# Find an element in array such that sum of left array is equal to sum of right array

• Difficulty Level : Easy
• Last Updated : 21 Apr, 2022

Given, an array of size n. Find an element that divides the array into two sub-arrays with equal sums.
Examples:

```Input: 1 4 2 5
Output: 2
Explanation: If 2 is the partition,
subarrays are : {1, 4} and {5}

Input: 2 3 4 1 4 5
Output: 1
Explanation: If 1 is the partition,
Subarrays are : {2, 3, 4} and {4, 5}```

Method 1 (Simple)
Consider every element starting from the second element. Compute the sum of elements on its left and the sum of elements on its right. If these two sums are the same, return the element.

Method 2 (Using Prefix and Suffix Arrays :

```We form a prefix and suffix sum arrays

Given array: 1 4 2 5
Prefix Sum:  1  5 7 12
Suffix Sum:  12 11 7 5

Now, we will traverse both prefix arrays.
The index at which they yield equal result,
is the index where the array is partitioned
with equal sum.```

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `int` `findElement(``int` `arr[], ``int` `n)``{``    ``// Forming prefix sum array from 0``    ``int` `prefixSum[n];``    ``prefixSum[0] = arr[0];``    ``for` `(``int` `i = 1; i < n; i++)``        ``prefixSum[i] = prefixSum[i - 1] + arr[i];` `    ``// Forming suffix sum array from n-1``    ``int` `suffixSum[n];``    ``suffixSum[n - 1] = arr[n - 1];``    ``for` `(``int` `i = n - 2; i >= 0; i--)``        ``suffixSum[i] = suffixSum[i + 1] + arr[i];` `    ``// Find the point where prefix and suffix``    ``// sums are same.``    ``for` `(``int` `i = 1; i < n - 1; i++)``        ``if` `(prefixSum[i] == suffixSum[i])``            ``return` `arr[i];` `    ``return` `-1;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 4, 2, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << findElement(arr, n);``    ``return` `0;``}`

## Java

 `// Java program to find an element``// such that sum of right side element``// is equal to sum of left side``public` `class` `GFG {``    ` `    ``// Finds an element in an array such that``    ``// left and right side sums are equal``    ``static` `int` `findElement(``int` `arr[], ``int` `n)``    ``{``        ``// Forming prefix sum array from 0``        ``int``[] prefixSum = ``new` `int``[n];``        ``prefixSum[``0``] = arr[``0``];``        ``for` `(``int` `i = ``1``; i < n; i++)``            ``prefixSum[i] = prefixSum[i - ``1``] + arr[i];``     ` `        ``// Forming suffix sum array from n-1``        ``int``[] suffixSum = ``new` `int``[n];``        ``suffixSum[n - ``1``] = arr[n - ``1``];``        ``for` `(``int` `i = n - ``2``; i >= ``0``; i--)``            ``suffixSum[i] = suffixSum[i + ``1``] + arr[i];``     ` `        ``// Find the point where prefix and suffix``        ``// sums are same.``        ``for` `(``int` `i = ``1``; i < n - ``1``; i++)``            ``if` `(prefixSum[i] == suffixSum[i])``                ``return` `arr[i];``     ` `        ``return` `-``1``;``    ``}``     ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `arr[] = { ``1``, ``4``, ``2``, ``5` `};``        ``int` `n = arr.length;``        ``System.out.println(findElement(arr, n));``    ``}``}``// This code is contributed by Sumit Ghosh`

## Python 3

 `# Python 3 Program to find an element``# such that sum of right side element``# is equal to sum of left side` `# Function for Finds an element in``# an array such that left and right``# side sums are equal``def` `findElement(arr, n) :``    ` `    ``# Forming prefix sum array from 0``    ``prefixSum ``=` `[``0``] ``*` `n``    ``prefixSum[``0``] ``=` `arr[``0``]``    ``for` `i ``in` `range``(``1``, n) :``        ``prefixSum[i] ``=` `prefixSum[i ``-` `1``] ``+` `arr[i]` `    ``# Forming suffix sum array from n-1``    ``suffixSum ``=` `[``0``] ``*` `n``    ``suffixSum[n ``-` `1``] ``=` `arr[n ``-` `1``]``    ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``) :``        ``suffixSum[i] ``=` `suffixSum[i ``+` `1``] ``+` `arr[i]` `    ``# Find the point where prefix``    ``# and suffix sums are same.``    ``for` `i ``in` `range``(``1``, n ``-` `1``, ``1``) :``        ``if` `prefixSum[i] ``=``=` `suffixSum[i] :``            ``return` `arr[i]``        ` `    ``return` `-``1` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``arr ``=` `[ ``1``, ``4``, ``2``, ``5``]``    ``n ``=` `len``(arr)``    ``print``(findElement(arr, n))` `# This code is contributed by ANKITRAI1`

