查找与-1相乘使数组总和为0的元素

给定一个由N个整数组成的数组。我们的任务是找到元素的最小索引，以便将其乘以-1时，整个数组的总和将变为0。如果没有这样的索引，则返回-1。

例子：

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

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

天真的方法：
一个简单的解决方案是将每个元素乘以-1，然后检查新的和是否为0。此算法适用于O(N ² ) 。

高效的方法：
如果我们将S作为数组的初始总和，然后将当前元素A _i乘以-1，则新总和将变为S – 2 * A _i，并且该总和应等于0。因此，当第一次S = 2时* A _i，则当前索引为我们的要求，如果没有元素满足条件，则我们的答案为-1。该算法的时间复杂度为O(N) 。

以下是上述想法的实现：

C++

// C++ program to find minimum index
// such that sum becomes 0 when the
// element is multiplied by -1
  
#include 
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));
    }
}

输出：

2