# Minimizing array sum by subtracting larger elements from smaller ones

Given an array of n elements perform, we need to minimize array sum. We are allowed to perform below operation any number of times.

- Choose any two elements from the array say A and B where A > B and then subtract B from A.

**Examples:**

Input : 1 arr[] = 1 Output : 1 There is no need to apply the above operation because there is only a single number that is 1. Hence, the minimum sum of the array is 1. Input : 3 arr[] = 2 4 6 Output : 6 Perform the following operations:- subtract 2 from 4 then the array becomes 2 2 6 subtract 2 (at 2nd position) from 6 the array becomes 2 2 4 subtract 2 (at 2nd position) from 4 the array becomes 2 2 2 Now the sum of the array will be 6.

**Approach : **After applying all the operations, all the values in the given array will be equal otherwise we can still choose two numbers A and B such that A > B and can reduce the sum further. Value of each element after applying all the operation will be equal to gcd of the array i.e. ans.

Therefore, minimum possible sum will be equal to n * ans.

Below is the implementation of above approach:

## C++

`// CPP program to find the minimum` `// sum of the array.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// returns gcd of two numbers` `int` `gcd(` `int` `a, ` `int` `b)` `{` ` ` `if` `(b == 0)` ` ` `return` `a; ` ` ` `return` `gcd(b, a % b);` `}` `// returns the gcd of the array.` `int` `gcdofArray(` `int` `arr[], ` `int` `n)` `{` ` ` `int` `ans = arr[0];` ` ` `for` `(` `int` `i = 1; i < n; i++)` ` ` `ans = gcd(ans, arr[i]); ` ` ` `return` `ans;` `}` `// Driver Function` `int` `main()` `{` ` ` `int` `arr[] = { 2, 4, 6 }, n;` ` ` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << n * gcdofArray(arr, n)` ` ` `<< endl;` ` ` `return` `0;` `}` |

## Java

`// java program to find the minimum` `// sum of the array.` `import` `java.io.*;` `class` `GFG {` ` ` ` ` `// returns gcd of two numbers` ` ` `static` `int` `gcd(` `int` `a, ` `int` `b)` ` ` `{` ` ` `if` `(b == ` `0` `)` ` ` `return` `a;` ` ` `return` `gcd(b, a % b);` ` ` `}` ` ` ` ` `// returns the gcd of the array.` ` ` `static` `int` `gcdofArray(` `int` `arr[], ` `int` `n)` ` ` `{` ` ` `int` `ans = arr[` `0` `];` ` ` `for` `(` `int` `i = ` `1` `; i < n; i++)` ` ` `ans = gcd(ans, arr[i]);` ` ` `return` `ans;` ` ` `}` ` ` ` ` `// Driver Function` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `arr[] = { ` `2` `, ` `4` `, ` `6` `}, n;` ` ` `n = arr.length;` ` ` `System.out.println( n * gcdofArray(arr, n)) ;` ` ` ` ` `}` `}` `// This code is contributed by vt_m` |

## Python3

`# Python3 code to find the minimum` `# sum of the array.` `# returns gcd of two numbers` `def` `gcd(a, b):` ` ` `if` `b ` `=` `=` `0` `:` ` ` `return` `a` ` ` `return` `gcd(b, a ` `%` `b)` ` ` `# returns the gcd of the array.` `def` `gcdofArray(arr, n):` ` ` `ans ` `=` `arr[` `0` `]` ` ` `for` `i ` `in` `range` `(n):` ` ` `ans ` `=` `gcd(ans, arr[i])` ` ` `return` `ans` ` ` `# Driver Code` `arr ` `=` `[ ` `2` `, ` `4` `, ` `6` `]` `n ` `=` `len` `(arr)` `print` `(n ` `*` `gcdofArray(arr, n))` `# This code is contributed by "Sharad_Bhardwaj".` |

## C#

`// C# program to find the minimum` `// sum of the array.` `using` `System;` `class` `GFG` `{` ` ` ` ` `// returns gcd of two numbers` ` ` `static` `int` `gcd(` `int` `a, ` `int` `b)` ` ` `{` ` ` `if` `(b == 0)` ` ` `return` `a;` ` ` `return` `gcd(b, a % b);` ` ` `}` ` ` ` ` `// returns the gcd of the array.` ` ` `static` `int` `gcdofArray(` `int` `[]arr, ` `int` `n)` ` ` `{` ` ` `int` `ans = arr[0];` ` ` `for` `(` `int` `i = 1; i < n; i++)` ` ` `ans = gcd(ans, arr[i]);` ` ` `return` `ans;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `int` `[]arr = {2, 4, 6};` ` ` `int` `n;` ` ` `n = arr.Length;` ` ` `Console.WriteLine(n * gcdofArray(arr, n)) ;` ` ` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP program to find the minimum` `// sum of the array.` `// returns gcd of two numbers` `function` `gcd(` `$a` `, ` `$b` `)` `{` ` ` `if` `(` `$b` `== 0)` ` ` `return` `$a` `;` ` ` `return` `gcd(` `$b` `, ` `$a` `% ` `$b` `);` `}` `// returns the gcd of the array.` `function` `gcdofArray(` `$arr` `, ` `$n` `)` `{` ` ` `$ans` `= ` `$arr` `[0];` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `$ans` `= gcd(` `$ans` `, ` `$arr` `[` `$i` `]);` ` ` `return` `$ans` `;` `}` `// Driver Code` `$arr` `= ` `array` `( 2, 4, 6 );` `$n` `= ` `count` `(` `$arr` `);` `echo` `$n` `* gcdofArray(` `$arr` `, ` `$n` `), ` `"\n"` `;` `// This code is contributed by Sach` `?>` |

## Javascript

`<script>` ` ` `// Javascript program to find the minimum` ` ` `// sum of the array.` ` ` ` ` `// returns gcd of two numbers` ` ` `function` `gcd(a, b)` ` ` `{` ` ` `if` `(b == 0)` ` ` `return` `a; ` ` ` `return` `gcd(b, a % b);` ` ` `}` ` ` `// returns the gcd of the array.` ` ` `function` `gcdofArray(arr, n)` ` ` `{` ` ` `let ans = arr[0];` ` ` `for` `(let i = 1; i < n; i++)` ` ` `ans = gcd(ans, arr[i]); ` ` ` `return` `ans;` ` ` `}` ` ` `let arr = [ 2, 4, 6 ], n;` ` ` `n = arr.length;` ` ` `document.write(n * gcdofArray(arr, n));` `</script>` |

**Output: **

6

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