## C#

 `// C# program to find an element``// such that sum of right side element``// is equal to sum of left side``using` `System;` `class` `GFG``{``    ` `    ``// Finds an element in an``    ``// array such that left``    ``// and right side sums``    ``// are equal``    ``static` `int` `findElement(``int` `[]arr,``                           ``int` `n)``    ``{``        ``// Forming prefix sum``        ``// array from 0``        ``int``[] prefixSum = ``new` `int``[n];``        ``prefixSum[0] = arr[0];``        ``for` `(``int` `i = 1; i < n; i++)``            ``prefixSum[i] = prefixSum[i - 1] +``                                     ``arr[i];``    ` `        ``// Forming suffix sum``        ``// array from n-1``        ``int``[] suffixSum = ``new` `int``[n];``        ``suffixSum[n - 1] = arr[n - 1];``        ``for` `(``int` `i = n - 2; i >= 0; i--)``            ``suffixSum[i] = suffixSum[i + 1] +``                                     ``arr[i];``    ` `        ``// Find the point where prefix``        ``// and suffix sums are same.``        ``for` `(``int` `i = 1; i < n - 1; i++)``            ``if` `(prefixSum[i] == suffixSum[i])``                ``return` `arr[i];``    ` `        ``return` `-1;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 1, 4, 2, 5 };``        ``int` `n = arr.Length;``        ``Console.WriteLine(findElement(arr, n));``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

 `= 0; ``\$i``--)``        ``\$suffixSum``[``\$i``] = ``\$suffixSum``[``\$i` `+ 1] +``                                    ``\$arr``[``\$i``];` `    ``// Find the point where prefix``    ``// and suffix sums are same.``    ``for` `(``\$i` `= 1; ``\$i` `< ``\$n` `- 1; ``\$i``++)``        ``if` `(``\$prefixSum``[``\$i``] == ``\$suffixSum``[``\$i``])``            ``return` `\$arr``[``\$i``];` `    ``return` `-1;``}` `// Driver code``\$arr` `= ``array``( 1, 4, 2, 5 );``\$n` `= sizeof(``\$arr``);``echo` `findElement(``\$arr``, ``\$n``);` `// This code is contributed``// by ChitraNayal``?>`

## Javascript

 ``
Output
`2`

Method 3 (Space efficient)
We calculate the sum of the whole array except the first element in right_sum, considering it to be the partitioning element. Now, we traverse the array from left to right, subtracting an element from right_sum and adding an element to left_sum. At the point where right_sum equals left_sum, we get the partition.

Below is the implementation :

## C++

 `#include ``using` `namespace` `std;` `// Function to compute partition``int` `findElement(``int` `arr[], ``int` `size)``{``    ``int` `right_sum = 0, left_sum = 0;` `    ``// Computing right_sum``    ``for` `(``int` `i = 1; i < size; i++)``        ``right_sum += arr[i];` `    ``// Checking the point of partition``    ``// i.e. left_Sum == right_sum``    ``for` `(``int` `i = 0, j = 1; j < size; i++, j++) {``        ``right_sum -= arr[j];``        ``left_sum += arr[i];` `        ``if` `(left_sum == right_sum)``            ``return` `arr[i + 1];``    ``}` `    ``return` `-1;``}` `// Driver``int` `main()``{``    ``int` `arr[] = { 2, 3, 4, 1, 4, 5 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << findElement(arr, size);``    ``return` `0;``}`

