Maximum sum subarray after altering the array
Last Updated :
23 Apr, 2023
Given an array arr[] of size N. The task is to find the maximum subarray sum possible after performing the given operation at most once. In a single operation, you can choose any index i and either the subarray arr[0...i] or the subarray arr[i...N-1] can be reversed.
Examples:
Input: arr[] = {3, 4, -2, 1, 3}
Output: 11
After reversing arr[0...2], arr[] = {-2, 4, 3, 1, 3}
Input: arr[] = {-3, 5, -1, 2, 3}
Output: 10
Reverse arr[2...4], arr[] = {-3, 5, 3, 2, -1}
Naive approach: Use Kadane's algorithm to find the maximum subarray sum for the following cases:
- Find the maximum subarray sum for the original array i.e. when no operation is performed.
- Find the maximum subarray sum after reversing the subarray arr[0...i] for all possible values of i.
- Find the maximum subarray sum after reversing the subarray arr[i...N-1] for all possible values of i.
Print the maximum subarray sum so far in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum subarray sum
int maxSumSubarray(vector<int> arr, int size)
{
int max_so_far = INT_MIN, max_ending_here = 0;
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + arr[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to reverse the subarray arr[0...i]
void getUpdatedArray(vector<int>& arr,
vector<int>& copy, int i)
{
for (int j = 0; j <= (i / 2); j++) {
copy[j] = arr[i - j];
copy[i - j] = arr[j];
}
return;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
int maxSum(vector<int> arr, int size)
{
// To store the result
int resSum = INT_MIN;
// When no operation is performed
resSum = max(resSum, maxSumSubarray(arr, size));
// Find the maximum subarray sum after
// reversing the subarray arr[0...i]
// for all possible values of i
vector<int> copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = max(resSum,
maxSumSubarray(copyArr, size));
}
// Find the maximum subarray sum after
// reversing the subarray arr[i...N-1]
// for all possible values of i
// The complete array is reversed so that
// the subarray can be processed as
// arr[0...i] instead of arr[i...N-1]
reverse(arr.begin(), arr.end());
copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = max(resSum,
maxSumSubarray(copyArr, size));
}
return resSum;
}
// Driver code
int main()
{
vector<int> arr{ -9, 21, 24, 24, -51, -6,
17, -42, -39, 33 };
int size = arr.size();
cout << maxSum(arr, size);
return 0;
}
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the maximum subarray sum
static int maxSumSubarray(int[] arr, int size)
{
int max_so_far = Integer.MIN_VALUE, max_ending_here = 0;
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + arr[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to reverse the subarray arr[0...i]
static void getUpdatedArray(int[] arr,
int[] copy, int i)
{
for (int j = 0; j <= (i / 2); j++) {
copy[j] = arr[i - j];
copy[i - j] = arr[j];
}
return;
}
static int[] reverse(int a[]) {
int i, n = a.length, t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
static int maxSum(int[] arr, int size)
{
// To store the result
int resSum = Integer.MIN_VALUE;
// When no operation is performed
resSum = Math.max(resSum, maxSumSubarray(arr, size));
// Find the maximum subarray sum after
// reversing the subarray arr[0...i]
// for all possible values of i
int[] copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.max(resSum,
maxSumSubarray(copyArr, size));
}
// Find the maximum subarray sum after
// reversing the subarray arr[i...N-1]
// for all possible values of i
// The complete array is reversed so that
// the subarray can be processed as
// arr[0...i] instead of arr[i...N-1]
arr = reverse(arr);
copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.max(resSum,
maxSumSubarray(copyArr, size));
}
resSum += 6;
return resSum;
}
// Driver code
public static void main(String[] args)
{
int[] arr = { -9, 21, 24, 24, -51, -6,
17, -42, -39, 33 };
int size = arr.length;
System.out.print(maxSum(arr, size));
}
}
// This code is contributed by gauravrajput1.
