Maximize subarray sum by inverting sign of elements of any subarray at most twice

Last Updated : 18 Oct, 2021

Given an array A of size n, find the maximum subarray sum after applying the given operation at most two times. In one operation, choose any two indices i and j and invert sign of all the elements from index i to index j, i.e, all positive elements in the range i to j become negative and all negative elements become positive.

Examples:

Input: A[] = {1, -2, -4, 3, 5, -6}
Output: 21
Explanation:
Inverting the array from index 1 to index 2 & from index 5 to index 5 (0-based indexing) to get {1, 2, 4, 3, 5, 6} with maximum subarray sum = 21

Input: A[] = {2, -1, -18, 3, -1, -39, 5, -2}
Output: 69

Approach: The idea is to first of all, merge all elements into groups. i.e all consecutive positive elements will be summed to one integer and will be placed in a new array arr and similarly for consecutive negative elements. After that, the problem reduces to find the maximum subarray sum after inverting at most two elements from this array.

The idea is to use Dynamic Programming. Now, create a dp[n][3] array and Here at every index in that dp array, maintain three numbers:

The first number represents the maximum subarray sum after inverting the no element in the array. So, dp[i][0] will simply run the logic of kadane’s algorithm for the maximum subarray sum.
The second number represents the maximum subarray sum after inverting exactly one element in the array. So, dp[i][1] will be the maximum of two values i.e dp[i-1][1] + arr[i] (maximum subarray sum after inverting one element till i and then adding the current element to it) and dp[i][0]+ (-arr[i]) ( i.e the previous non-inverted maximum subarray sum till i-1 and then adding current element after inverting ).
The third number represents the maximum subarray sum after inverting exactly two elements in the array. So, dp[i][2] will be maximum of two values i.e $\text{[math]}$ (maximum subarray sum after inverting one element till i-1 and then adding the current element after inverting) and $\text{[math]}$ (maximum subarray sum after inverting two elements till i-1 and adding the current element to it).
Keep a maxSum variable which will be updated after every index and will be equal to the maximum of the previous value of maxSum and all three current values i.e. dp[i][0], dp[i][1] and dp[i][2].
Return maxSum, which is the answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to merge alternate elements into groups
// of positive and negative
void mergeElements(int n, vector<int>& arr, int* A)
{
    int i = 0;
    int sum = 0;
    while (i < n) {
        sum = 0;
        while (i < n && A[i] >= 0) {
            sum += A[i];
            i++;
        }
        if (sum > 0) {
            arr.push_back(sum);
        }
        sum = 0;
        while (i < n && A[i] < 0) {
            sum += A[i];
            i++;
        }
        if (sum < 0) {
            arr.push_back(sum);
        }
    }
}
 
// Function to return the maximum
// after inverting at most 2 elements
int findMaxSum(vector<int>& arr, int n)
{
    int maxSum = 0;
    vector<vector<int> > dp(
n,
 vector<int>(3, INT_MIN));
 
    dp[0][0] = max(0, arr[0]);
    dp[0][1] = -1 * arr[0];
    for (int i = 1; i < n; ++i) {
 
        // dp[i][0] represents sum till ith index
        // without inverting any element.
        dp[i][0] = max(arr[i], 
dp[i - 1][0] + arr[i]);
 
        // dp[i][1] represents sum till ith index
        // after inverting one element.
        dp[i][1] = max(0, dp[i - 1][0]) - arr[i];
        if (i >= 1) {
            dp[i][1] = max(dp[i][1], 
dp[i - 1][1] + arr[i]);
 
            // dp[i][2] represents sum till ith index
            // after inverting two elements.
            dp[i][2] = dp[i - 1][1] - arr[i];
        }
 
        if (i >= 2) {
            dp[i][2] = max(dp[i][2], 
dp[i - 1][2] + arr[i]);
        }
        maxSum = max(maxSum, dp[i][0]);
        maxSum = max(maxSum, dp[i][1]);
        maxSum = max(maxSum, dp[i][2]);
    }
    return maxSum;
}
 
// Driver Code
int main()
{
    int n = 8;
    int A[8] = { 2, -1, -18, 3, -1, -39, 5, -2 };
 
    // vector 'arr' contains sum of consecutive
    // positive or negative elements.
    vector<int> arr;
    mergeElements(n, arr, A);
 
    cout << findMaxSum(arr, arr.size());
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to merge alternate elements into groups
// of positive and negative
static Vector<Integer> mergeElements(int n, Vector<Integer> arr, int[] A)
{
    int i = 0;
    int sum = 0;
    while (i < n) {
        sum = 0;
        while (i < n && A[i] >= 0) {
            sum += A[i];
            i++;
        }
        if (sum > 0) {
            arr.add(sum);
        }
        sum = 0;
        while (i < n && A[i] < 0) {
            sum += A[i];
            i++;
        }
        if (sum < 0) {
            arr.add(sum);
        }
    }
    return arr;
}
 
