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

• Difficulty Level : Medium
• Last Updated : 01 Jun, 2021

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+1
1+1+5
1+5+1
5+1+1
1+6
6+1

Input : 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 initialize count[N + 1] = {0}
count = 1
for i = 1 to N
for j = 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 ` `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 = 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);``    ``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 = 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

 `= 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

 ``

Output:

`Total number of ways = 6`

Time Complexity: O(N*m)
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