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

Last Updated : 10 Mar, 2022

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++

// 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# 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

Time Complexity: O(n)
Auxiliary Space: O(n)

Number of positions such that adding K to the element is greater than sum of all other elements

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

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