## Java

 `// Java program to find an element``// such that sum of right side element``// is equal to sum of left side``public` `class` `GFG {``    ` `    ``// Function to compute partition``    ``static` `int` `findElement(``int` `arr[], ``int` `size)``    ``{``        ``int` `right_sum = ``0``, left_sum = ``0``;``     ` `        ``// Computing right_sum``        ``for` `(``int` `i = ``1``; i < size; i++)``            ``right_sum += arr[i];``     ` `        ``// Checking the point of partition``        ``// i.e. left_Sum == right_sum``        ``for` `(``int` `i = ``0``, j = ``1``; j < size; i++, j++) {``            ``right_sum -= arr[j];``            ``left_sum += arr[i];``     ` `            ``if` `(left_sum == right_sum)``                ``return` `arr[i + ``1``];``        ``}``     ` `        ``return` `-``1``;``    ``}``     ` `    ``// Driver``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `arr[] = { ``2``, ``3``, ``4``, ``1``, ``4``, ``5` `};``        ``int` `size = arr.length;``        ``System.out.println(findElement(arr, size));``    ``}``}``// This code is contributed by Sumit Ghosh`

## Python 3

 `# Python 3 Program to find an element``# such that sum of right side element``# is equal to sum of left side` `# Function to compute partition``def` `findElement(arr, size) :` `    ``right_sum, left_sum ``=` `0``, ``0` `    ``# Computing right_sum``    ``for` `i ``in` `range``(``1``, size) :``        ``right_sum ``+``=` `arr[i]` `    ``i, j ``=` `0``, ``1` `    ``# Checking the point of partition``    ``# i.e. left_Sum == right_sum``    ``while` `j < size :``        ``right_sum ``-``=` `arr[j]``        ``left_sum ``+``=` `arr[i]` `        ``if` `left_sum ``=``=` `right_sum :``            ``return` `arr[i ``+` `1``]` `        ``j ``+``=` `1``        ``i ``+``=` `1` `    ``return` `-``1` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``arr ``=` `[ ``2``, ``3``, ``4``, ``1``, ``4``, ``5``]``    ``n ``=` `len``(arr)``    ``print``(findElement(arr, n))` `# This code is contributed by ANKITRAI1`

## C#

 `// C# program to find an``// element such that sum``// of right side element``// is equal to sum of``// left side``using` `System;` `class` `GFG``{``    ``// Function to compute``    ``// partition``    ``static` `int` `findElement(``int` `[]arr,``                           ``int` `size)``    ``{``        ``int` `right_sum = 0,``            ``left_sum = 0;``    ` `        ``// Computing right_sum``        ``for` `(``int` `i = 1; i < size; i++)``            ``right_sum += arr[i];``    ` `        ``// Checking the point``        ``// of partition i.e.``        ``// left_Sum == right_sum``        ``for` `(``int` `i = 0, j = 1;``                 ``j < size; i++, j++)``        ``{``            ``right_sum -= arr[j];``            ``left_sum += arr[i];``    ` `            ``if` `(left_sum == right_sum)``                ``return` `arr[i + 1];``        ``}``    ` `        ``return` `-1;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = {2, 3, 4, 1, 4, 5};``        ``int` `size = arr.Length;``        ``Console.WriteLine(findElement(arr, size));``    ``}``}` `// This code is contributed``// by anuj_67.`

