Position of Elements which are equal to sum of all Preceding elements

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

Examples:

Input : Arr[] = {1, 2, 3, 6, 3, 15, 5}
Output :3 4 6

Here, the element at index “3” i.e. 3 is equal to the sum of preceding elements (1 + 2).
Similarly, at index 4, 6 = 1+2+3 (sum of all preceding elements).
And element at index 6 i.e. 15 = 1 + 2 + 3 + 6 + 3.

Input: Arr[] = {7, 5, 17, 25}
Output: -1



Approach:
While traversing the array Arr[], maintain a sum variable that store the sum of elements till i – 1. Compare the sum with current element Arr[i]. If it is equal, push the index of this element into the answer vector.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation
#include <bits/stdc++.h>
using namespace std;
  
// function to return valid indexes
vector<int> find_idx(int ar[], int n)
{
  
    // Vector to store the answer
    vector<int> answer;
  
    // Initial sum would always
    // be first element
    int sum = ar[0];
  
    for (int i = 1; i < n; i++) {
  
        // Check if sum till now
        // is equal to current element
        if (sum == ar[i]) {
            answer.push_back(i + 1);
        }
  
        // Updating the sum by
        // adding the current
        // element in each
        // iteration.
        sum += ar[i];
    }
  
    return answer;
}
  
// Driver code
int main()
{
    int ar[] = { 1, 2, 3, 6, 3, 15, 5 };
    int n = sizeof(ar) / sizeof(int);
  
    vector<int> ans = find_idx(ar, n);
  
    if (ans.size() != 0) {
        for (int i : ans) {
            cout << i << " ";
        }
    }
    else {
        cout << "-1";
    }
  
    cout << endl;
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
      
// function to return valid indexes
static Vector<Integer> find_idx(int ar[], int n)
{
  
    // Vector to store the answer
    Vector<Integer> answer = new Vector<Integer>();
  
    // Initial sum would always
    // be first element
    int sum = ar[0];
  
    for (int i = 1; i < n; i++) 
    {
  
        // Check if sum till now
        // is equal to current element
        if (sum == ar[i]) 
        {
            answer.add(i + 1);
        }
  
        // Updating the sum by adding the 
        // current element in each iteration.
        sum += ar[i];
    }
    return answer;
}
  
// Driver code
public static void main(String[] args) 
{
    int ar[] = { 1, 2, 3, 6, 3, 15, 5 };
    int n = ar.length;
  
    Vector<Integer> ans = find_idx(ar, n);
  
    if (ans.size() != 0
    {
        for (int i : ans) 
        {
            System.out.print(i + " ");
        }
    }
    else 
    {
        System.out.println("-1");
    }
}
  
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the above approach
  
# function to return valid indexes 
def find_idx(ar, n) : 
  
    # Vector to store the answer 
    answer = []; 
  
    # Initial sum would always 
    # be first element 
    sum = ar[0]; 
  
    for i in range(1, n) :
  
        # Check if sum till now 
        # is equal to current element 
        if (sum == ar[i]) :
            answer.append(i + 1); 
  
        # Updating the sum by 
        # adding the current 
        # element in each 
        # iteration. 
        sum += ar[i];
  
    return answer; 
  
# Driver code 
if __name__ == "__main__"
  
    ar = [ 1, 2, 3, 6, 3, 15, 5 ]; 
    n = len(ar); 
  
    ans = find_idx(ar, n); 
  
    if (len(ans) != 0) :
          
        for i in ans :
            print(i, end = " "); 
              
    else :
          
        print("-1"); 
  
    print(); 
  
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
using System.Collections.Generic;
      
class GFG 
{
      
// function to return valid indexes
static List<int> find_idx(int []ar, int n)
{
  
    // Vector to store the answer
    List<int> answer = new List<int>();
  
    // Initial sum would always
    // be first element
    int sum = ar[0];
  
    for (int i = 1; i < n; i++) 
    {
  
        // Check if sum till now
        // is equal to current element
        if (sum == ar[i]) 
        {
            answer.Add(i + 1);
        }
  
        // Updating the sum by adding the 
        // current element in each iteration.
        sum += ar[i];
    }
    return answer;
}
  
// Driver code
public static void Main(String[] args) 
{
    int []ar = { 1, 2, 3, 6, 3, 15, 5 };
    int n = ar.Length;
  
    List<int> ans = find_idx(ar, n);
  
    if (ans.Count != 0) 
    {
        foreach (int i in ans) 
        {
            Console.Write(i + " ");
        }
    }
    else
    {
        Console.WriteLine("-1");
    }
}
}
  
// This code is contributed by Princi Singh

chevron_right


Output:

3 4 6

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.