Given an array arr[]. Find the maximum value of prefix sum which is also suffix sum for index i in arr[].
Examples :
Input : arr[] = {-1, 2, 3, 0, 3, 2, -1}
Output : 4
Prefix sum of arr[0..3] =
Suffix sum of arr[3..6]
Input : arr[] = {-2, 5, 3, 1, 2, 6, -4, 2}
Output : 7
Prefix sum of arr[0..3] =
Suffix sum of arr[3..7]A Simple Solution is to one by one check the given condition (prefix sum equal to suffix sum) for every element and returns the element that satisfies the given condition with maximum value.
C++
#include <bits/stdc++.h>
using namespace std;
int findMaxSum(int arr[], int n)
{
int res = INT_MIN;
for (int i = 0; i < n; i++)
{
int prefix_sum = arr[i];
for (int j = 0; j < i; j++)
prefix_sum += arr[j];
int suffix_sum = arr[i];
for (int j = n - 1; j > i; j--)
suffix_sum += arr[j];
if (prefix_sum == suffix_sum)
res = max(res, prefix_sum);
}
return res;
}
int main()
{
int arr[] = {-2, 5, 3, 1,
2, 6, -4, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findMaxSum(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int findMaxSum(int []arr, int n)
{
int res = Integer.MIN_VALUE;
for (int i = 0; i < n; i++)
{
int prefix_sum = arr[i];
for (int j = 0; j < i; j++)
prefix_sum += arr[j];
int suffix_sum = arr[i];
for (int j = n - 1; j > i; j--)
suffix_sum += arr[j];
if (prefix_sum == suffix_sum)
res = Math.max(res, prefix_sum);
}
return res;
}
public static void main (String[] args)
{
int arr[] = {-2, 5, 3, 1, 2, 6, -4, 2 };
int n = arr.length;
System.out.println(findMaxSum(arr, n));
}
}
|
Python3
import sys
def findMaxSum(arr, n):
res = -sys.maxsize - 1
for i in range(n):
prefix_sum = arr[i]
for j in range(i):
prefix_sum += arr[j]
suffix_sum = arr[i]
j = n - 1
while(j > i):
suffix_sum += arr[j]
j -= 1
if (prefix_sum == suffix_sum):
res = max(res, prefix_sum)
return res
if __name__ == '__main__':
arr = [-2, 5, 3, 1, 2, 6, -4, 2]
n = len(arr)
print(findMaxSum(arr, n))
|
C#
using System;
class GFG {
static int findMaxSum(int []arr, int n)
{
int res = int.MinValue;
for (int i = 0; i < n; i++)
{
int prefix_sum = arr[i];
for (int j = 0; j < i; j++)
prefix_sum += arr[j];
int suffix_sum = arr[i];
for (int j = n - 1; j > i; j--)
suffix_sum += arr[j];
if (prefix_sum == suffix_sum)
res = Math.Max(res, prefix_sum);
}
return res;
}
public static void Main ()
{
int []arr = {-2, 5, 3, 1, 2, 6, -4, 2 };
int n = arr.Length;
Console.WriteLine(findMaxSum(arr, n));
}
}
|
PHP
<?php
function findMaxSum( $arr, $n)
{
$res = PHP_INT_MIN;
for ( $i = 0; $i < $n; $i++)
{
$prefix_sum = $arr[$i];
for ( $j = 0; $j < $i; $j++)
$prefix_sum += $arr[$j];
$suffix_sum = $arr[$i];
for ( $j = $n - 1; $j > $i; $j--)
$suffix_sum += $arr[$j];
if ($prefix_sum == $suffix_sum)
$res = max($res, $prefix_sum);
}
return $res;
}
$arr = array(-2, 5, 3, 1,
2, 6, -4, 2 );
$n = count($arr);
echo findMaxSum($arr, $n);
?>
|
Javascript
<script>
function findMaxSum(arr, n)
{
var res = -1000000000;
for (var i = 0; i < n; i++)
{
var prefix_sum = arr[i];
for (var j = 0; j < i; j++)
prefix_sum += arr[j];
var suffix_sum = arr[i];
for (var j = n - 1; j > i; j--)
suffix_sum += arr[j];
if (prefix_sum == suffix_sum)
res = Math.max(res, prefix_sum);
}
return res;
}
var arr = [-2, 5, 3, 1,
2, 6, -4, 2 ];
var n = arr.length;
document.write( findMaxSum(arr, n));
</script>
|
Time Complexity: O(n2)
Auxiliary Space: O(n)
A Better Approach is to traverse the array and store prefix sum for each index in array presum[], in which presum[i] stores sum of subarray arr[0..i]. Do another traversal of the array and store suffix sum in another array suffsum[], in which suffsum[i] stores sum of subarray arr[i..n-1]. After this for each index check if presum[i] is equal to suffsum[i] and if they are equal then compare their value with the overall maximum so far.