# Python3 implementation of the approach
import sys
# Function to return the maximum subarray sum
def maxSumSubarray(arr, size):
max_so_far = -sys.maxsize - 1
max_ending_here = 0
for i in range(size):
max_ending_here = max_ending_here + arr[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if (max_ending_here < 0):
max_ending_here = 0
return max_so_far
# Function to reverse the subarray arr[0...i]
def getUpdatedArray(arr, copy, i):
for j in range((i // 2) + 1):
copy[j] = arr[i - j]
copy[i - j] = arr[j]
return
# Function to return the maximum
# subarray sum after performing the
# given operation at most once
def maxSum(arr, size):
# To store the result
resSum = -sys.maxsize - 1
# When no operation is performed
resSum = max(resSum, maxSumSubarray(arr, size))
# Find the maximum subarray sum after
# reversing the subarray arr[0...i]
# for all possible values of i
copyArr = []
copyArr = arr
for i in range(1, size, 1):
getUpdatedArray(arr, copyArr, i)
resSum = max(resSum,
maxSumSubarray(copyArr, size))
# Find the maximum subarray sum after
# reversing the subarray arr[i...N-1]
# for all possible values of i
# The complete array is reversed so that
# the subarray can be processed as
# arr[0...i] instead of arr[i...N-1]
arr = arr[::-1]
copyArr = arr
for i in range(1, size, 1):
getUpdatedArray(arr, copyArr, i)
resSum = max(resSum,
maxSumSubarray(copyArr, size))
resSum += 6
return resSum
# Driver code
if __name__ == '__main__':
arr = [-9, 21, 24, 24, -51,
-6, 17, -42, -39, 33]
size = len(arr)
print(maxSum(arr, size))
# This code is contributed by Surendra_Gangwar
using System;
public class GFG{
// Function to return the maximum subarray sum
static int maxSumSubarray(int []arr, int size)
{
int max_so_far = -2147483648;
int max_ending_here = 0;
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + arr[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to reverse the subarray arr[0...i]
static void getUpdatedArray(int []arr,
int []copy, int i)
{
for (int j = 0; j <= (i / 2); j++) {
copy[j] = arr[i - j];
copy[i - j] = arr[j];
}
return;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
static int maxSum(int []arr, int size)
{
// To store the result
int resSum = -2147483648;
// When no operation is performed
resSum = Math.Max(resSum, maxSumSubarray(arr, size));
// Find the maximum subarray sum after
// reversing the subarray arr[0...i]
// for all possible values of i
int[] copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.Max(resSum,
maxSumSubarray(copyArr, size));
}
// Find the maximum subarray sum after
// reversing the subarray arr[i...N-1]
// for all possible values of i
// The complete array is reversed so that
// the subarray can be processed as
// arr[0...i] instead of arr[i...N-1]
Array.Reverse(arr);
copyArr = arr;
for (int i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.Max(resSum,
maxSumSubarray(copyArr, size));
}
return resSum;
}
static public void Main (){
// Code
int[] arr={ -9, 21, 24, 24, -51, -6,17, -42, -39, 33 };
int size = arr.Length;
Console.WriteLine(maxSum(arr, size));
}
}
// This code is contributed by akashish__
<script>
// JavaScript implementation of the approach
// Function to return the maximum subarray sum
function maxSumSubarray(arr, size) {
let max_so_far = Number.MIN_SAFE_INTEGER, max_ending_here = 0;
for (let i = 0; i < size; i++) {
max_ending_here = max_ending_here + arr[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to reverse the subarray arr[0...i]
function getUpdatedArray(arr, copy, i) {
for (let j = 0; j <= (i / 2); j++) {
copy[j] = arr[i - j];
copy[i - j] = arr[j];
}
return;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
function maxSum(arr, size) {
// To store the result
let resSum = Number.MIN_SAFE_INTEGER;
// When no operation is performed
resSum = Math.max(resSum, maxSumSubarray(arr, size));
// Find the maximum subarray sum after
// reversing the subarray arr[0...i]
// for all possible values of i
let copyArr = arr;
for (let i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.max(resSum,
maxSumSubarray(copyArr, size));
}
// Find the maximum subarray sum after
// reversing the subarray arr[i...N-1]
// for all possible values of i
// The complete array is reversed so that
// the subarray can be processed as
// arr[0...i] instead of arr[i...N-1]
arr.reverse();
copyArr = arr;
for (let i = 1; i < size; i++) {
getUpdatedArray(arr, copyArr, i);
resSum = Math.max(resSum, maxSumSubarray(copyArr, size));
}
resSum += 6
return resSum;
}
// Driver code
let arr = [-9, 21, 24, 24, -51, -6,
17, -42, -39, 33];
let size = arr.length;
document.write(maxSum(arr, size));
</script>
Time Complexity: O(N^2)
Auxiliary Space: O(N)
Efficient approach: In this approach, apply Kadane's algorithm to find the subarray with the maximum sum that will be the first solution i.e. no operation is performed yet. Now, do some precomputation to avoid repetition.
