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# Count of indices up to which prefix and suffix sum is equal for given Array

Given an array arr[] of integers, the task is to find the number of indices up to which prefix sum and suffix sum are equal.

Example:

Input: arr = [9, 0, 0, -1, 11, -1]
Output: 2
Explanation:  The indices up to which prefix and suffix sum are equal are given below:
At index 1 prefix and suffix sum are 9
At index 2 prefix and suffix sum are 9

Input: arr = [5, 0, 4, -1, -3, 0, 2, -2, 0, 3, 2]
Output: 3
Explanation:  The prefix subarrays and suffix subarrays with equal sum are given below:
At index 1 prefix and suffix sum are 5
At index 5 prefix and suffix sum are 5
At index 8 prefix and suffix sum are 5

Naive Approach: The given problem can be solved by traversing the array arr from left to right and calculating prefix sum till that index then iterating the array arr from right to left and calculating the suffix sum then checking if prefix and suffix sum are equal.

## C++

 `// C++ code for the above approach``#include ``using` `namespace` `std;` `// Function to calculate number of``// equal prefix and suffix sums``// till the same indices``int` `equalSumPreSuf(``int` `arr[], ``int` `n)``{``    ``int` `count = 0;` `    ``// Iterate over the array``    ``for` `(``int` `i = 0; i < n; i++) {``        ``int` `sum1 = 0, sum2 = 0;` `        ``// Find the left subarray sum``        ``for` `(``int` `j = 0; j < i; j++) {``            ``sum1 += arr[j];``        ``}` `        ``// Find the right subarray sum``        ``for` `(``int` `j = i + 1; j < n; j++) {``            ``sum2 += arr[j];``        ``}` `        ``// Check if both are equal``        ``if` `(sum1 == sum2)``            ``count++;``    ``}` `    ``// Return the count``    ``return` `count;``}` `// Driver code``int` `main()``{` `    ``// Initialize the array``    ``int` `arr[] = { 9, 0, 0, -1, 11, -1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``// Call the function and``    ``// print its result``    ``cout << (equalSumPreSuf(arr, n));``}`

## Java

 `import` `java.io.*;``import` `java.util.*;` `public` `class` `Gfg {` `    ``public` `static` `int` `equalSumPreSuf(``int``[] arr, ``int` `n)``    ``{``        ``int` `count = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``int` `sum1 = ``0``, sum2 = ``0``;` `            ``for` `(``int` `j = ``0``; j < i; j++) {``                ``sum1 += arr[j];``            ``}` `            ``for` `(``int` `j = i + ``1``; j < n; j++) {``                ``sum2 += arr[j];``            ``}` `            ``if` `(sum1 == sum2)``                ``count++;``        ``}` `        ``return` `count;``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Initialize the array``        ``int``[] arr = { ``9``, ``0``, ``0``, -``1``, ``11``, -``1` `};``        ``int` `n = arr.length;``        ``// Call the function and``        ``// print its result``        ``System.out.println(equalSumPreSuf(arr, n));``    ``}``}`

## Python3

 `# Function to calculate number of``# equal prefix and suffix sums``# till the same indices``def` `equalSumPreSuf(arr, n):``    ``count ``=` `0``    ` `    ``# Iterate over the array``    ``for` `i ``in` `range``(n):``        ``sum1 ``=` `0``        ``sum2 ``=` `0``        ` `        ``# Find the left subarray sum``        ``for` `j ``in` `range``(i):``            ``sum1 ``+``=` `arr[j]``            ` `        ``# Find the right subarray sum``        ``for` `j ``in` `range``(i ``+` `1``, n):``            ``sum2 ``+``=` `arr[j]``            ` `        ``# Check if both are equal``        ``if` `sum1 ``=``=` `sum2:``            ``count ``+``=` `1``            ` `    ``# Return the count``    ``return` `count` `# Driver code``arr ``=` `[``9``, ``0``, ``0``, ``-``1``, ``11``, ``-``1``]``n ``=` `len``(arr)` `# Call the function and``# print its result``print``(equalSumPreSuf(arr, n))` `# This code is contributed by divya_p123.`