## PHP

 ``

## Javascript

 ``
Output
`1`

Method 4 (Both Time and Space Efficient)

```We define two pointers i and j to traverse the array from left and right,
left_sum and right_sum to store sum from right and left respectively

If left_sum is lesser than increment i and if right_sum is lesser than decrement j
and, find a position where left_sum == right_sum and i and j are next to each other```

Note: This solution is only applicable if the array contains only positive elements.

Below is the implementation of the above approach:

## C++

 `// C++ program to find an element``// such that sum of right side element``// is equal to sum of left side``#include``using` `namespace` `std;``    ` `// Function to compute``// partition``int` `findElement(``int` `arr[],``                ``int` `size)``{``  ``int` `right_sum = 0,``      ``left_sum = 0;``  ` `  ``// Maintains left``  ``// cumulative sum``  ``left_sum = 0;` `  ``// Maintains right``  ``// cumulative sum``  ``right_sum = 0;``  ``int` `i = -1, j = -1;` `  ``for``(i = 0, j = size - 1;``      ``i < j; i++, j--)``  ``{``    ``left_sum += arr[i];``    ``right_sum += arr[j];` `    ``// Keep moving i towards``    ``// center until left_sum``    ``//is found lesser than right_sum``    ``while``(left_sum < right_sum &&``          ``i < j)``    ``{``      ``i++;``      ``left_sum += arr[i];``    ``}``    ` `    ``// Keep moving j towards``    ``// center until right_sum is``    ``// found lesser than left_sum``    ``while``(right_sum < left_sum &&``          ``i < j)``    ``{``      ``j--;``      ``right_sum += arr[j];``    ``}``  ``}``  ` `  ``if``(left_sum == right_sum && i == j)``    ``return` `arr[i];``  ``else``    ``return` `-1;``}` `// Driver code``int` `main()``{``  ``int` `arr[] = {2, 3, 4,``               ``1, 4, 5};``  ``int` `size = ``sizeof``(arr) /``             ``sizeof``(arr[0]);``  ``cout << (findElement(arr, size));``}` `// This code is contributed by shikhasingrajput`

## Java

 `// Java program to find an element``// such that sum of right side element``// is equal to sum of left side``public` `class` `Gfg {``    ` `    ``// Function to compute partition``    ``static` `int` `findElement(``int` `arr[], ``int` `size)``    ``{``        ``int` `right_sum = ``0``, left_sum = ``0``;``        ``// Maintains left cumulative sum``        ``left_sum = ``0``;``        ` `        ``// Maintains right cumulative sum``        ``right_sum=``0``;``        ``int` `i = -``1``, j = -``1``;``        ` `        ``for``( i = ``0``, j = size-``1` `; i < j ; i++, j-- ){``            ``left_sum += arr[i];``            ``right_sum += arr[j];``            ` `            ``// Keep moving i towards center until``            ``// left_sum is found lesser than right_sum``            ``while``(left_sum < right_sum && i < j){``                ``i++;``                ``left_sum += arr[i];``            ``}``            ``// Keep moving j towards center until``            ``// right_sum is found lesser than left_sum``            ``while``(right_sum < left_sum && i < j){``                ``j--;``                ``right_sum += arr[j];``            ``}``        ``}``        ``if``(left_sum == right_sum && i == j)``            ``return` `arr[i];``        ``else``            ``return` `-``1``;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `arr[] = {``2``, ``3``, ``4``, ``1``, ``4``, ``5``};``        ``int` `size = arr.length;``        ``System.out.println(findElement(arr, size));``    ``}``}`

## Python3

 `# Python3 program to find an element``# such that sum of right side element``# is equal to sum of left side` `# Function to compute partition``def` `findElement(arr, size):``  ` `    ``# Maintains left cumulative sum``    ``left_sum ``=` `0``;` `    ``# Maintains right cumulative sum``    ``right_sum ``=` `0``;``    ``i ``=` `0``; j ``=` `-``1``;``    ``j ``=` `size ``-` `1``;``    ` `    ``while``(i < j):``        ``if``(i < j):``            ``left_sum ``+``=` `arr[i];``            ``right_sum ``+``=` `arr[j];` `            ``# Keep moving i towards center``            ``# until left_sum is found``            ``# lesser than right_sum``            ``while` `(left_sum < right_sum ``and``                   ``i < j):``                ``i ``+``=` `1``;``                ``left_sum ``+``=` `arr[i];` `            ``# Keep moving j towards center``            ``# until right_sum is found``            ``# lesser than left_sum``            ``while` `(right_sum < left_sum ``and``                   ``i < j):``                ``j ``-``=` `1``;``                ``right_sum ``+``=` `arr[j];``            ``j ``-``=` `1``            ``i ``+``=` `1``    ``if` `(left_sum ``=``=` `right_sum && i ``=``=` `j):``        ``return` `arr[i];``    ``else``:``        ``return` `-``1``;` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``  ` `    ``arr ``=` `[``2``, ``3``, ``4``,``           ``1``, ``4``, ``5``];``    ``size ``=` `len``(arr);``    ``print``(findElement(arr, size));` `# This code is contributed by shikhasingrajput`

