Given an array arr[] of integer elements, the task is to find maximum possible sub-array sum after changing the signs of at most two elements.
Examples:
Input: arr[] = {-5, 3, 2, 7, -8, 3, 7, -9, 10, 12, -6}
Output: 61
We can get 61 from index 0 to 10 by
changing the sign of elements at 4th and 7th indices i.e.
-8 and -9. We could have chosen -5 and -6 but this gives us
smaller sum 48.
Input: arr[] = {-5, -3, -18, 0, -4}
Output: 22
Approach: This problem can be solved using Dynamic Programming. Let’s suppose there are n elements in the array. We build our solution from smallest length to largest length.
At each step, we change the solution for length i to i+1.
For each step we have three cases:
- (Maximum sub-array sum) by altering sign of at most 0 element.
- (Maximum sub-array sum) by altering sign of at most 1 element.
- (Maximum sub-array sum) by altering sign of at most 2 element.
These cases use each others previous values.
- Case 1: We have two choices either to take current element or to add current value into previous value of same case.we store whichever is larger.
- Case 2: We have two choices here
- We alter the sign of current element and then add it to 0 or previous case 1 value. we store whichever is larger.
- we take the current element of array and add it to previous case 2 value.If this value is larger than value we get in (a) case then we update else not.
- Case 3: We again have two choices here
- We alter the sign of current element and add it to previous case 2 value.
- We add current element into previous case 3 value. Larger value obtained from (a) and (b) is stored for current case.
We update the max value out of these 3 cases and store it in a variable.
For each case of each step we take Two dimensional array dp[n+1][3] if given array contains n elements.
Recurrence Relation:
Case 1: dp[i][0] = max(dp[i – 1][0] + arr[i], arr[i])
Case 2: dp[i][1] = max(max(0, dp[i – 1][0]) – arr[i], dp[i – 1][1] + arr[i])
Case 3: dp[i][2] = max(dp[i – 1][1] – arr[i], dp[i – 1][2] + arr[i])
solution = max(solution, max(dp[i][0], dp[i][1], dp[i][2]))
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <algorithm>#include <iostream>using namespace std;// Function to return the maximum required sub-array sumint maxSum(int a[], int n){ int ans = 0; int* arr = new int[n + 1]; // Creating one based indexing for (int i = 1; i <= n; i++) arr[i] = a[i - 1]; // 2d array to contain solution for each step int** dp = new int*[n + 1]; for (int i = 0; i <= n; i++) dp[i] = new int[3]; for (int i = 1; i <= n; ++i) { // Case 1: Choosing current or (current + previous) // whichever is smaller dp[i][0] = max(arr[i], dp[i - 1][0] + arr[i]); // Case 2:(a) Altering sign and add to previous case 1 or // value 0 dp[i][1] = max(0, dp[i - 1][0]) - arr[i]; // Case 2:(b) Adding current element with previous case 2 // and updating the maximum if (i >= 2) dp[i][1] = max(dp[i][1], dp[i - 1][1] + arr[i]); // Case 3:(a) Altering sign and add to previous case 2 if (i >= 2) dp[i][2] = dp[i - 1][1] - arr[i]; // Case 3:(b) Adding current element with previous case 3 if (i >= 3) dp[i][2] = max(dp[i][2], dp[i - 1][2] + arr[i]); // Updating the maximum value of variable ans ans = max(ans, dp[i][0]); ans = max(ans, dp[i][1]); ans = max(ans, dp[i][2]); } // Return the final solution return ans;}// Driver codeint main(){ int arr[] = { -5, 3, 2, 7, -8, 3, 7, -9, 10, 12, -6 