Maximum subarray sum by flipping signs of at most K array elements

Last Updated : 11 Jul, 2025

Given an array arr[] of N integers and an integer K, The task is to find the maximum sub-array sum by flipping signs of at most K array elements.

Examples:

Input: arr[] = {-6, 2, -1, -1000, 2}, k = 2
Output: 1009
We can flip the signs of -6 and -1000, to get maximum subarray sum as 1009

Input: arr[] = {-1, -2, -100, -10}, k = 1
Output: 100
We can only flip the sign of -100 to get 100

Input: {1, 2, 100, 10}, k = 1
Output: 113
We do not need to flip any elements

Approach: The problem can be solved using Dynamic Programming. Let dp[i][j] be the maximum sub-array sum from index i with j flips. A recursive function can be written in order to solve the problem and we can use memoization to avoid multiple function calls. The recursive DP function (findSubarraySum(ind, flips)) will be called from every index with the number of initial flips as 0.

ans = max(0, a[ind] + findSubarraySum(ind + 1, flips, n, a, k))
ans = max(ans, -a[ind] + findSubarraySum(ind + 1, flips + 1, n, a, k))
If the value is negative, we are replacing it by 0, similarly as we do in Kadane's algorithm.

The recursive function will have two states, one will be if we flip the i-th index. The second one is if we don't flip the i-th index. The base cases are if the ind==n when we have completed a traversal till the last index. We can use memoization in order to store the results which can be used later to avoid multiple same-function calls. The maximum of all dp[i][0] will be our answer.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define right 2
#define left 4
int dp[left][right];

// Function to find the maximum subarray sum with flips
// starting from index i
int findSubarraySum(int ind, int flips, int n, int a[],
                    int k)
{

    // If the number of flips have exceeded
    if (flips > k)
        return -1e9;

    // Complete traversal
    if (ind == n)
        return 0;

    // If the state has previously been visited
    if (dp[ind][flips] != -1)
        return dp[ind][flips];

    // Initially
    int ans = 0;

    // Use Kadane's algorithm and call two states
    ans = max(
        0,
        a[ind] + findSubarraySum(ind + 1, flips, n, a, k));
    ans = max(ans, -a[ind]
                       + findSubarraySum(ind + 1, flips + 1,
                                         n, a, k));

    // Memoize the answer and return it
    return dp[ind][flips] = ans;
}

// Utility function to call flips from index and
// return the answer
int findMaxSubarraySum(int a[], int n, int k)
{
    // Create DP array
    // int dp[n][k+1];
    memset(dp, -1, sizeof(dp));

    int ans = -1e9;

    // Iterate and call recursive function
    // from every index to get the maximum subarray sum
    for (int i = 0; i < n; i++)
        ans = max(ans, findSubarraySum(i, 0, n, a, k));

    // corner case
    if (ans == 0 && k == 0)
        return *max_element(a, a + n);

    return ans;
}

// Driver Code
int main()
{
    int a[] = { -1, -2, -100, -10 };
    int n = sizeof(a) / sizeof(a[0]);
    int k = 1;

    cout << findMaxSubarraySum(a, n, k);

    return 0;
}

Java

// Java implementation of the approach
import java.util.Arrays;
class GFG {

    static int right = 2;
    static int left = 4;
    static int[][] dp = new int[left][right];

    // Function to find the maximum subarray sum with flips
    // starting from index i
    static int findSubarraySum(int ind, int flips, int n,
                               int[] a, int k)
    {

        // If the number of flips have exceeded
        if (flips > k)
            return (int)(-1e9);

        // Complete traversal
        if (ind == n)
            return 0;

        // If the state has previously been visited
        if (dp[ind][flips] != -1)
            return dp[ind][flips];

        // Initially
        int ans = 0;

        // Use Kadane's algorithm and call two states
        ans = Math.max(0, a[ind]
                              + findSubarraySum(
                                  ind + 1, flips, n, a, k));
        ans = Math.max(ans, -a[ind]
                                + findSubarraySum(ind + 1,
                                                  flips + 1,
                                                  n, a, k));

        // Memoize the answer and return it
        return dp[ind][flips] = ans;
    }

    // Utility function to call flips from index and
    // return the answer
    static int findMaxSubarraySum(int[] a, int n, int k)
    {
        // Create DP array
        // int dp[n,k+1];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < k + 1; j++)
                dp[i][j] = -1;

        int ans = (int)(-1e9);

        // Iterate and call recursive function
        // from every index to get the maximum subarray sum
        for (int i = 0; i < n; i++)
            ans = Math.max(ans,
                           findSubarraySum(i, 0, n, a, k));

        // corner case
        if (ans == 0 && k == 0)
            return Arrays.stream(a).max().getAsInt();

        return ans;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] a = { -1, -2, -100, -10 };
        int n = a.length;
        int k = 1;

        System.out.println(findMaxSubarraySum(a, n, k));
    }
}

// This code is contributed by mits

Python3

# Python3 implementation of the approach 
import numpy as np

right = 3; 
left = 6; 

dp = np.ones((left, right))
dp = -1 * dp

# Function to find the maximum 
# subarray sum with flips starting 
# from index i 
def findSubarraySum(ind, flips, n, a, k) : 