C++
#include <bits/stdc++.h>
using namespace std;
int findMaxSum(int arr[], int n)
{
int preSum[n];
int suffSum[n];
int ans = INT_MIN;
preSum[0] = arr[0];
for (int i = 1; i < n; i++)
preSum[i] = preSum[i - 1] + arr[i];
suffSum[n - 1] = arr[n - 1];
if (preSum[n - 1] == suffSum[n - 1])
ans = max(ans, preSum[n - 1]);
for (int i = n - 2; i >= 0; i--)
{
suffSum[i] = suffSum[i + 1] + arr[i];
if (suffSum[i] == preSum[i])
ans = max(ans, preSum[i]);
}
return ans;
}
int main()
{
int arr[] = { -2, 5, 3, 1,
2, 6, -4, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findMaxSum(arr, n);
return 0;
}
|
Java
import java.io.*;
public class GFG {
static int findMaxSum(int []arr, int n)
{
int []preSum = new int[n];
int []suffSum = new int[n];
int ans = Integer.MIN_VALUE;
preSum[0] = arr[0];
for (int i = 1; i < n; i++)
preSum[i] = preSum[i - 1] + arr[i];
suffSum[n - 1] = arr[n - 1];
if (preSum[n - 1] == suffSum[n - 1])
ans = Math.max(ans, preSum[n - 1]);
for (int i = n - 2; i >= 0; i--)
{
suffSum[i] = suffSum[i + 1] + arr[i];
if (suffSum[i] == preSum[i])
ans = Math.max(ans, preSum[i]);
}
return ans;
}
static public void main (String[] args)
{
int []arr = { -2, 5, 3, 1, 2, 6, -4, 2 };
int n = arr.length;
System.out.println( findMaxSum(arr, n));
}
}
|
Python3
def findMaxSum(arr, n):
preSum = [0 for i in range(n)]
suffSum = [0 for i in range(n)]
ans = -10000000
preSum[0] = arr[0]
for i in range(1, n):
preSum[i] = preSum[i - 1] + arr[i]
suffSum[n - 1] = arr[n - 1]
if (preSum[n - 1] == suffSum[n - 1]):
ans = max(ans, preSum[n - 1])
for i in range(n - 2, -1, -1):
suffSum[i] = suffSum[i + 1] + arr[i]
if (suffSum[i] == preSum[i]):
ans = max(ans, preSum[i])
return ans
if __name__=='__main__':
arr = [-2, 5, 3, 1,2, 6, -4, 2]
n = len(arr)
print(findMaxSum(arr, n))
|
C#
using System;
public class GFG {
static int findMaxSum(int []arr, int n)
{
int []preSum = new int[n];
int []suffSum = new int[n];
int ans = int.MinValue;
preSum[0] = arr[0];
for (int i = 1; i < n; i++)
preSum[i] = preSum[i - 1] + arr[i];
suffSum[n - 1] = arr[n - 1];
if (preSum[n - 1] == suffSum[n - 1])
ans = Math.Max(ans, preSum[n - 1]);
for (int i = n - 2; i >= 0; i--)
{
suffSum[i] = suffSum[i + 1] + arr[i];
if (suffSum[i] == preSum[i])
ans = Math.Max(ans, preSum[i]);
}
return ans;
}
static public void Main ()
{
int []arr = { -2, 5, 3, 1, 2, 6, -4, 2 };
int n = arr.Length;
Console.WriteLine( findMaxSum(arr, n));
}
}
|
PHP
<?php
function findMaxSum($arr, $n)
{
$preSum[$n] = array();
$suffSum[$n] = array();
$ans = PHP_INT_MIN;
$preSum[0] = $arr[0];
for ($i = 1; $i < $n; $i++)
$preSum[$i] = $preSum[$i - 1] + $arr[$i];
$suffSum[$n - 1] = $arr[$n - 1];
if ($preSum[$n - 1] == $suffSum[$n - 1])
$ans = max($ans, $preSum[$n - 1]);
for ($i = $n - 2; $i >= 0; $i--)
{
$suffSum[$i] = $suffSum[$i + 1] + $arr[$i];
if ($suffSum[$i] == $preSum[$i])
$ans = max($ans, $preSum[$i]);
}
return $ans;
}
$arr = array( -2, 5, 3, 1, 2, 6, -4, 2 );
$n = sizeof($arr);
echo findMaxSum($arr, $n);
?>
|
Javascript
<script>
