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 indexesvector<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 codeint 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 approachimport java.util.*;class GFG { // function to return valid indexesstatic 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 codepublic 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 approachusing System;using System.Collections.Generic; class GFG { // function to return valid indexesstatic 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 codepublic 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 indexesfunction 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 codelet 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)
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