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxSum(arr, n); return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function to return the maximum required sub-array sum static int maxSum(int []a, int n) { int ans = 0; int [] arr = new int[n + 1]; // Creating one based indexing for (int i = 1; i <= n; i++) arr[i] = a[i - 1]; // 2d array to contain solution for each step int [][] dp = new int [n + 1][3]; for (int i = 1; i <= n; ++i) { // Case 1: Choosing current or (current + previous) // whichever is smaller dp[i][0] = Math.max(arr[i], dp[i - 1][0] + arr[i]); // Case 2:(a) Altering sign and add to previous case 1 or // value 0 dp[i][1] = Math.max(0, dp[i - 1][0]) - arr[i]; // Case 2:(b) Adding current element with previous case 2 // and updating the maximum if (i >= 2) dp[i][1] = Math.max(dp[i][1], dp[i - 1][1] + arr[i]); // Case 3:(a) Altering sign and add to previous case 2 if (i >= 2) dp[i][2] = dp[i - 1][1] - arr[i]; // Case 3:(b) Adding current element with previous case 3 if (i >= 3) dp[i][2] = Math.max(dp[i][2], dp[i - 1][2] + arr[i]); // Updating the maximum value of variable ans ans = Math.max(ans, dp[i][0]); ans = Math.max(ans, dp[i][1]); ans = Math.max(ans, dp[i][2]); } // Return the final solution return ans; } // Driver code public static void main (String[] args) { int arr[] = { -5, 3, 2, 7, -8, 3, 7, -9, 10, 12, -6 }; int n = arr.length; System.out.println(maxSum(arr, n)); }}// This code is contributed by ihritik |
Python3
# Python3 implementation of the approach # Function to return the maximum # required sub-array sum def maxSum(a, n): ans = 0 arr = [0] * (n + 1) # Creating one based indexing for i in range(1, n + 1): arr[i] = a[i - 1] # 2d array to contain solution for each step dp = [[0 for i in range(3)] for j in range(n + 1)] for i in range(0, n + 1): # Case 1: Choosing current or # (current + previous) whichever is smaller dp[i][0] = max(arr[i], dp[i - 1][0] + arr[i]) # Case 2:(a) Altering sign and add to # previous case 1 or value 0 dp[i][1] = max(0, dp[i - 1][0]) - arr[i] # Case 2:(b) Adding current element with # previous case 2 and updating the maximum if i >= 2: dp[i][1] = max(dp[i][1], dp[i - 1][1] + arr[i]) # Case 3:(a) Altering sign and # add to previous case 2 if i >= 2: dp[i][2] = dp[i - 1][1] - arr[i] # Case 3:(b) Adding current element # with previous case 3 if i >= 3: dp[i][2] = max(dp[i][2], dp[i - 1][2] + arr[i]) # Updating the maximum value # of variable ans ans = max(ans, dp[i][0]) ans = max(ans, dp[i][1]) ans = max(ans, dp[i][2]) # Return the final solution return ans # Driver code if __name__ == "__main__": arr = [-5, 3, 2, 7, -8, 3, 7, -9, 10, 12, -6] n = len(arr) print(maxSum(arr, n)) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the maximum required sub-array sum static int maxSum(int [] a, int n) { int ans = 0; int [] arr = new int[n + 1]; // Creating one based indexing for (int i = 1; i <= n; i++) arr[i] = a[i - 1]; // 2d array to contain solution for each step int [, ] dp = new int [n + 1, 3]; for (int i = 1; i <= n; ++i) { // Case 1: Choosing current or (current + previous) // whichever is smaller dp[i, 0] = Math.Max(arr[i], dp[i - 1, 0] + arr[i]); // Case 2:(a) Altering sign and add to previous case 1 or // value 0 dp[i, 1] = Math.Max(0, dp[i - 1, 0]) - arr[i]; // Case 2:(b) Adding current element with previous case 2 // and updating the maximum if (i >= 2) dp[i, 1] = Math.Max(dp[i, 1], dp[i - 1, 1] + arr[i]); // Case 