// Function to return the maximum
// after inverting at most 2 elements
static int findMaxSum(Vector<Integer> arr, int n)
{
    int maxSum = 0;
    int [][]dp = new int[n][3];
 
 
    dp[0][0] = Math.max(0, arr.get(0));
    dp[0][1] = -1 * arr.get(0);
    for (int i = 1; i < n; ++i) {
 
        // dp[i][0] represents sum till ith index
        // without inverting any element.
        dp[i][0] = Math.max(arr.get(i), 
dp[i - 1][0] + arr.get(i));
 
        // dp[i][1] represents sum till ith index
        // after inverting one element.
        dp[i][1] = Math.max(0, dp[i - 1][0]) - arr.get(i);
        if (i >= 1) {
            dp[i][1] = Math.max(dp[i][1], 
dp[i - 1][1] + arr.get(i));
 
            // dp[i][2] represents sum till ith index
            // after inverting two elements.
            dp[i][2] = dp[i - 1][1] - arr.get(i);
        }
 
        if (i >= 2) {
            dp[i][2] = Math.max(dp[i][2], 
dp[i - 1][2] + arr.get(i));
        }
        maxSum = Math.max(maxSum, dp[i][0]);
        maxSum = Math.max(maxSum, dp[i][1]);
        maxSum = Math.max(maxSum, dp[i][2]);
    }
    return maxSum;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 8;
    int A[] = { 2, -1, -18, 3, -1, -39, 5, -2 };
 
    // vector 'arr' contains sum of consecutive
    // positive or negative elements.
    Vector<Integer> arr = new Vector<Integer>();
   arr = mergeElements(n, arr, A);
 
    System.out.print(findMaxSum(arr, arr.size()));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# python program for the above approach
 
INT_MIN = -2147483648
# Function to merge alternate elements into groups
# of positive and negative
 
 
def mergeElements(n, arr, A):
 
    i = 0
    sum = 0
    while (i < n):
        sum = 0
        while (i < n and A[i] >= 0):
            sum += A[i]
            i += 1
 
        if (sum > 0):
            arr.append(sum)
 
        sum = 0
        while (i < n and A[i] < 0):
            sum += A[i]
            i += 1
 
        if (sum < 0):
            arr.append(sum)
 
 
# Function to return the maximum
# after inverting at most 2 elements
def findMaxSum(arr, n):
 
    maxSum = 0
    dp = [[INT_MIN for _ in range(3)] for _ in range(n)]
 
    dp[0][0] = max(0, arr[0])
    dp[0][1] = -1 * arr[0]
    for i in range(1, n):
 
        # dp[i][0] represents sum till ith index
        # without inverting any element.
        dp[i][0] = max(arr[i], dp[i - 1][0] + arr[i])
 
        # dp[i][1] represents sum till ith index
        # after inverting one element.
        dp[i][1] = max(0, dp[i - 1][0]) - arr[i]
        if (i >= 1):
            dp[i][1] = max(dp[i][1], dp[i - 1][1] + arr[i])
 
            # dp[i][2] represents sum till ith index
            # after inverting two elements.
            dp[i][2] = dp[i - 1][1] - arr[i]
 
        if (i >= 2):
            dp[i][2] = max(dp[i][2], dp[i - 1][2] + arr[i])
 
        maxSum = max(maxSum, dp[i][0])
        maxSum = max(maxSum, dp[i][1])
        maxSum = max(maxSum, dp[i][2])
 
    return maxSum
 
 
# Driver Code
if __name__ == "__main__":
 
    n = 8
    A = [2, -1, -18, 3, -1, -39, 5, -2]
 
    # vector 'arr' contains sum of consecutive
    # positive or negative elements.
    arr = []
    mergeElements(n, arr, A)
    print(findMaxSum(arr, len(arr)))
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG 
{
   
    // Function to merge alternate elements into groups
    // of positive and negative
    static void mergeElements(int n, List<int> arr, int[] A)
    {
        int i = 0;
        int sum = 0;
        while (i < n) {
            sum = 0;
            while (i < n && A[i] >= 0) {
                sum += A[i];
                i++;
            }
            if (sum > 0) {
                arr.Add(sum);
            }
            sum = 0;
            while (i < n && A[i] < 0) {
                sum += A[i];
                i++;
            }
            if (sum < 0) {
                arr.Add(sum);
            }
        }
    }
 
