# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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; ` ` `  `    ``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 find_idx(``int` `ar[], ``int` `n) ` `{ ` ` `  `    ``// Vector to store the answer ` `    ``Vector answer = ``new` `Vector(); ` ` `  `    ``// 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 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; ` ` `  `    ``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 `

Output:

```3 4 6
```

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