## C#

 `using` `System;` `class` `Gfg {``    ``// Function to calculate number of``    ``// equal prefix and suffix sums``    ``// till the same indices``    ``static` `int` `equalSumPreSuf(``int``[] arr, ``int` `n)``    ``{``        ``int` `count = 0;``        ``for` `(``int` `i = 0; i < n; i++) {``            ``int` `sum1 = 0, sum2 = 0;` `            ``for` `(``int` `j = 0; j < i; j++) {``                ``sum1 += arr[j];``            ``}` `            ``for` `(``int` `j = i + 1; j < n; j++) {``                ``sum2 += arr[j];``            ``}` `            ``if` `(sum1 == sum2)``                ``count++;``        ``}` `        ``return` `count;``    ``}` `    ``// Driver code``    ``static` `void` `Main()``    ``{` `        ``// Initialize the array``        ``int``[] arr = { 9, 0, 0, -1, 11, -1 };``        ``int` `n = arr.Length;``        ``// Call the function and``        ``// print its result``        ``Console.WriteLine(equalSumPreSuf(arr, n));``    ``}``}`

## Javascript

 `// Javascript code for the above approach` `// Function to calculate number of``// equal prefix and suffix sums``// till the same indices``function` `equalSumPreSuf( arr,  n)``{``    ``let count = 0;` `    ``// Iterate over the array``    ``for` `(let i = 0; i < n; i++) {``        ``let sum1 = 0, sum2 = 0;` `        ``// Find the left subarray sum``        ``for` `(let j = 0; j < i; j++) {``            ``sum1 += arr[j];``        ``}` `        ``// Find the right subarray sum``        ``for` `(let j = i + 1; j < n; j++) {``            ``sum2 += arr[j];``        ``}` `        ``// Check if both are equal``        ``if` `(sum1 == sum2)``            ``count++;``    ``}` `    ``// Return the count``    ``return` `count;``}` `// Driver code``// Initialize the array``let arr = [ 9, 0, 0, -1, 11, -1 ];``let n = arr.length;` `// Call the function and``// print its result``console.log(equalSumPreSuf(arr, n));` `// This code is contributed by agarwalpoojaa976.`

Output

```2

```

Time Complexity: O(N^2)
Auxiliary Space: O(1)

Better Approach:

This approach to solve the problem is to precompute the prefix and suffix sum and store the result for both in different arrays. Now count the number of indices where prefix[i] == suffix[i]. This count will be our answer.

Algorithm:

1. Initialize an array arr of n integers.
2. Calculate the prefix sum of the array and store it in a vector prefix. The prefix sum of an element at index i is the sum of all the elements from index 0 to i.
3. Calculate the suffix sum of the array and store it in a vector suffix. The suffix sum of an element at index i is the sum of all the elements from index i to n-1.
4. Initialize a variable count to 0.
5. Traverse the array and check if prefix[i] is equal to suffix[i]. If it is, then increment count.
6. Return the count as the answer.

Below is the implementation of the approach:

## C++

 `// C++ code for the above approach` `#include ``using` `namespace` `std;` `// Function to calculate number of``// equal prefix and suffix sums``// till the same indices``int` `equalSumPreSuf(``int` `arr[], ``int` `n) {``    ``// Calculate prefix sum``    ``vector<``int``> prefix(n);``    ``prefix[0] = arr[0];``    ``for` `(``int` `i = 1; i < n; i++) {``        ``prefix[i] = prefix[i - 1] + arr[i];``    ``}` `    ``// Calculate suffix sum``    ``vector<``int``> suffix(n);``    ``suffix[n - 1] = arr[n - 1];``    ``for` `(``int` `i = n - 2; i >= 0; i--) {``        ``suffix[i] = suffix[i + 1] + arr[i];``    ``}` `    ``// Count the number of indices where prefix[i] == suffix[i]``    ``int` `count = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(prefix[i] == suffix[i]) {``            ``count++;``        ``}``    ``}` `    ``return` `count;``}` `int` `main() {``    ``// Initialize the array``    ``int` `arr[] = {5, 0, 4, -1, -3, 0,``                 ``2, -2, 0, 3, 2};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``  ` `    ``// Call the function and``    ``// print its result``    ``cout << (equalSumPreSuf(arr, n));` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `Main {``    ``// Function to calculate number of``    ``// equal prefix and suffix sums``    ``// till the same indices``    ``static` `int` `equalSumPreSuf(``int` `arr[], ``int` `n)``    ``{``        ``// Calculate prefix sum``        ``int``[] prefix = ``new` `int``[n];``        ``prefix[``0``] = arr[``0``];``        ``for` `(``int` `i = ``1``; i < n; i++) {``            ``prefix[i] = prefix[i - ``1``] + arr[i];``        ``}` `        ``// Calculate suffix sum``        ``int``[] suffix = ``new` `int``[n];``        ``suffix[n - ``1``] = arr[n - ``1``];``        ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) {``            ``suffix[i] = suffix[i + ``1``] + arr[i];``        ``}` `        ``// Count the number of indices where prefix[i] ==``        ``// suffix[i]``        ``int` `count = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(prefix[i] == suffix[i]) {``                ``count++;``            ``}``        ``}` `        ``return` `count;``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Initialize the array``        ``int` `arr[] = { ``5``, ``0``, ``4``, -``1``, -``3``, ``0``, ``2``, -``2``, ``0``, ``3``, ``2` `};``        ``int` `n = arr.length;` `        ``// Call the function and``        ``// print its result``        ``System.out.println(equalSumPreSuf(arr, n));``    ``}``}`

## Python3

 `# Function to calculate number of``# equal prefix and suffix sums``# till the same indices``def` `equalSumPreSuf(arr, n):``    ``# Calculate prefix sum``    ``prefix ``=` `[``0``] ``*` `n``    ``prefix[``0``] ``=` `arr[``0``]``    ``for` `i ``in` `range``(``1``, n):``        ``prefix[i] ``=` `prefix[i ``-` `1``] ``+` `arr[i]` `    ``# Calculate suffix sum   ``    ``suffix ``=` `[``0``] ``*` `n``    ``suffix[n ``-` `1``] ``=` `arr[n ``-` `1``]``    ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``):``        ``suffix[i] ``=` `suffix[i ``+` `1``] ``+` `arr[i]` `    ``# Count the number of indices where prefix[i] == suffix[i]   ``    ``count ``=` `0``    ``for` `i ``in` `range``(n):``        ``if` `prefix[i] ``=``=` `suffix[i]:``            ``count ``+``=` `1``    ``# Returning result``    ``return` `count` `# test case``arr ``=` `[``5``, ``0``, ``4``, ``-``1``, ``-``3``, ``0``, ``2``, ``-``2``, ``0``, ``3``, ``2``]``n ``=` `len``(arr)` `print``(equalSumPreSuf(arr, n))`

## C#

 `// C# code for the above approach` `using` `System;` `class` `GFG{` `    ``// Function to calculate number of``    ``// equal prefix and suffix sums``    ``// till the same indices``    ``static` `int` `equalSumPreSuf(``int``[] arr, ``int` `n) {``        ``// Calculate prefix sum``        ``int``[] prefix = ``new` `int``[n];``        ``prefix[0] = arr[0];``        ``for` `(``int` `i = 1; i < n; i++) {``            ``prefix[i] = prefix[i - 1] + arr[i];``        ``}``    ` `        ``// Calculate suffix sum``        ``int``[] suffix = ``new` `int``[n];``        ``suffix[n - 1] = arr[n - 1];``        ``for` `(``int` `i = n - 2; i >= 0; i--) {``            ``suffix[i] = suffix[i + 1] + arr[i];``        ``}``    ` `        ``// Count the number of indices where prefix[i] == suffix[i]``        ``int` `count = 0;``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(prefix[i] == suffix[i]) {``                ``count++;``            ``}``        ``}``    ` `        ``return` `count;``    ``}``    ` `    ``static` `void` `Main(``string``[] args){``        ``// Initialize the array``        ``int``[] arr = {5, 0, 4, -1, -3, 0,``                     ``2, -2, 0, 3, 2};``        ``int` `n = arr.Length;``      ` `        ``// Call the function and``        ``// print its result``        ``Console.WriteLine(equalSumPreSuf(arr, n));``    ` `    ``}``}`