    # If the number of flips
    # have exceeded 
    if (flips > k) :
        return -1e9; 

    # Complete traversal 
    if (ind == n) : 
        return 0; 

    # If the state has previously 
    # been visited 
    if (dp[ind][flips] != -1) :
        return dp[ind][flips];
    
    # Initially 
    ans = 0; 

    # Use Kadane's algorithm and
    # call two states 
    ans = max(0, a[ind] + 
              findSubarraySum(ind + 1,
                              flips, n, a, k)); 
    ans = max(ans, -a[ind] + 
              findSubarraySum(ind + 1, flips + 1, 
                                       n, a, k)); 

    # Memoize the answer and return it 
    dp[ind][flips] = ans; 
    
    return dp[ind][flips] ;

# Utility function to call flips 
# from index and return the answer 
def findMaxSubarraySum(a, n, k) : 

    ans = -1e9;
    
    # Iterate and call recursive 
    # function from every index to
    # get the maximum subarray sum 
    for i in range(n) : 
        ans = max(ans, findSubarraySum(i, 0, n, a, k)); 
    
    # corner casae
    if ans == 0 and k == 0:
      return max(a);

    return ans; 

# Driver Code 
if __name__ == "__main__" :

    a = [-1, -2, -100, -10]; 
    n = len(a) ;
    k = 1; 

    print(findMaxSubarraySum(a, n, k));
    
# This code is contributed by Ryuga

// C# implementation of the approach
using System;
using System.Linq;

class GFG {

    static int right = 2;
    static int left = 4;
    static int[, ] dp = new int[left + 1, right + 1];

    // Function to find the maximum subarray sum
    // with flips starting from index i
    static int findSubarraySum(int ind, int flips, int n,
                               int[] a, int k)
    {

        // If the number of flips have exceeded
        if (flips > k)
            return -(int)1e9;

        // Complete traversal
        if (ind == n)
            return 0;

        // If the state has previously been visited
        if (dp[ind, flips] != -1)
            return dp[ind, flips];

        // Initially
        int ans = 0;

        // Use Kadane's algorithm and call two states
        ans = Math.Max(0, a[ind]
                              + findSubarraySum(
                                  ind + 1, flips, n, a, k));
        ans = Math.Max(ans, -a[ind]
                                + findSubarraySum(ind + 1,
                                                  flips + 1,
                                                  n, a, k));

        // Memoize the answer and return it
        return dp[ind, flips] = ans;
    }

    // Utility function to call flips from
    // index and return the answer
    static int findMaxSubarraySum(int[] a, int n, int k)
    {
        // Create DP array
        // int dp[n][k+1];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < k + 1; j++)
                dp[i, j] = -1;

        int ans = -(int)1e9;

        // Iterate and call recursive function
        // from every index to get the maximum subarray sum
        for (int i = 0; i < n; i++)
            ans = Math.Max(ans,
                           findSubarraySum(i, 0, n, a, k));

        // corner case
        if (ans == 0 && k == 0)
            return a.Max();

        return ans;
    }

    // Driver Code
    static void Main()
    {
        int[] a = { -1, -2, -100, -10 };
        int n = a.Length;
        int k = 1;

        Console.WriteLine(findMaxSubarraySum(a, n, k));
    }
}

// This code is contributed by mits

JavaScript

<script>
    // Javascript implementation of the approach
    
    let right = 2;
    let left = 4;
    let dp = new Array(left);
 
    // Function to find the maximum subarray sum with flips
    // starting from index i
    function findSubarraySum(ind, flips, n, a, k)
    {
 
        // If the number of flips have exceeded
        if (flips > k)
            return (-1e9);
 
        // Complete traversal
        if (ind == n)
            return 0;
 
        // If the state has previously been visited
        if (dp[ind][flips] != -1)
            return dp[ind][flips];
 
        // Initially
        let ans = 0;
 
        // Use Kadane's algorithm and call two states
        ans = Math.max(0, a[ind]
                              + findSubarraySum(
                                  ind + 1, flips, n, a, k));
        ans = Math.max(ans, -a[ind]
                                + findSubarraySum(ind + 1,
                                                  flips + 1,
                                                  n, a, k));
 
        // Memoize the answer and return it
        return dp[ind][flips] = ans;
    }
 
    // Utility function to call flips from index and
    // return the answer
    function findMaxSubarraySum(a, n, k)
    {
        // Create DP array
        // int dp[n,k+1];
        for (let i = 0; i < n; i++)
        {
            dp[i] = new Array(k);
            for (let j = 0; j < k + 1; j++)
            {
                dp[i][j] = -1;
            }
        }
 
        let ans = (-1e9);
 
        // Iterate and call recursive function
        // from every index to get the maximum subarray sum
        for (let i = 0; i < n; i++)
            ans = Math.max(ans,
                           findSubarraySum(i, 0, n, a, k));
 
        // corner case
        if (ans == 0 && k == 0)
        {
            let max = Number.MIN_VALUE;
            for(let i = 0; i < a.length; i++)
            {
                max = Math.max(max, a[i]);
            }
            return max;
        }
 
        return ans;
    }
    
    let a = [ -1, -2, -100, -10 ];
    let n = a.length;
    let k = 1;

    document.write(findMaxSubarraySum(a, n, k));
    
</script>

Output

Time Complexity: O(N * K), as we are using a loop to traverse N times and we are recursively calling the function K times which will cost O(K). Where N is the number of elements in the array and K is the limit of elements.
Auxiliary Space: O(N * K), as we are using extra space for memorization. Where N is the number of elements in the array and K is the limit of elements.

Analysis of Algorithms

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