## C#

 `// C# program to find an element``// such that sum of right side element``// is equal to sum of left side``using` `System;` `class` `GFG{``    ` `// Function to compute partition``static` `int` `findElement(``int` `[]arr, ``int` `size)``{``    ``int` `right_sum = 0, left_sum = 0;``    ` `    ``// Maintains left cumulative sum``    ``left_sum = 0;``    ` `    ``// Maintains right cumulative sum``    ``right_sum = 0;``    ``int` `i = -1, j = -1;``    ` `    ``for``(i = 0, j = size - 1; i < j; i++, j--)``    ``{``        ``left_sum += arr[i];``        ``right_sum += arr[j];``        ` `        ``// Keep moving i towards center until``        ``// left_sum is found lesser than right_sum``        ``while` `(left_sum < right_sum && i < j)``        ``{``            ``i++;``            ``left_sum += arr[i];``        ``}``        ` `        ``// Keep moving j towards center until``        ``// right_sum is found lesser than left_sum``        ``while` `(right_sum < left_sum && i < j)``        ``{``            ``j--;``            ``right_sum += arr[j];``        ``}``    ``}``    ``if` `(left_sum == right_sum && i == j)``        ``return` `arr[i];``    ``else``        ``return` `-1;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]arr = { 2, 3, 4, 1, 4, 5 };``    ``int` `size = arr.Length;``    ` `    ``Console.WriteLine(findElement(arr, size));``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
Output
`1`

Since all loops start traversing from the last updated i and j pointers and do not cross each other, they run n times in the end.

Time Complexity – O(n)
Auxiliary Space – O(1)

Method 5 (The efficient algorithm):

1. Here we define two pointers to the array -> start = 0, end = n-1
2. Two variables to take care of sum -> left_sum = 0, right_sum = 0

Here our algorithm goes like this:

1. We initialize for loop till the entire size of the array
2. Basically we check if left_sum > right_sum => add the current end element to the right_sum and decrement end
3. If right_sum < left_sum => add the current start element to the left_sum and increment start
4. By these two conditions, we make sure that left_sum and right_sum are nearly balanced, so we can arrive at our solution easily
5. To make this work during all test cases we need to add a couple of conditional statements to our logic:
6. If the equilibrium element is found our start will be equal to the end variable and left_sum will be equal right_sum => return the equilibrium element (here we say start == end because we increment/ decrement the pointer after adding the current start/ end value to respective sum. So if, equilibrium element is found start and end should be at the same location)
7. If start is equal to end variable but left_sum is not equal right_sum => no equilibrium element return -1
8. If left_sum is equal right_sum, but start is not equal to end => we are still in the middle of the algorithm even though we found that left_sum is equal right_sum we haven’t got that one required equilibrium element (So, in this case, add the current end element to the right_sum and decrement end (or) add the current start element to the left_sum and increment start, to make our algorithm continue further).
9. Even here there is one test case that needs to be handled:
10. When there is only one element in the array our algorithm exits without entering for a loop.
11. So we can check if our functions enter the loop if not we can directly return the value as the answer.

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `// Function to find equilibrium point``// a: input array``// n: size of array``int` `equilibriumPoint(``int` `a[], ``int` `n)``{``    ``// Here we define two pointers to the array -> start =``    ``// 0, end = n-1 Two variables to take care of sum ->``    ``// left_sum = 0, right_sum = 0``    ``int` `i = 0, start = 0, end = n - 1, left_sum = 0,``        ``right_sum = 0;` `    ``for` `(i = 0; i < n; i++) {` `        ``// if the equilibrium element is found our start``        ``// will be equal to end variable and  left_sum will``        ``// be equal right_sum => return the equilibrium``        ``// element``        ``if` `(start == end && right_sum == left_sum)``            ``return` `a[start];` `        ``// if start is equal to end variable but left_sum is``        ``// not equal right_sum => no equilibrium element``        ``// return -1``        ``if` `(start == end)``            ``return` `-1;` `        ``// if left_sum  > right_sum => add the current end``        ``// element to the right_sum and