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++
#include <bits/stdc++.h>
using namespace std;
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;
}
int main()
{
int arr[] = { 9, 0, 0, -1, 11, -1 };
int n = sizeof(arr) / sizeof(arr[0]);
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)
{
int[] arr = { 9, 0, 0, -1, 11, -1 };
int n = arr.length;
System.out.println(equalSumPreSuf(arr, n));
}
}
|
Python3
def equalSumPreSuf(arr, n):
count = 0
for i in range(n):
sum1 = 0
sum2 = 0
for j in range(i):
sum1 += arr[j]
for j in range(i + 1, n):
sum2 += arr[j]
if sum1 == sum2:
count += 1
return count
arr = [9, 0, 0, -1, 11, -1]
n = len(arr)
print(equalSumPreSuf(arr, n))
|
C#
using System;
class Gfg {
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;
}
static void Main()
{
int[] arr = { 9, 0, 0, -1, 11, -1 };
int n = arr.Length;
Console.WriteLine(equalSumPreSuf(arr, n));
}
}
|
Javascript
function equalSumPreSuf( arr, n)
{
let count = 0;
for (let i = 0; i < n; i++) {
let sum1 = 0, sum2 = 0;
for (let j = 0; j < i; j++) {
sum1 += arr[j];
}
for (let j = i + 1; j < n; j++) {
sum2 += arr[j];
}
if (sum1 == sum2)
count++;
}
return count;
}
let arr = [ 9, 0, 0, -1, 11, -1 ];
let n = arr.length;
console.log(equalSumPreSuf(arr, n));
|
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:
- Initialize an array arr of n integers.
- 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.
- 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.
- Initialize a variable count to 0.
- Traverse the array and check if prefix[i] is equal to suffix[i]. If it is, then increment count.
- Return the count as the answer.
Below is the implementation of the approach:
C++
#include <bits/stdc++.h>
using namespace std;
int equalSumPreSuf(int arr[], int n) {
vector<int> prefix(n);
prefix[0] = arr[0];
for (int i = 1; i < n; i++) {
prefix[i] = prefix[i - 1] + arr[i];
}
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];
}
int count = 0;
for (int i = 0; i < n; i++) {
if (prefix[i] == suffix[i]) {
count++;
}
}
return count;
}
int main() {
int arr[] = {5, 0, 4, -1, -3, 0,
2, -2, 0, 3, 2};
int n = sizeof(arr) / sizeof(arr[0]);
cout << (equalSumPreSuf(arr, n));
return 0;
}
|
Java
import java.util.*;
public class Main {
static int equalSumPreSuf(int arr[], int n)
{
int[] prefix = new int[n];
prefix[0] = arr[0];
for (int i = 1; i < n; i++) {
prefix[i] = prefix[i - 1] + arr[i];
}
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];
}
int count = 0;
for (int i = 0; i < n; i++) {
if (prefix[i] == suffix[i]) {
count++;
}
}
return count;
}
public static void main(String[] args)
{
int arr[] = { 5, 0, 4, -1, -3, 0, 2, -2, 0, 3, 2 };
int n = arr.length;
System.out.println(equalSumPreSuf(arr, n));
}
}
|
Python3
def equalSumPreSuf(arr, n):
prefix = [0] * n
prefix[0] = arr[0]
for i in range(1, n):
prefix[i] = prefix[i - 1] + arr[i]
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 = 0
for i in range(n):
if prefix[i] == suffix[i]:
count += 1
return count
arr = [5, 0, 4, -1, -3, 0, 2, -2, 0, 3, 2]
n = len(arr)
print(equalSumPreSuf(arr, n))
|
C#
using System;
class GFG{
static int equalSumPreSuf(int[] arr, int n) {