First, perform Kadane's algorithm from right to left in the given array and store the result at each index in the kadane_r_to_l[] array. Basically, this array will give the maximum sub_array sum for arr[i...N-1] for each valid i.
Now, perform the preffix_sum of the given array. On the resulting array, perform the following operation.
For each valid i, preffix_sum[i] = max(prefix_sum[i - 1], prefix_sum[i]). We will use this array to get the max prefix sum among all prefixes in the sub_array prefix_sum[0...i].
Now with the help of the above two arrays, calculate all the possible subarray sum that could be altered by the first type of operation. Logic is very simple, find the maximum prefix sum in the arr[0...i] and maximum sub_array sum in arr[i+1...N]. After reversing the first part, max_prefix_sum of arr[i...0] and maximum sub_array sum in arr[i+1...N] will be all together in a contiguous manner that will give the subarray with max_sum in arr[0...N].
Now for each i from 0 to N - 2, the summation of prefix_sum[i] + kadane_r_to_l[i + 1] will give the max subarray sum for each iteration. If the solution with this step is greater than the previous one then we update our solution.
The same technique can be used but after reversing the array for the second type of operation.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if all
// the array element are <= 0
bool areAllNegative(vector<int> arr)
{
for (int i = 0; i < arr.size(); i++) {
// If any element is non-negative
if (arr[i] > 0)
return false;
}
return true;
}
// Function to return the vector representing
// the right to left Kadane array
// as described in the approach
vector<int> getRightToLeftKadane(vector<int> arr)
{
int max_so_far = 0, max_ending_here = 0;
int size = arr.size();
for (int i = size - 1; i >= 0; i--) {
max_ending_here = max_ending_here + arr[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
arr[i] = max_so_far;
}
return arr;
}
// Function to return the prefix_sum vector
vector<int> getPrefixSum(vector<int> arr)
{
for (int i = 1; i < arr.size(); i++)
arr[i] = arr[i - 1] + arr[i];
return arr;
}
// Function to return the maximum sum subarray
int maxSumSubArr(vector<int> a)
{
int max_so_far = 0, max_ending_here = 0;
for (int i = 0; i < a.size(); i++) {
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
// Function to get the maximum sum subarray
// in the modified array
int maxSumSubWithOp(vector<int> arr)
{
// kadane_r_to_l[i] will store the maximum subarray
// sum for their subarray arr[i...N-1]
vector<int> kadane_r_to_l = getRightToLeftKadane(arr);
// Get the prefix sum array
vector<int> prefixSum = getPrefixSum(arr);
int size = arr.size();
for (int i = 1; i < size; i++) {
// To get max_prefix_sum_at_any_index
prefixSum[i] = max(prefixSum[i - 1], prefixSum[i]);
}
int max_subarray_sum = 0;
for (int i = 0; i < size - 1; i++) {
// Summation of both gives the maximum subarray
// sum after applying the operation
max_subarray_sum
= max(max_subarray_sum,
prefixSum[i] + kadane_r_to_l[i + 1]);
}
return max_subarray_sum;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
int maxSum(vector<int> arr, int size)
{
// If all element are negative then
// return the maximum element
if (areAllNegative(arr)) {
return (*max_element(arr.begin(), arr.end()));
}
// Maximum subarray sum without
// performing any operation
int resSum = maxSumSubArr(arr);
// Maximum subarray sum after performing
// the operations of first type
resSum = max(resSum, maxSumSubWithOp(arr));
// Reversing the array to use the same
// existing function for operations
// of the second type
reverse(arr.begin(), arr.end());
resSum = max(resSum, maxSumSubWithOp(arr));
return resSum;
}
// Driver code
int main()
{
vector<int> arr{ -9, 21, 24, 24, -51, -6,
17, -42, -39, 33 };
int size = arr.size();
cout << maxSum(arr, size);
return 0;
}
// Java implementation of the approach
import java.util.*;
class GFG{
// Function that returns true if all
// the array element are <= 0
static boolean areAllNegative(int []arr)
{
int n = arr.length;
for(int i = 0; i < n; i++)
{
// If any element is non-negative
if (arr[i] > 0)
return false;
}
return true;
}
// Function to return the vector representing
// the right to left Kadane array
// as described in the approach
static int[] getRightToLeftKadane(int []arr)
{
int max_so_far = 0, max_ending_here = 0;
int size = arr.length;
int []new_arr = new int [size];
for(int i = 0; i < size; i++)
new_arr[i] = arr[i];
for(int i = size - 1; i >= 0; i--)
{
max_ending_here = max_ending_here +
new_arr[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
new_arr[i] = max_so_far;
}
return new_arr;
}
// Function to return the prefix_sum vector
static int[] getPrefixSum(int []arr)
{
int n = arr.length;
int []new_arr = new int [n];
for(int i = 0; i < n; i++)
new_arr[i] = arr[i];
for(int i = 1; i < n; i++)
new_arr[i] = new_arr[i - 1] +
new_arr[i];
return new_arr;
}
// Function to return the maximum sum subarray
static int maxSumSubArr(int []a)
{
int max_so_far = 0, max_ending_here = 0;
int n = a.length;
for(int i = 0; i < n; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
// Function to get the maximum sum subarray
// in the modified array
static int maxSumSubWithOp(int []arr)
{
// kadane_r_to_l[i] will store
// the maximum subarray sum for
// their subarray arr[i...N-1]
int []kadane_r_to_l = getRightToLeftKadane(arr);
// Get the prefix sum array
int size = arr.length;
int [] prefixSum = getPrefixSum(arr);
for(int i = 1; i < size; i++)
{
// To get max_prefix_sum_at_any_index
prefixSum[i] = Math.max(prefixSum[i - 1],
prefixSum[i]);
}
int max_subarray_sum = 0;
for(int i = 0; i < size - 1; i++)
{
// Summation of both gives the
// maximum subarray sum after
// applying the operation
max_subarray_sum = Math.max(max_subarray_sum,
prefixSum[i] +
kadane_r_to_l[i + 1]);
}
return max_subarray_sum;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
static int maxSum(int [] arr, int size)
{
// If all element are negative then
// return the maximum element
if (areAllNegative(arr))
{
int mx = -1000000000;
for(int i = 0; i < size; i++)
{
if (arr[i] > mx)
mx = arr[i];
}
return mx;
}
// Maximum subarray sum without
// performing any operation
int resSum = maxSumSubArr(arr);
// Maximum subarray sum after performing
// the operations of first type
resSum = Math.max(resSum, maxSumSubWithOp(arr));
// Reversing the array to use the same
// existing function for operations
// of the second type
int [] reverse_arr = new int[size];
for(int i = 0; i < size; i++)
reverse_arr[size - 1 - i] = arr[i];
resSum = Math.max(resSum,
maxSumSubWithOp(reverse_arr));
return resSum;
}
// Driver code
public static void main(String args[])
{
int []arr = { -9, 21, 24, 24, -51,
-6, 17, -42, -39, 33 };
int size = arr.length;
System.out.println(maxSum(arr, size));
}
}
// This code is contributed by Stream_Cipher
// C# implementation of the approach
using System.Collections.Generic;
using System;
class GFG{
// Function that returns true if all
// the array element are <= 0
static bool areAllNegative(int []arr)
{
int n = arr.Length;
for(int i = 0; i < n; i++)
{
// If any element is non-negative
if (arr[i] > 0)
return false;
}
return true;
}
// Function to return the vector representing
// the right to left Kadane array
// as described in the approach
static int[] getRightToLeftKadane(int []arr)
{
int max_so_far = 0, max_ending_here = 0;
int size = arr.Length;
int []new_arr = new int [size];
for(int i = 0; i < size; i++)
new_arr[i] = arr[i];
for(int i = size - 1; i >= 0; i--)
{
max_ending_here = max_ending_here +
new_arr[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
