Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 04 Jun, 2021

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 
 

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

 

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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

Javascript




<script>
// Javascript implementation
 
// function to return valid indexes
function find_idx(ar, n) {
 
    // Vector to store the answer
    let answer = [];
 
    // Initial sum would always
    // be first element
    let sum = ar[0];
 
    for (let i = 1; i < n; i++) {
 
        // Check if sum till now
        // is equal to current element
        if (sum == ar[i]) {
            answer.push(i + 1);
        }
 
        // Updating the sum by
        // adding the current
        // element in each
        // iteration.
        sum += ar[i];
    }
 
    return answer;
}
 
// Driver code
 
let ar = [1, 2, 3, 6, 3, 15, 5];
let n = ar.length;
 
let ans = find_idx(ar, n);
 
if (ans.length != 0) {
    for (let i of ans) {
        document.write(i + " ");
    }
}
else {
    document.write("-1");
}
 
document.write("<br>");
 
 
// This code is contributed by gfgking.
</script>
Output: 
3 4 6

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :