Given an array of N integers. The task is to find the smallest index of an element such that when multiplied by -1 the sum of whole array becomes 0. If there is no such index return -1.

**Examples:**

Input :arr[] = {1, 3, -5, 3, 4}Output :2Input :arr[] = {5, 3, 6, -7, -4}Output :-1

**Naive Approach :**

The simple solution will be to take each element, multiply it by -1 and check if the new sum is 0. This algorithm works in **O(N ^{2})**.

**Efficient Approach :**

If we take S as our initial sum of the array and we multiply current element A_{i} by -1 then the new sum will become S – 2*A_{i} and this should be equal to 0. So when for the first time** S = 2*A _{i}** then the current index is our required and if no element satisfies the condition then our answer will be -1. The time complexity of this algorithm is

**O(N)**.

Below is the implementation of the above idea :

## C++

`// C++ program to find minimum index ` `// such that sum becomes 0 when the ` `// element is multiplied by -1 ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find minimum index ` `// such that sum becomes 0 when the ` `// element is multiplied by -1 ` `int` `minIndex(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `// Find array sum ` ` ` `int` `sum = 0; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `sum += arr[i]; ` ` ` ` ` `// Find element with value equal to sum/2 ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// when sum is equal to 2*element ` ` ` `// then this is our required element ` ` ` `if` `(2 * arr[i] == sum) ` ` ` `return` `(i + 1); ` ` ` `} ` ` ` ` ` `return` `-1; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 1, 3, -5, 3, 4 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `cout << minIndex(arr, n) << endl; ` ` ` `return` `0; ` `} ` |

## Java

`// Java program to find minimum index ` `// such that sum becomes 0 when the ` `// element is multiplied by -1 ` ` ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` `// Function to find minimum index ` `// such that sum becomes 0 when the ` `// element is multiplied by -1 ` ` ` `static` `int` `minIndex(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `// Find array sum ` ` ` `int` `sum = ` `0` `; ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `sum += arr[i]; ` ` ` ` ` `// Find element with value equal to sum/2 ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) { ` ` ` ` ` `// when sum is equal to 2*element ` ` ` `// then this is our required element ` ` ` `if` `(` `2` `* arr[i] == sum) ` ` ` `return` `(i + ` `1` `); ` ` ` `} ` ` ` ` ` `return` `-` `1` `; ` `} ` ` ` `// Driver code ` ` ` ` ` ` ` `public` `static` `void` `main (String[] args) { ` ` ` `int` `[]arr = { ` `1` `, ` `3` `, -` `5` `, ` `3` `, ` `4` `}; ` ` ` `int` `n =arr.length; ` ` ` `System.out.print( minIndex(arr, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by anuj_67.. ` |

## Python 3

`# Python 3 program to find minimum index ` `# such that sum becomes 0 when the ` `# element is multiplied by -1 ` ` ` `# Function to find minimum index ` `# such that sum becomes 0 when the ` `# element is multiplied by -1 ` `def` `minIndex(arr, n): ` ` ` ` ` `# Find array sum ` ` ` `sum` `=` `0` ` ` `for` `i ` `in` `range` `(` `0` `, n): ` ` ` `sum` `+` `=` `arr[i] ` ` ` ` ` `# Find element with value ` ` ` `# equal to sum/2 ` ` ` `for` `i ` `in` `range` `(` `0` `, n): ` ` ` ` ` `# when sum is equal to 2*element ` ` ` `# then this is our required element ` ` ` `if` `(` `2` `*` `arr[i] ` `=` `=` `sum` `): ` ` ` `return` `(i ` `+` `1` `) ` ` ` ` ` `return` `-` `1` ` ` `# Driver code ` `arr ` `=` `[ ` `1` `, ` `3` `, ` `-` `5` `, ` `3` `, ` `4` `]; ` `n ` `=` `len` `(arr); ` `print` `(minIndex(arr, n)) ` ` ` `# This code is contributed ` `# by Akanksha Rai ` |

## C#

`// C# program to find minimum index ` `// such that sum becomes 0 when the ` `// element is multiplied by -1 ` ` ` `using` `System; ` ` ` `class` `GFG { ` ` ` `// Function to find minimum index ` `// such that sum becomes 0 when the ` `// element is multiplied by -1 ` ` ` `static` `int` `minIndex(` `int` `[] arr, ` `int` `n) ` `{ ` ` ` `// Find array sum ` ` ` `int` `sum = 0; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `sum += arr[i]; ` ` ` ` ` `// Find element with value equal to sum/2 ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// when sum is equal to 2*element ` ` ` `// then this is our required element ` ` ` `if` `(2 * arr[i] == sum) ` ` ` `return` `(i + 1); ` ` ` `} ` ` ` ` ` `return` `-1; ` `} ` ` ` `// Driver code ` ` ` ` ` ` ` `public` `static` `void` `Main () { ` ` ` `int` `[] arr = { 1, 3, -5, 3, 4 }; ` ` ` `int` `n =arr.Length; ` ` ` `Console.Write( minIndex(arr, n)); ` ` ` `} ` `} ` |

## PHP

`<?php ` `// PHP program to find minimum index ` `// such that sum becomes 0 when the ` `// element is multiplied by -1 ` ` ` `// Function to find minimum index ` `// such that sum becomes 0 when the ` `// element is multiplied by -1 ` `function` `minIndex(&` `$arr` `, ` `$n` `) ` `{ ` ` ` `// Find array sum ` ` ` `$sum` `= 0; ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `$sum` `+= ` `$arr` `[` `$i` `]; ` ` ` ` ` `// Find element with value equal to sum/2 ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `{ ` ` ` ` ` `// when sum is equal to 2*element ` ` ` `// then this is our required element ` ` ` `if` `(2 * ` `$arr` `[` `$i` `] == ` `$sum` `) ` ` ` `return` `(` `$i` `+ 1); ` ` ` `} ` ` ` `return` `-1; ` `} ` ` ` `// Driver code ` `$arr` `= ` `array` `(1, 3, -5, 3, 4 ); ` `$n` `= sizeof(` `$arr` `); ` `echo` `(minIndex(` `$arr` `, ` `$n` `)); ` ` ` `// This code is contributed ` `// by Shivi_Aggarwal ` `?> ` |

**Output:**

2

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