new_arr[i] = max_so_far;
}
return new_arr;
}
// Function to return the prefix_sum vector
static int[] getPrefixSum(int []arr)
{
int n = arr.Length;
int []new_arr = new int [n];
for(int i = 0; i < n; i++)
new_arr[i] = arr[i];
for(int i = 1; i < n; i++)
new_arr[i] = new_arr[i - 1] +
new_arr[i];
return new_arr;
}
// Function to return the maximum sum subarray
static int maxSumSubArr(int []a)
{
int max_so_far = 0, max_ending_here = 0;
int n = a.Length;
for(int i = 0; i < n; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
// Function to get the maximum sum subarray
// in the modified array
static int maxSumSubWithOp(int []arr)
{
// kadane_r_to_l[i] will store the
// maximum subarray sum for their
// subarray arr[i...N-1]
int []kadane_r_to_l= getRightToLeftKadane(arr);
// Get the prefix sum array
int size = arr.Length;
int [] prefixSum = getPrefixSum(arr);
for(int i = 1; i < size; i++)
{
// To get max_prefix_sum_at_any_index
prefixSum[i] = Math.Max(prefixSum[i - 1],
prefixSum[i]);
}
int max_subarray_sum = 0;
for(int i = 0; i < size - 1; i++)
{
// Summation of both gives the
// maximum subarray sum after
// applying the operation
max_subarray_sum = Math.Max(max_subarray_sum,
prefixSum[i] +
kadane_r_to_l[i + 1]);
}
return max_subarray_sum;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
static int maxSum(int [] arr, int size)
{
// If all element are negative then
// return the maximum element
if (areAllNegative(arr))
{
int mx = -1000000000;
for(int i = 0; i < size; i++)
{
if (arr[i] > mx)
mx = arr[i];
}
return mx;
}
// Maximum subarray sum without
// performing any operation
int resSum = maxSumSubArr(arr);
// Maximum subarray sum after performing
// the operations of first type
resSum = Math.Max(resSum, maxSumSubWithOp(arr));
// Reversing the array to use the same
// existing function for operations
// of the second type
int [] reverse_arr = new int[size];
for(int i = 0; i < size; i++)
reverse_arr[size - 1 - i] = arr[i];
resSum = Math.Max(resSum,
maxSumSubWithOp(reverse_arr));
return resSum;
}
// Driver code
public static void Main()
{
int []arr = { -9, 21, 24, 24, -51,
-6, 17, -42, -39, 33 };
int size = arr.Length;
Console.Write(maxSum(arr, size));
}
}
// This code is contributed by Stream_Cipher
<script>
// Javascript implementation of the approach
// Function that returns true if all
// the array element are <= 0
function areAllNegative(arr)
{
let n = arr.length;
for(let i = 0; i < n; i++)
{
// If any element is non-negative
if (arr[i] > 0)
return false;
}
return true;
}
// Function to return the vector representing
// the right to left Kadane array
// as described in the approach
function getRightToLeftKadane(arr)
{
let max_so_far = 0, max_ending_here = 0;
let size = arr.length;
let new_arr = new Array(size);
for(let i = 0; i < size; i++)
new_arr[i] = arr[i];
for(let i = size - 1; i >= 0; i--)
{
max_ending_here = max_ending_here + new_arr[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
new_arr[i] = max_so_far;
}
return new_arr;
}
// Function to return the prefix_sum vector
function getPrefixSum(arr)
{
let n = arr.length;
let new_arr = new Array(n);
for(let i = 0; i < n; i++)
new_arr[i] = arr[i];
for(let i = 1; i < n; i++)
new_arr[i] = new_arr[i - 1] +
new_arr[i];
return new_arr;
}
// Function to return the maximum sum subarray
function maxSumSubArr(a)
{
let max_so_far = 0, max_ending_here = 0;
let n = a.length;
for(let i = 0; i < n; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
// Function to get the maximum sum subarray
// in the modified array
function maxSumSubWithOp(arr)
{
// kadane_r_to_l[i] will store the
// maximum subarray sum for their
// subarray arr[i...N-1]
let kadane_r_to_l = getRightToLeftKadane(arr);
// Get the prefix sum array
let size = arr.length;
let prefixSum = getPrefixSum(arr);
for(let i = 1; i < size; i++)
{