3:(a) Altering sign and add to previous case 2 if (i >= 2) dp[i, 2] = dp[i - 1, 1] - arr[i]; // Case 3:(b) Adding current element with previous case 3 if (i >= 3) dp[i, 2] = Math.Max(dp[i, 2], dp[i - 1, 2] + arr[i]); // Updating the maximum value of variable ans ans = Math.Max(ans, dp[i, 0]); ans = Math.Max(ans, dp[i, 1]); ans = Math.Max(ans, dp[i, 2]); } // Return the final solution return ans; } // Driver code public static void Main () { int [] arr = { -5, 3, 2, 7, -8, 3, 7, -9, 10, 12, -6 }; int n = arr.Length; Console.WriteLine(maxSum(arr, n)); }}// This code is contributed by ihritik |
PHP
<?php// PHP implementation of the approach // Function to return the maximum// required sub-array sum function maxSum($a, $n) { $ans = 0; $arr = array(); // Creating one based indexing for ($i = 1; $i <= $n; $i++) $arr[$i] = $a[$i - 1]; // 2d array to contain solution // for each step $dp = array(array()); for ($i = 1; $i <= $n; ++$i) { // Case 1: Choosing current or (current + // previous) whichever is smaller $dp[$i][0] = max($arr[$i], $dp[$i - 1][0] + $arr[$i]); // Case 2:(a) Altering sign and add to // previous case 1 or value 0 $dp[$i][1] = max(0, $dp[$i - 1][0]) - $arr[$i]; // Case 2:(b) Adding current element with // previous case 2 and updating the maximum if ($i >= 2) $dp[$i][1] = max($dp[$i][1], $dp[$i - 1][1] + $arr[$i]); // Case 3:(a) Altering sign and // add to previous case 2 if ($i >= 2) $dp[$i][2] = $dp[$i - 1][1] - $arr[$i]; // Case 3:(b) Adding current element // with previous case 3 if ($i >= 3) $dp[$i][2] = max($dp[$i][2], $dp[$i - 1][2] + $arr[$i]); // Updating the maximum value of variable ans $ans = max($ans, $dp[$i][0]); $ans = max($ans, $dp[$i][1]); $ans = max($ans, $dp[$i][2]); } // Return the final solution return $ans; } // Driver code $arr = array( -5, 3, 2, 7, -8, 3, 7, -9, 10, 12, -6 ); $n = count($arr) ; echo maxSum($arr, $n); // This code is contributed by Ryuga?> |
Javascript
<script> // JavaScript implementation of the approach // Function to return the maximum required sub-array sum function maxSum(a, n) { let ans = 0; let arr = new Array(n + 1); // Creating one based indexing for (let i = 1; i <= n; i++) arr[i] = a[i - 1]; // 2d array to contain solution for each step let dp = new Array(n + 1); for (let i = 0; i <= n; ++i) { dp[i] = new Array(3); for (let j = 0; j < 3; ++j) { dp[i][j] = 0; } } for (let i = 1; i <= n; ++i) { // Case 1: Choosing current or (current + previous) // whichever is smaller dp[i][0] = Math.max(arr[i], dp[i - 1][0] + arr[i]); // Case 2:(a) Altering sign and add to previous case 1 or // value 0 dp[i][1] = Math.max(0, dp[i - 1][0]) - arr[i]; // Case 2:(b) Adding current element with previous case 2 // and updating the maximum if (i >= 2) dp[i][1] = Math.max(dp[i][1], dp[i - 1][1] + arr[i]); // Case 3:(a) Altering sign and add to previous case 2 if (i >= 2) dp[i][2] = dp[i - 1][1] - arr[i]; // Case 3:(b) Adding current element with previous case 3 if (i >= 3) dp[i][2] = Math.max(dp[i][2], dp[i - 1][2] + arr[i]); // Updating the maximum value of variable ans ans = Math.max(ans, dp[i][0]); ans = Math.max(ans, dp[i][1]); ans = Math.max(ans, dp[i][2]); } // Return the final solution return ans; } let arr = [ -5, 3, 2, 7, -8, 3, 7, -9, 10, 12, -6 ]; let n = arr.length; document.write(maxSum(arr, n));</script> |
Output:
61
Time Complexity : 
Space Complexity : 
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