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

• Last Updated : 03 Dec, 2021

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.
Time Complexity: O(N^2)

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 indicesint 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 codeint 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 indicesfrom 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 codeif __name__ == '__main__':       # Initialize the array    arr = [5, 0, 4, -1, -3, 0, 2, -2, 0, 3, 2];     # Call the function and    # prits result    print(equalSumPreSuf(arr));     # This code is contributed by 29AjayKumar

## C#

 // C# implementation for the above approachusing 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)

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