decrement end``        ``if` `(left_sum > right_sum) {``            ``right_sum += a[end];``            ``end--;``        ``}` `        ``// if right_sum < left_sum => add the current start``        ``// element to the left_sum and increment start``        ``else` `if` `(right_sum > left_sum) {``            ``left_sum += a[start];``            ``start++;``        ``}``        ``/*``            ``if  left_sum is equal right_sum but start is not``           ``equal to end => we are still in the middle of``           ``algorithm even though we found that left_sum is``           ``equal right_sum we haven't got that one required``           ``equilibrium element (So, in this case add the``           ``current end element to the right_sum and``           ``decrement end (or) add the current start element``           ``to the left_sum and increment start, to make our``           ``algorithm continue further)``        ``*/` `        ``else` `{``            ``right_sum += a[end];``            ``end--;``        ``}``    ``}` `    ``// When there is only one element in array our algorithm``    ``// exits without entering for loop So we can check if our``    ``// functions enters the loop if not we can directly``    ``// return the value as answer``    ``if` `(!i) {``        ``return` `a[0];``    ``}``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 3, 4, 1, 4, 5 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << (equilibriumPoint(arr, size));``}`

## Java

 `/*package whatever //do not write package name here */``import` `java.io.*;``class` `GFG``{` `  ``// Function to find equilibrium point``  ``// a: input array``  ``// n: size of array``  ``static` `int` `equilibriumPoint(``int` `a[], ``int` `n)``  ``{` `    ``// Here we define two pointers to the array -> start =``    ``// 0, end = n-1 Two variables to take care of sum ->``    ``// left_sum = 0, right_sum = 0``    ``int` `i = ``0``, start = ``0``, end = n - ``1``, left_sum = ``0``,``    ``right_sum = ``0``;` `    ``for` `(i = ``0``; i < n; i++)``    ``{` `      ``// if the equilibrium element is found our start``      ``// will be equal to end variable and  left_sum will``      ``// be equal right_sum => return the equilibrium``      ``// element``      ``if` `(start == end && right_sum == left_sum)``        ``return` `a[start];` `      ``// if start is equal to end variable but left_sum is``      ``// not equal right_sum => no equilibrium element``      ``// return -1``      ``if` `(start == end)``        ``return` `-``1``;` `      ``// if left_sum  > right_sum => add the current end``      ``// element to the right_sum and decrement end``      ``if` `(left_sum > right_sum)``      ``{``        ``right_sum += a[end];``        ``end--;``      ``}` `      ``// if right_sum < left_sum => add the current start``      ``// element to the left_sum and increment start``      ``else` `if` `(right_sum > left_sum)``      ``{``        ``left_sum += a[start];``        ``start++;``      ``}``      ``/*``                ``if  left_sum is equal right_sum but start is not``               ``equal to end => we are still in the middle of``               ``algorithm even though we found that left_sum is``               ``equal right_sum we haven't got that one required``               ``equilibrium element (So, in this case add the``               ``current end element to the right_sum and``               ``decrement end (or) add the current start element``               ``to the left_sum and increment start, to make our``               ``algorithm continue further)``            ``*/` `      ``else` `{``        ``right_sum += a[end];``        ``end--;``      ``}``    ``}` `    ``// When there is only one element in array our algorithm``    ``// exits without entering for loop So we can check if our``    ``// functions enters the loop if not we can directly``    ``// return the value as answer``    ``if` `(i == ``0``)``    ``{``      ``return` `a[``0``];``    ``}``    ``return` `-``1``;``  ``}` `  ``// Driver code``  ``public` `static` `void` `main (String[] args)``  ``{``    ``int` `arr[] = { ``2``, ``3``, ``4``, ``1``, ``4``, ``5` `};``    ``int` `size = arr.length;``    ``System.out.println(equilibriumPoint(arr, size));``  ``}``}` `// This code is contributed by avanitrachhadiya2155`

