Find the element whose multiplication with -1 makes array sum 0

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 : 2

Input : 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(N2).



Efficient Approach :
If we take S as our initial sum of the array and we multiply current element Ai by -1 then the new sum will become S – 2*Ai and this should be equal to 0. So when for the first time S = 2*Ai 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 :

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// 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;
}
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// 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..
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# 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
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// 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));
    }
}
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<?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 
?>
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Output:
2



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