## Javascript

 `// Function to calculate number of``// equal prefix and suffix sums``// till the same indices``function` `equalSumPreSuf(arr, n) {` `    ``// Calculate prefix sum``    ``let prefix = [];``    ``prefix[0] = arr[0];``    ``for` `(let i = 1; i < n; i++) {``        ``prefix[i] = prefix[i - 1] + arr[i];``    ``}`  `    ``// Calculate suffix sum``    ``let suffix = [];``    ``suffix[n - 1] = arr[n - 1];``    ``for` `(let i = n - 2; i >= 0; i--) {``        ``suffix[i] = suffix[i + 1] + arr[i];``    ``}`  `    ``// Count the number of indices where prefix[i] == suffix[i]``    ``let count = 0;``    ``for` `(let i = 0; i < n; i++) {``        ``if` `(prefix[i] == suffix[i]) {``            ``count++;``        ``}``    ``}``    ``return` `count;``}`  `// Test case``let arr = [5, 0, 4, -1, -3, 0, 2, -2, 0, 3, 2];``let n = arr.length;` `console.log(equalSumPreSuf(arr, n));`

Output

```3

```

Time Complexity: O(N) where N is size of input array. This is because a for loop runs from 1 to N.

Space Complexity: O(N) as vectors prefix and suffix are created of size N where N is size of input array.

Approach: The above approach can be optimized by iterating the array arr twice. The idea is to precompute the suffix sum as the total subarray sum. Then iterate the array a second time to calculate prefix sum at every index then comparing the prefix and suffix sums and update the suffix sum. Follow the steps below to solve the problem:

• Initialize a variable res to zero to calculate the answer
• Initialize a variable sufSum to store the suffix sum
• Initialize a variable preSum to store the prefix sum
• Traverse the array arr and add every element arr[i] to sufSum
• Iterate the array arr again at every iteration:
• Add the current element arr[i] into preSum
• If preSum and sufSum are equal then increment the value of res by 1
• Subtract the current element arr[i] from sufSum
• Return the answer stored in res

Below is the implementation of the above approach:

## C++

 `// C++ code for the above approach``#include ``using` `namespace` `std;` `// Function to calculate number of``// equal prefix and suffix sums``// till the same indices``int` `equalSumPreSuf(``int` `arr[], ``int` `n)``{` `    ``// Initialize a variable``    ``// to store the result``    ``int` `res = 0;` `    ``// Initialize variables to``    ``// calculate prefix and suffix sums``    ``int` `preSum = 0, sufSum = 0;` `    ``// Length of array arr``    ``int` `len = n;` `    ``// Traverse the array from right to left``    ``for` `(``int` `i = len - 1; i >= 0; i--)``    ``{` `        ``// Add the current element``        ``// into sufSum``        ``sufSum += arr[i];``    ``}` `    ``// Iterate the array from left to right``    ``for` `(``int` `i = 0; i < len; i++)``    ``{` `        ``// Add the current element``        ``// into preSum``        ``preSum += arr[i];` `        ``// If prefix sum is equal to``        ``// suffix sum then increment res by 1``        ``if` `(preSum == sufSum)``        ``{` `            ``// Increment the result``            ``res++;``        ``}` `        ``// Subtract the value of current``        ``// element arr[i] from suffix sum``        ``sufSum -= arr[i];``    ``}` `    ``// Return the answer``    ``return` `res;``}` `// Driver code``int` `main()``{` `    ``// Initialize the array``    ``int` `arr[] = {5, 0, 4, -1, -3, 0,``                 ``2, -2, 0, 3, 2};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``// Call the function and``    ``// print its result``    ``cout << (equalSumPreSuf(arr, n));``}` `// This code is contributed by Potta Lokesh`