int[] prefix = new int[n];
prefix[0] = arr[0];
for (int i = 1; i < n; i++) {
prefix[i] = prefix[i - 1] + arr[i];
}
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];
}
int count = 0;
for (int i = 0; i < n; i++) {
if (prefix[i] == suffix[i]) {
count++;
}
}
return count;
}
static void Main(string[] args){
int[] arr = {5, 0, 4, -1, -3, 0,
2, -2, 0, 3, 2};
int n = arr.Length;
Console.WriteLine(equalSumPreSuf(arr, n));
}
}
|
Javascript
function equalSumPreSuf(arr, n) {
let prefix = [];
prefix[0] = arr[0];
for (let i = 1; i < n; i++) {
prefix[i] = prefix[i - 1] + arr[i];
}
let suffix = [];
suffix[n - 1] = arr[n - 1];
for (let i = n - 2; i >= 0; i--) {
suffix[i] = suffix[i + 1] + arr[i];
}
let count = 0;
for (let i = 0; i < n; i++) {
if (prefix[i] == suffix[i]) {
count++;
}
}
return count;
}
let arr = [5, 0, 4, -1, -3, 0, 2, -2, 0, 3, 2];
let n = arr.length;
console.log(equalSumPreSuf(arr, n));
|
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++
#include <bits/stdc++.h>
using namespace std;
int equalSumPreSuf(int arr[], int n)
{
int res = 0;
int preSum = 0, sufSum = 0;
int len = n;
for (int i = len - 1; i >= 0; i--)
{
sufSum += arr[i];
}
for (int i = 0; i < len; i++)
{
preSum += arr[i];
if (preSum == sufSum)
{
res++;
}
sufSum -= arr[i];
}
return res;
}
int main()
{
int arr[] = {5, 0, 4, -1, -3, 0,
2, -2, 0, 3, 2};
int n = sizeof(arr) / sizeof(arr[0]);
cout << (equalSumPreSuf(arr, n));
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
public static int equalSumPreSuf(int[] arr)
{
int res = 0;
int preSum = 0, sufSum = 0;
int len = arr.length;
for (int i = len - 1; i >= 0; i--) {
sufSum += arr[i];
}
for (int i = 0; i < len; i++) {
preSum += arr[i];
if (preSum == sufSum) {
res++;
}
sufSum -= arr[i];
}
return res;
}
public static void main(String[] args)
{
int[] arr = { 5, 0, 4, -1, -3, 0,
2, -2, 0, 3, 2 };
System.out.println(equalSumPreSuf(arr));
}
}
|
Python3
from builtins import range
def equalSumPreSuf(arr):
res = 0;
preSum = 0;
sufSum = 0;
length = len(arr);
for i in range(length - 1,-1,-1):
sufSum += arr[i];
for i in range(length):
preSum += arr[i];
if (preSum == sufSum):
res += 1;
sufSum -= arr[i];
return res;
if __name__ == '__main__':
arr = [5, 0, 4, -1, -3, 0, 2, -2, 0, 3, 2];
print(equalSumPreSuf(arr));
|
C#
using System;
class GFG
{
static int equalSumPreSuf(int[] arr)
{
int res = 0;
int preSum = 0, sufSum = 0;
int len = arr.Length;
for (int i = len - 1; i >= 0; i--) {
sufSum += arr[i];
}
for (int i = 0; i < len; i++) {
preSum += arr[i];
if (preSum == sufSum) {
res++;
}
sufSum -= arr[i];
}
return res;
}
public static void Main()
{
int[] arr = { 5, 0, 4, -1, -3, 0,
2, -2, 0, 3, 2 };
Console.Write(equalSumPreSuf(arr));
}
}
|
Javascript
<script>
function equalSumPreSuf(arr, n)
{
let res = 0;
let preSum = 0, sufSum = 0;
let len = n;
for (let i = len - 1; i >= 0; i--)
{
sufSum += arr[i];
}
for (let i = 0; i < len; i++)
{
preSum += arr[i];
if (preSum == sufSum)
{
res++;
}
sufSum -= arr[i];
}
return res;
}
let arr = [5, 0, 4, -1, -3, 0,
2, -2, 0, 3, 2];
let n = arr.length
document.write(equalSumPreSuf(arr, n));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)