## Python3

 `# Function to find equilibrium point``# a: input array``# n: size of array``def` `equilibriumPoint(a, n):` `    ``# Here we define two pointers to the array -> start =``    ``# 0, end = n-1 Two variables to take care of sum ->``    ``# left_sum = 0, right_sum = 0``    ``i,start,end,left_sum,right_sum ``=` `0``,``0``,n ``-` `1``,``0``,``0` `    ``for` `i ``in` `range``(n):` `        ``# if the equilibrium element is found our start``        ``# will be equal to end variable and left_sum will``        ``# be equal right_sum => return the equilibrium``        ``# element``        ``if` `(start ``=``=` `end ``and` `right_sum ``=``=` `left_sum):``            ``return` `a[start]` `        ``# if start is equal to end variable but left_sum is``        ``# not equal right_sum => no equilibrium element``        ``# return -1``        ``if` `(start ``=``=` `end):``            ``return` `-``1` `        ``# if left_sum > right_sum => add the current end``        ``# element to the right_sum and decrement end``        ``if` `(left_sum > right_sum):``            ``right_sum ``+``=` `a[end]``            ``end ``-``=` `1` `        ``# if right_sum < left_sum => add the current start``        ``# element to the left_sum and increment start``        ``elif` `(right_sum > left_sum):``            ``left_sum ``+``=` `a[start]``            ``start ``+``=` `1``        ` `        ``#     if left_sum is equal right_sum but start is not``        ``# equal to end => we are still in the middle of``        ``# algorithm even though we found that left_sum is``        ``# equal right_sum we haven't got that one required``        ``# equilibrium element (So, in this case add the``        ``# current end element to the right_sum and``        ``# decrement end (or) add the current start element``        ``# to the left_sum and increment start, to make our``        ``# algorithm continue further)``        ``else``:``            ``right_sum ``+``=` `a[end]``            ``end ``-``=` `1` `    ``# When there is only one element in array our algorithm``    ``# exits without entering for loop So we can check if our``    ``# functions enters the loop if not we can directly``    ``# return the value as answer``    ``if` `(``not` `i):``        ``return` `a[``0``]` `# Driver code``arr ``=` `[ ``2``, ``3``, ``4``, ``1``, ``4``, ``5` `]``size ``=` `len``(arr)``print``(equilibriumPoint(arr, size))` `# This code is contributed by Shinjanpatra`

## C#

 `using` `System;` `public` `class` `GFG{``    ` `    ``// Function to find equilibrium point``  ``// a: input array``  ``// n: size of array``  ``static` `int` `equilibriumPoint(``int``[] a, ``int` `n)``  ``{`` ` `    ``// Here we define two pointers to the array -> start =``    ``// 0, end = n-1 Two variables to take care of sum ->``    ``// left_sum = 0, right_sum = 0``    ``int` `i = 0, start = 0, end = n - 1, left_sum = 0,``    ``right_sum = 0;`` ` `    ``for` `(i = 0; i < n; i++)``    ``{`` ` `      ``// if the equilibrium element is found our start``      ``// will be equal to end variable and  left_sum will``      ``// be equal right_sum => return the equilibrium``      ``// element``      ``if` `(start == end && right_sum == left_sum)``        ``return` `a[start];`` ` `      ``// if start is equal to end variable but left_sum is``      ``// not equal right_sum => no equilibrium element``      ``// return -1``      ``if` `(start == end)``        ``return` `-1;`` ` `      ``// if left_sum  > right_sum => add the current end``      ``// element to the right_sum and decrement end``      ``if` `(left_sum > right_sum)``      ``{``        ``right_sum += a[end];``        ``end--;``      ``}`` ` `      ``// if right_sum < left_sum => add the current start``      ``// element to the left_sum and increment start``      ``else` `if` `(right_sum > left_sum)``      ``{``        ``left_sum += a[start];``        ``start++;``      ``}``      ``/*``                ``if  left_sum is equal right_sum but start is not``               ``equal to end => we are still in the middle of``               ``algorithm even though we found that left_sum is``               ``equal right_sum we haven't got that one required``               ``equilibrium element (So, in this case add the``               ``current end element to the right_sum and``               ``decrement end (or) add the current start element``               ``to the left_sum and increment start, to make our``               ``algorithm continue further)``            ``*/`` ` `      ``else` `{``        ``right_sum += a[end];``        ``end--;``      ``}``    ``}`` ` `    ``// When there is only one element in array our algorithm``    ``// exits without entering for loop So we can check if our``    ``// functions enters the loop if not we can directly``    ``// return the value as answer``    ``if` `(i == 0)``    ``{``      ``return` `a[0];``    ``}``    ``return` `-1;``  ``}`` ` `  ``// Driver code``    ` `    ``static` `public` `void` `Main (){``        ` `        ``int``[] arr = { 2, 3, 4, 1, 4, 5 };``    ``int` `size = arr.Length;``    ``Console.WriteLine(equilibriumPoint(arr, size));``        ` `    ``}``}`

## Javascript

 ``
Output
`1`

Time complexity: O(n)

Space complexity: O(1)

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