## Java

 `// Java implementation for the above approach` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `    ``// Function to calculate number of``    ``// equal prefix and suffix sums``    ``// till the same indices``    ``public` `static` `int` `equalSumPreSuf(``int``[] arr)``    ``{` `        ``// Initialize a variable``        ``// to store the result``        ``int` `res = ``0``;` `        ``// Initialize variables to``        ``// calculate prefix and suffix sums``        ``int` `preSum = ``0``, sufSum = ``0``;` `        ``// Length of array arr``        ``int` `len = arr.length;` `        ``// Traverse the array from right to left``        ``for` `(``int` `i = len - ``1``; i >= ``0``; i--) {` `            ``// Add the current element``            ``// into sufSum``            ``sufSum += arr[i];``        ``}` `        ``// Iterate the array from left to right``        ``for` `(``int` `i = ``0``; i < len; i++) {` `            ``// Add the current element``            ``// into preSum``            ``preSum += arr[i];` `            ``// If prefix sum is equal to``            ``// suffix sum then increment res by 1``            ``if` `(preSum == sufSum) {` `                ``// Increment the result``                ``res++;``            ``}` `            ``// Subtract the value of current``            ``// element arr[i] from suffix sum``            ``sufSum -= arr[i];``        ``}` `        ``// Return the answer``        ``return` `res;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``// Initialize the array``        ``int``[] arr = { ``5``, ``0``, ``4``, -``1``, -``3``, ``0``,``                      ``2``, -``2``, ``0``, ``3``, ``2` `};` `        ``// Call the function and``        ``// print its result``        ``System.out.println(equalSumPreSuf(arr));``    ``}``}`

## Python3

 `# Python implementation for the above approach` `# Function to calculate number of``# equal prefix and suffix sums``# till the same indices``from` `builtins ``import` `range` `def` `equalSumPreSuf(arr):``  ` `    ``# Initialize a variable``    ``# to store the result``    ``res ``=` `0``;` `    ``# Initialize variables to``    ``# calculate prefix and suffix sums``    ``preSum ``=` `0``;``    ``sufSum ``=` `0``;` `    ``# Length of array arr``    ``length ``=` `len``(arr);` `    ``# Traverse the array from right to left``    ``for` `i ``in` `range``(length ``-` `1``,``-``1``,``-``1``):``      ` `        ``# Add the current element``        ``# into sufSum``        ``sufSum ``+``=` `arr[i];` `    ``# Iterate the array from left to right``    ``for` `i ``in` `range``(length):` `        ``# Add the current element``        ``# into preSum``        ``preSum ``+``=` `arr[i];` `        ``# If prefix sum is equal to``        ``# suffix sum then increment res by 1``        ``if` `(preSum ``=``=` `sufSum):``          ` `            ``# Increment the result``            ``res ``+``=` `1``;` `        ``# Subtract the value of current``        ``# element arr[i] from suffix sum``        ``sufSum ``-``=` `arr[i];` `    ``# Return the answer``    ``return` `res;` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``  ` `    ``# Initialize the array``    ``arr ``=` `[``5``, ``0``, ``4``, ``-``1``, ``-``3``, ``0``, ``2``, ``-``2``, ``0``, ``3``, ``2``];` `    ``# Call the function and``    ``# print its result``    ``print``(equalSumPreSuf(arr));` `    ``# This code is contributed by 29AjayKumar`

## C#

 `// C# implementation for the above approach``using` `System;``class` `GFG``{` `    ``// Function to calculate number of``    ``// equal prefix and suffix sums``    ``// till the same indices``    ``static` `int` `equalSumPreSuf(``int``[] arr)``    ``{` `        ``// Initialize a variable``        ``// to store the result``        ``int` `res = 0;` `        ``// Initialize variables to``        ``// calculate prefix and suffix sums``        ``int` `preSum = 0, sufSum = 0;` `        ``// Length of array arr``        ``int` `len = arr.Length;` `        ``// Traverse the array from right to left``        ``for` `(``int` `i = len - 1; i >= 0; i--) {` `            ``// Add the current element``            ``// into sufSum``            ``sufSum += arr[i];``        ``}` `        ``// Iterate the array from left to right``        ``for` `(``int` `i = 0; i < len; i++) {` `            ``// Add the current element``            ``// into preSum``            ``preSum += arr[i];` `            ``// If prefix sum is equal to``            ``// suffix sum then increment res by 1``            ``if` `(preSum == sufSum) {` `                ``// Increment the result``                ``res++;``            ``}` `            ``// Subtract the value of current``            ``// element arr[i] from suffix sum``            ``sufSum -= arr[i];``        ``}` `        ``// Return the answer``        ``return` `res;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{` `        ``// Initialize the array``        ``int``[] arr = { 5, 0, 4, -1, -3, 0,``                      ``2, -2, 0, 3, 2 };` `        ``// Call the function and``        ``// print its result``        ``Console.Write(equalSumPreSuf(arr));``    ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Javascript

 ``

Output

```3

```

Time Complexity: O(N)
Auxiliary Space: O(1)