` ` ` `// of ways to sum up to 'N'` ` ` `function` `countWays(N)` ` ` `{` ` ` `let count = ` `new` `Array(N+1);` ` ` `count.fill(0);` ` ` ` ` `// base case` ` ` `count[0] = 1;` ` ` ` ` `// count ways for all values up` ` ` `// to 'N' and store the result` ` ` `for` `(let i = 1; i <= N; i++)` ` ` `for` `(let j = 0; j < arr.length; j++)` ` ` ` ` `// if i >= arr[j] then` ` ` `// accumulate count for value 'i' as` ` ` `// ways to form value 'i-arr[j]'` ` ` `if` `(i >= arr[j])` ` ` `count[i] += count[i - arr[j]];` ` ` ` ` `// required number of ways` ` ` `return` `count[N];` ` ` ` ` `}` ` ` ` ` `let N = 7;` ` ` `document.write(` `"Total number of ways = "` `+ countWays(N));` ` ` `</script>` |

**Output:**

Total number of ways = 6

**Time Complexity:** O(N*m)

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