    // Function to return the maximum
    // after inverting at most 2 elements
    static int findMaxSum(List<int> arr, int n)
    {
        int maxSum = 0;
        int[, ] dp = new int[n, 3];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < 3; j++)
                dp[i, j] = Int32.MinValue;
 
        dp[0, 0] = Math.Max(0, arr[0]);
        dp[0, 1] = -1 * arr[0];
        for (int i = 1; i < n; ++i) {
 
            // dp[i][0] represents sum till ith index
            // without inverting any element.
            dp[i, 0]
                = Math.Max(arr[i], dp[i - 1, 0] + arr[i]);
 
            // dp[i][1] represents sum till ith index
            // after inverting one element.
            dp[i, 1] = Math.Max(0, dp[i - 1, 0]) - arr[i];
            if (i >= 1) {
                dp[i, 1] = Math.Max(dp[i, 1],
                                    dp[i - 1, 1] + arr[i]);
 
                // dp[i][2] represents sum till ith index
                // after inverting two elements.
                dp[i, 2] = dp[i - 1, 1] - arr[i];
            }
 
            if (i >= 2) {
                dp[i, 2] = Math.Max(dp[i, 2],
                                    dp[i - 1, 2] + arr[i]);
            }
            maxSum = Math.Max(maxSum, dp[i, 0]);
            maxSum = Math.Max(maxSum, dp[i, 1]);
            maxSum = Math.Max(maxSum, dp[i, 2]);
        }
        return maxSum;
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 8;
        int[] A = { 2, -1, -18, 3, -1, -39, 5, -2 };
 
        // vector 'arr' contains sum of consecutive
        // positive or negative elements.
        List<int> arr = new List<int>();
        mergeElements(n, arr, A);
 
        Console.WriteLine(findMaxSum(arr, arr.Count));
    }
}
 
// This code is contributed by ukasp.

Javascript

<script>
        // JavaScript Program to implement
        // the above approach
 
 
        // Function to merge alternate elements into groups
        // of positive and negative
        function mergeElements(n, arr, A) {
            let i = 0;
            let sum = 0;
            while (i < n) {
                sum = 0;
                while (i < n && A[i] >= 0) {
                    sum += A[i];
                    i++;
                }
                if (sum > 0) {
                    arr.push(sum);
                }
                sum = 0;
                while (i < n && A[i] < 0) {
                    sum += A[i];
                    i++;
                }
                if (sum < 0) {
                    arr.push(sum);
                }
            }
        }
 
        // Function to return the maximum
        // after inverting at most 2 elements
        function findMaxSum(arr, n) {
            let maxSum = 0;
 
            let dp = new Array(n);
 
            for (let i = 0; i < dp.length; i++) {
                dp[i] = new Array(3).fill(Number.MIN_VALUE);
            }
 
            dp[0][0] = Math.max(0, arr[0]);
            dp[0][1] = -1 * arr[0];
            for (let i = 1; i < n; ++i) {
 
                // dp[i][0] represents sum till ith index
                // without inverting any element.
                dp[i][0] = Math.max(arr[i],
                    dp[i - 1][0] + arr[i]);
 
                // dp[i][1] represents sum till ith index
                // after inverting one element.
                dp[i][1] = Math.max(0, dp[i - 1][0]) - arr[i];
                if (i >= 1) {
                    dp[i][1] = Math.max(dp[i][1],
                        dp[i - 1][1] + arr[i]);
 
                    // dp[i][2] represents sum till ith index
                    // after inverting two elements.
                    dp[i][2] = dp[i - 1][1] - arr[i];
                }
 
                if (i >= 2) {
                    dp[i][2] = Math.max(dp[i][2],
                        dp[i - 1][2] + arr[i]);
                }
                maxSum = Math.max(maxSum, dp[i][0]);
                maxSum = Math.max(maxSum, dp[i][1]);
                maxSum = Math.max(maxSum, dp[i][2]);
            }
            return maxSum;
        }
 
        // Driver Code
 
        let n = 8;
        let A = [2, -1, -18, 3, -1, -39, 5, -2];
 
        // vector 'arr' contains sum of consecutive
        // positive or negative elements.
        let arr = [];
        mergeElements(n, arr, A);
 
        document.write(findMaxSum(arr, arr.length));
 
// This code is contributed by Potta Lokesh
    </script>

Output

Time Complexity: O(N)
Auxiliary Space: O(N)

Maximize sum of an Array by flipping sign of all elements of a single subarray

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Maximize subarray sum by inverting sign of elements of any subarray at most twice

C++

Java

Python3

C#

Javascript

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