Given an array arr[] of size N, the task is to find the maximum sum of the elements of the array after applying the given operation any number of times. In a single operation, choose an index 1 ? i < N and multiply both arr[i] and arr[i – 1] by -1.
Examples:
Input: arr[] = {-10, 5, -4}
Output: 19
Perform the operation for i = 1 and
the array becomes {10, -5, -4}
Perform the operation for i = 2 and
the array becomes {10, 5, 4}
10 + 5 + 4 = 19
Input: arr[] = {10, -4, -8, -11, 3}
Output: 30
Approach: This problem can be solved using dynamic programming. Since it is useless to choose and flip the same arr[i], we will consider flipping at most once for each element of the array in order from the left. Let dp[i][0] represents the maximum possible sum up to the ith index without flipping the ith index. dp[i][1] represents the maximum possible sum up to the ith index with flipping the ith index. So, dp(n, 0) is our required answer.
Base conditions:
- dp[0][0] = 0
- dp[0][1] = INT_MIN
Recurrence relation:
- dp[i + 1][0] = max(dp[i][0] + arr[i], dp[i][1] – arr[i])
- dp[i + 1][1] = max(dp[i][0] – arr[i], dp[i][1] + arr[i])
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 possible// sum after performing the given operationint max_sum(int a[], int n){ // Dp vector to store the answer vector<vector<int> > dp(n + 1, vector<int>(2, 0)); // Base value dp[0][0] = 0, dp[0][1] = -999999; for (int i = 0; i <= n - 1; i++) { dp[i + 1][0] = max(dp[i][0] + a[i], dp[i][1] - a[i]); dp[i + 1][1] = max(dp[i][0] - a[i], dp[i][1] + a[i]); } // Return the maximum sum return dp[n][0];}// Driver codeint main(){ int a[] = { -10, 5, -4 }; int n = sizeof(a) / sizeof(a[0]); cout << max_sum(a, n); return 0;} |
Java
// Java implementation of the approachclass GFG {// Function to return the maximum possible// sum after performing the given operationstatic int max_sum(int a[], int n){ // Dp vector to store the answer int [][]dp = new int[n + 1][2]; // Base value dp[0][0] = 0; dp[0][1] = -999999; for (int i = 0; i <= n - 1; i++) { dp[i + 1][0] = Math.max(dp[i][0] + a[i], dp[i][1] - a[i]); dp[i + 1][1] = Math.max(dp[i][0] - a[i], dp[i][1] + a[i]); } // Return the maximum sum return dp[n][0];}// Driver codepublic static void main(String[] args) { int a[] = { -10, 5, -4 }; int n = a.length; System.out.println(max_sum(a, n));}}// This code is contributed by 29AjayKumar |
Python3
# Python implementation of the approach# Function to return the maximum possible# sum after performing the given operationdef max_sum(a, n): # Dp vector to store the answer dp = [[0 for i in range(2)] for j in range(n+1)] # Base value dp[0][0] = 0; dp[0][1] = -999999; for i in range(0, n): dp[i + 1][0] = max(dp[i][0] + a[i], dp[i][1] - a[i]); dp[i + 1][1] = max(dp[i][0] - a[i], dp[i][1] + a[i]); # Return the maximum sum return dp[n][0];# Driver codeif __name__ == '__main__': a = [-10, 5, -4 ]; n = len(a); print(max_sum(a, n)); # This code is contributed by 29AjayKumar |
Javascript
<script>// javascript implementation of the approach // Function to return the maximum possible // sum after performing the given operation function max_sum(a , n) { // Dp vector to store the answer var dp = Array(n + 1).fill().map(()=>Array(2).fill(0)); // Base value dp[0][0] = 0; dp[0][1] = -999999; for (i = 0; i <= n - 1; i++) { dp[i + 1][0] = Math.max(dp[i][0] + a[i], dp[i][1] - a[i]); dp[i + 1][1] = Math.max(dp[i][0] - a[i], dp[i][1] + a[i]); } // Return the maximum sum return dp[n][0]; } // Driver code var a = [ -10, 5, -4 ]; var n = a.length; document.write(max_sum(a, n));// This code is contributed by todaysgaurav</script> |
C#
// C# implementation of the approachusing System; class GFG {// Function to return the maximum possible// sum after performing the given operationstatic int max_sum(int []a, int n){ // Dp vector to store the answer int [,]dp = new int[n + 1, 2]; // Base value dp[0, 0] = 0; dp[0, 1] = -999999; for (int i = 0; i <= n - 1; i++) { dp[i + 1, 0] = Math.Max(dp[i, 0] + a[i], dp[i, 1] - a[i]); dp[i + 1, 1] = Math.Max(dp[i, 0] - a[i], dp[i, 1] + a[i]); } // Return the maximum sum return dp[n, 0];}// Driver codepublic static void Main(String[] args) { int []a = { -10, 5, -4 }; int n = a.Length; Console.WriteLine(max_sum(a, n));}}// This code is contributed by PrinciRaj1992 |
Output:
19
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time.