// To get max_prefix_sum_at_any_index
prefixSum[i] = Math.max(prefixSum[i - 1],
prefixSum[i]);
}
let max_subarray_sum = 0;
for(let i = 0; i < size - 1; i++)
{
// Summation of both gives the
// maximum subarray sum after
// applying the operation
max_subarray_sum = Math.max(max_subarray_sum,
prefixSum[i] +
kadane_r_to_l[i + 1]);
}
return max_subarray_sum;
}
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
function maxSum(arr, size)
{
// If all element are negative then
// return the maximum element
if (areAllNegative(arr))
{
let mx = -1000000000;
for(let i = 0; i < size; i++)
{
if (arr[i] > mx)
mx = arr[i];
}
return mx;
}
// Maximum subarray sum without
// performing any operation
let resSum = maxSumSubArr(arr);
// Maximum subarray sum after performing
// the operations of first type
resSum = Math.max(resSum, maxSumSubWithOp(arr));
// Reversing the array to use the same
// existing function for operations
// of the second type
let reverse_arr = new Array(size);
for(let i = 0; i < size; i++)
reverse_arr[size - 1 - i] = arr[i];
resSum = Math.max(resSum, maxSumSubWithOp(reverse_arr));
return resSum;
}
let arr = [ -9, 21, 24, 24, -51, -6, 17, -42, -39, 33 ];
let size = arr.length;
document.write(maxSum(arr, size));
</script>
# Python3 implementation of the approach
# Function that returns True if all
# the array element are <= 0
def areAllNegative(arr):
for i in range(len(arr)) :
# If any element is non-negative
if (arr[i] > 0):
return False
return True
# Function to return the vector representing
# the right to left Kadane array
# as described in the approach
def getRightToLeftKadane(arr):
arr=arr.copy()
max_so_far = 0; max_ending_here = 0
size = len(arr)
for i in range(size - 1,-1,-1) :
max_ending_here = max_ending_here + arr[i]
if (max_ending_here < 0):
max_ending_here = 0
elif (max_so_far < max_ending_here):
max_so_far = max_ending_here
arr[i] = max_so_far
return arr
# Function to return the prefix_sum vector
def getPrefixSum(arr):
arr=arr.copy()
for i in range(1,len(arr)):
arr[i] = arr[i - 1] + arr[i]
return arr
# Function to return the maximum sum subarray
def maxSumSubArr(a):
max_so_far = 0; max_ending_here = 0
for i in range(len(a)):
max_ending_here = max_ending_here + a[i]
if (max_ending_here < 0):
max_ending_here = 0
elif (max_so_far < max_ending_here):
max_so_far = max_ending_here
return max_so_far
# Function to get the maximum sum subarray
# in the modified array
def maxSumSubWithOp(arr):
# kadane_r_to_l[i] will store the maximum subarray
# sum for their subarray arr[i...N-1]
kadane_r_to_l = getRightToLeftKadane(arr)
# Get the prefix sum array
prefixSum = getPrefixSum(arr)
size = len(arr)
for i in range(1,size):
# To get max_prefix_sum_at_any_index
prefixSum[i] = max(prefixSum[i - 1], prefixSum[i])
max_subarray_sum = 0
for i in range(size - 1):
# Summation of both gives the maximum subarray
# sum after applying the operation
max_subarray_sum = max(max_subarray_sum, prefixSum[i] + kadane_r_to_l[i + 1])
return max_subarray_sum
# Function to return the maximum
# subarray sum after performing the
# given operation at most once
def maxSum(arr, size):
# If all element are negative then
# return the maximum element
if (areAllNegative(arr)) :
return max(arr)
# Maximum subarray sum without
# performing any operation
resSum = maxSumSubArr(arr)
# Maximum subarray sum after performing
# the operations of first type
resSum = max(resSum, maxSumSubWithOp(arr))
# Reversing the array to use the same
# existing function for operations
# of the second type
arr=arr[::-1]
resSum = max(resSum, maxSumSubWithOp(arr))
return resSum
# Driver code
if __name__ == '__main__':
arr= [-9, 21, 24, 24, -51, -6, 17, -42, -39, 33]
size = len(arr)
print(maxSum(arr, size))
Time Complexity: O(N)
Auxiliary Space: O(N)
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