Index of the elements which are equal to the sum of all succeeding elements

Given an array arr[] of N positive integers. The task is to find the index of the elements which are equal to the sum of all succeeding elements. If no such element exists then print -1.

Examples:

Input: arr[] = { 36, 2, 17, 6, 6, 5 }
Output: 0 2
arr[0] = arr[1] + arr[2] + arr[3] + arr[4] + arr[5]
arr[2] = arr[3] + arr[4] + arr[5]



Input: arr[] = {7, 25, 17, 7}
Output: -1

Approach: While traversing the given array arr[] from last index, maintain a sum variable that stores the sum of elements traversed till now. Compare the sum with the current element arr[i]. If it is equal, push the index of this element into the answer vector. If the size of the answer vector in the end is 0 then print -1 else print its content.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the valid indices
void find_idx(int arr[], int n)
{
  
    // Vector to store the indices
    vector<int> answer;
  
    // Initialise sum with 0
    int sum = 0;
  
    // Starting from the last element
    for (int i = n - 1; i >= 0; i--) {
  
        // If sum till now is equal to
        // the current element
        if (sum == arr[i]) {
            answer.push_back(i);
        }
  
        // Add current element to the sum
        sum += arr[i];
    }
  
    if (answer.size() == 0) {
        cout << "-1";
        return;
    }
  
    for (int i = answer.size() - 1; i >= 0; i--)
        cout << answer[i] << " ";
}
  
// Driver code
int main()
{
    int arr[] = { 36, 2, 17, 6, 6, 5 };
    int n = sizeof(arr) / sizeof(int);
  
    find_idx(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach 
import java.util.*;
  
class GFG 
{
      
    // Function to find the valid indices 
    static void find_idx(int arr[], int n) 
    
      
        // Vector to store the indices 
        Vector answer = new Vector();
      
        // Initialise sum with 0 
        int sum = 0
      
        // Starting from the last element 
        for (int i = n - 1; i >= 0; i--)
        
      
            // If sum till now is equal to 
            // the current element 
            if (sum == arr[i]) 
            
                answer.add(i); 
            
      
            // Add current element to the sum 
            sum += arr[i]; 
        
      
        if (answer.size() == 0)
        
            System.out.println("-1"); 
            return
        
      
        for (int i = answer.size() - 1; i >= 0; i--) 
            System.out.print(answer.get(i) + " "); 
    
      
    // Driver code 
    public static void main (String[] args) 
    
        int arr[] = { 36, 2, 17, 6, 6, 5 }; 
        int n = arr.length; 
      
        find_idx(arr, n); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach 
  
# Function to find the valid indices 
def find_idx(arr, n): 
  
    # List to store the indices 
    answer=[] 
  
    # Initialise sum with 0 
    _sum = 0
  
    # Starting from the last element 
    for i in range(n - 1, -1, -1):
  
        # If sum till now is equal to 
        # the current element 
        if (_sum == arr[i]) : 
            answer.append(i) 
  
        # Add current element to the sum 
        _sum += arr[i] 
  
    if (len(answer) == 0) : 
        print(-1
        return
  
    for i in range(len(answer) - 1, -1, -1): 
        print(answer[i], end = " "
  
# Driver code 
arr = [ 36, 2, 17, 6, 6, 5
n = len(arr) 
  
find_idx(arr, n)
  
# This code is contributed by
# divyamohan123

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C#

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// C# implementation of the approach 
using System;
using System.Collections.Generic;
  
class GFG 
{
      
    // Function to find the valid indices 
    static void find_idx(int[] arr, int n) 
    
      
        // List to store the indices 
        List<int> answer = new List<int>();
      
        // Initialise sum with 0 
        int sum = 0; 
      
        // Starting from the last element 
        for (int i = n - 1; i >= 0; i--)
        
      
            // If sum till now is equal to 
            // the current element 
            if (sum == arr[i]) 
            
                answer.Add(i); 
            
      
            // Add current element to the sum 
            sum += arr[i]; 
        
      
        if (answer.Count == 0)
        
            Console.WriteLine("-1"); 
            return
        
      
        for (int i = answer.Count - 1; i >= 0; i--) 
            Console.Write(answer[i] + " "); 
    
      
    // Driver code 
    public static void Main (String[] args) 
    
        int[] arr = { 36, 2, 17, 6, 6, 5 }; 
        int n = arr.Length; 
      
        find_idx(arr, n); 
    
}
  
// This code is contributed by Ashutosh450

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Output:

0 2


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