Auxiliary Space: O(N), as we are using extra space for DP.
Efficient approach : Space Optimization O(1)
To optimize the space complexity in previous code we are only using the previous row of the dp table to compute the values of the current row. Therefore, we can maintain just two variables to keep track of the values of the previous row instead of using a 2D vector. This will reduce the space complexity from O(N) to O(1).
Implementation Steps:
- Initialize two variables prev0 and prev1 to 0 and -999999 respectively.
- Loop through the input array from index 0 to n-1.
- Compute two new variables curr0 and curr1 as follows.
- curr0 is the maximum of (prev0 + a[i]) and (prev1 – a[i])
- curr1 is the maximum of (prev0 – a[i]) and (prev1 + a[i])
- Update prev0 to curr0 and prev1 to curr1 for further iteration.
- At last return final answer stored in prev0.
Implementation:
C++
// C++ code for above approach#include <bits/stdc++.h>using namespace std;// Function to return the maximum possible// sum after performing the given operationint max_sum(int a[], int n){ // Variables to store the values of the // previous row of the DP table int prev0 = 0, prev1 = -999999; for (int i = 0; i <= n - 1; i++) { int curr0 = max(prev0 + a[i], prev1 - a[i]); int curr1 = max(prev0 - a[i], prev1 + a[i]); // assigning values for further iterations prev0 = curr0; prev1 = curr1; } // Return the maximum sum return prev0;}// Driver codeint main(){ int a[] = { -10, 5, -4 }; int n = sizeof(a) / sizeof(a[0]); cout << max_sum(a, n); return 0;} |
Java
// Java code for above approachimport java.util.*;class Main { // Function to return the maximum possible // sum after performing the given operation static int maxSum(int[] a, int n) { // Variables to store the values of the // previous row of the DP table int prev0 = 0, prev1 = -999999; for (int i = 0; i <= n - 1; i++) { int curr0 = Math.max(prev0 + a[i], prev1 - a[i]); int curr1 = Math.max(prev0 - a[i], prev1 + a[i]); // assigning values for further iterations prev0 = curr0; prev1 = curr1; } // Return the maximum sum return prev0; } // Driver code public static void main(String[] args) { int[] a = { -10, 5, -4 }; int n = a.length; System.out.println(maxSum(a, n)); }} |
Python3
# Function to return the maximum possible sum after performing the given operationdef max_sum(a, n): # Variables to store the values of the previous row of the DP table prev0 = 0 prev1 = -999999 for i in range(n): curr0 = max(prev0 + a[i], prev1 - a[i]) curr1 = max(prev0 - a[i], prev1 + a[i]) # assigning values for further iterations prev0 = curr0 prev1 = curr1 # Return the maximum sum return prev0# Driver codea = [-10, 5, -4]n = len(a)print(max_sum(a, n)) |
C#
using System;class MainClass{ // Function to return the maximum possible // sum after performing the given operation static int maxSum(int[] a, int n) { // Variables to store the values of the // previous row of the DP table int prev0 = 0, prev1 = -999999; for (int i = 0; i <= n - 1; i++) { int curr0 = Math.Max(prev0 + a[i], prev1 - a[i]); int curr1 = Math.Max(prev0 - a[i], prev1 + a[i]); // assigning values for further iterations prev0 = curr0; prev1 = curr1; } // Return the maximum sum return prev0; } // Driver code public static void Main() { int[] a = { -10, 5, -4 }; int n = a.Length; Console.WriteLine(maxSum(a, n)); }} |
Javascript
// Function to return the maximum possible// sum after performing the given operationfunction max_sum(a, n) { // Variables to store the values of the // previous row of the DP table let prev0 = 0, prev1 = -999999; for (let i = 0; i <= n - 1; i++) { let curr0 = Math.max(prev0 + a[i], prev1 - a[i]); let curr1 = Math.max(prev0 - a[i], prev1 + a[i]); // assigning values for further iterations prev0 = curr0; prev1 = curr1; } // Return the maximum sum return prev0;}// Driver codelet a = [-10, 5, -4];let n = a.length;console.log(max_sum(a, n)); |
Output
19
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time.
Auxiliary Space: O(1)
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