function findMaxSum(arr, n)
{
let preSum = new Array(n);
preSum.fill(0);
let suffSum = new Array(n);
suffSum.fill(0);
let ans = Number.MIN_VALUE;
preSum[0] = arr[0];
for(let i = 1; i < n; i++)
preSum[i] = preSum[i - 1] + arr[i];
suffSum[n - 1] = arr[n - 1];
if (preSum[n - 1] == suffSum[n - 1])
ans = Math.max(ans, preSum[n - 1]);
for(let i = n - 2; i >= 0; i--)
{
suffSum[i] = suffSum[i + 1] + arr[i];
if (suffSum[i] == preSum[i])
ans = Math.max(ans, preSum[i]);
}
return ans;
}
let arr = [ -2, 5, 3, 1, 2, 6, -4, 2 ];
let n = arr.length;
document.write(findMaxSum(arr, n));
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(n)
Further Optimization :
We can avoid the use of extra space by first computing the total sum, then using it to find the current prefix and suffix sums.
C++
#include <bits/stdc++.h>
using namespace std;
int findMaxSum(int arr[], int n)
{
int sum = accumulate(arr, arr + n, 0);
int prefix_sum = 0, res = INT_MIN;
for (int i = 0; i < n; i++)
{
prefix_sum += arr[i];
if (prefix_sum == sum)
res = max(res, prefix_sum);
sum -= arr[i];
}
return res;
}
int main()
{
int arr[] = { -2, 5, 3, 1,
2, 6, -4, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findMaxSum(arr, n);
return 0;
}
|
Java
import java.lang.Math.*;
import java.util.stream.*;
class GFG {
static int findMaxSum(int arr[], int n)
{
int sum = IntStream.of(arr).sum();
int prefix_sum = 0,
res = Integer.MIN_VALUE;
for (int i = 0; i < n; i++)
{
prefix_sum += arr[i];
if (prefix_sum == sum)
res = Math.max(res, prefix_sum);
sum -= arr[i];
}
return res;
}
public static void main(String[] args)
{
int arr[] = { -2, 5, 3, 1,
2, 6, -4, 2 };
int n = arr.length;
System.out.print(findMaxSum(arr, n));
}
}
|
Python3
import sys
def findMaxSum(arr,n):
ss = sum(arr)
prefix_sum = 0
res = -sys.maxsize
for i in range(n):
prefix_sum += arr[i]
if prefix_sum == ss:
res = max(res, prefix_sum);
ss -= arr[i];
return res
if __name__=="__main__":
arr = [ -2, 5, 3, 1,
2, 6, -4, 2 ]
n = len(arr)
print(findMaxSum(arr, n))
|
C#
using System.Linq;
using System;
class GFG {
static int Add(int x, int y) {
return x + y;
}
static int findMaxSum(int []arr, int n)
{
int sum = arr.Aggregate(func:Add);
int prefix_sum = 0,
res = int.MinValue;
for (int i = 0; i < n; i++)
{
prefix_sum += arr[i];
if (prefix_sum == sum)
res = Math.Max(res, prefix_sum);
sum -= arr[i];
}
return res;
}
public static void Main()
{
int []arr = { -2, 5, 3, 1,
2, 6, -4, 2 };
int n = arr.Length;
Console.Write(findMaxSum(arr, n));
}
}
|
Javascript
<script>
function findMaxSum(arr,n)
{
let sum = 0;
for(let i=0; i < n; i++)
{
sum = sum + arr[i];
}
let prefix_sum = 0, res = Number.MIN_VALUE;
for (let i = 0; i < n; i++)
{
prefix_sum += arr[i];
if (prefix_sum == sum)
res = Math.max(res, prefix_sum);
sum -= arr[i];
}
return res;
}
let arr = [ -2, 5, 3, 1,
2, 6, -4, 2 ];
let n = arr.length;
document.write(findMaxSum(arr, n));
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
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