[Dynamic Programming] Maximum Sum of Subarray with at Most Two Deletions - Hard

Finding the maximum sum of a subarray with at most two deletions is a dynamic programming challenge. We need to think carefully about the elements involved. Our solution will keep track of the maximum sum of valid subarrays. We allow the removal of up to two elements. We can solve this problem efficiently using dynamic programming techniques. These techniques help us track cumulative sums and the best outcomes through iterations.

In this article, we will look into the details of the problem. First, we will understand the problem statement for the maximum sum of a subarray with two deletions. Then, we will explore different dynamic programming methods in Java, Python, and C++. We will also talk about how to make space usage better. We will introduce another method called sliding window. We will handle edge cases and check how well different methods perform. At the end, we will answer some common questions about this interesting topic.

  • [Dynamic Programming] Maximum Sum of Subarray with Up to Two Deletions - Hard
  • Understanding the Problem Statement for Maximum Sum of Subarray with Two Deletions
  • Dynamic Programming Approach for Maximum Sum of Subarray with Two Deletions in Java
  • Dynamic Programming Approach for Maximum Sum of Subarray with Two Deletions in Python
  • Dynamic Programming Approach for Maximum Sum of Subarray with Two Deletions in C++
  • Optimizing Space Complexity in Maximum Sum of Subarray with Two Deletions
  • Alternative Sliding Window Technique for Maximum Sum of Subarray with Two Deletions
  • Handling Edge Cases in Maximum Sum of Subarray with Two Deletions
  • Performance Analysis of Different Approaches for Maximum Sum of Subarray with Two Deletions
  • Frequently Asked Questions

If we want to learn more about dynamic programming, we can read these articles on related topics. They are helpful: Dynamic Programming: Maximum Subarray Sum Using Kadane’s Algorithm and Dynamic Programming: Minimum Path Sum in a Grid.

Understanding the Problem Statement for Maximum Sum of Subarray with Two Deletions

We have a problem. We need to find the maximum sum of a subarray. We can delete up to two elements from the array to help us get this maximum sum.

Problem Definition

We are given an integer array called nums. Our goal is to find the maximum sum of a group of numbers that are next to each other after we delete at most two elements. The tricky part is to do this quickly while thinking about how deleting elements will change the sum.

Example

  • Input: nums = [1, -2, 0, 3]

  • Output: 4 because the subarray [1, 0, 3] gives us the highest sum.

  • Input: nums = [1, -2, -2, 3]

  • Output: 3 since the subarray [3] has the highest sum.

Constraints

  • The array can have both positive and negative numbers.
  • The length of the array can be from 1 to 10^5.
  • We can always make the sum of the remaining numbers bigger by deleting some elements.

Key Considerations

  • We must keep the order of the numbers in the subarray, but we can delete elements from any spot.
  • Our solution should work well even with big input sizes. We want it to run in linear time, O(n).
  • The algorithm needs to keep track of the maximum sums as we go through the array. We need to think about cases with zero, one, or two deletions.

Next, we will create a dynamic programming plan to calculate the maximum sum based on the rules we have.

Dynamic Programming Approach for Maximum Sum of Subarray with Two Deletions in Java

We want to find the maximum sum of a subarray with at most two deletions using dynamic programming in Java. We will use a DP table to track the maximum sums while looking at different deletion cases.

Approach

  1. Initialization:
    • We create a DP array dp. Here, dp[i][j] shows the maximum sum of the subarray ending at index i with j deletions (0, 1, 2).
  2. Base Cases:
    • We start with the first element. For j = 0, the maximum sum is nums[0].
    • For j = 1 and j = 2, we set the values as needed.
  3. DP Transition:
    • For each element in the array, we update the DP table:
      • Without deletion: dp[i][0] = max(dp[i-1][0] + nums[i], nums[i])
      • With one deletion: dp[i][1] = max(dp[i-1][1] + nums[i], dp[i-1][0])
      • With two deletions: dp[i][2] = max(dp[i-1][2] + nums[i], dp[i-1][1])
  4. Final Result:
    • The final result will be the highest value in dp[n-1][0], dp[n-1][1], and dp[n-1][2].

Java Implementation

Here is the Java code that shows the above approach:

public class MaximumSumSubarray {
    public static int maxSumWithTwoDeletions(int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;

        int[][] dp = new int[n][3];

        // Base cases
        dp[0][0] = nums[0];
        dp[0][1] = Integer.MIN_VALUE;
        dp[0][2] = Integer.MIN_VALUE;

        for (int i = 1; i < n; i++) {
            dp[i][0] = Math.max(dp[i-1][0] + nums[i], nums[i]);
            dp[i][1] = Math.max(dp[i-1][1] + nums[i], dp[i-1][0]);
            dp[i][2] = Math.max(dp[i-1][2] + nums[i], dp[i-1][1]);
        }

        return Math.max(dp[n-1][0], Math.max(dp[n-1][1], dp[n-1][2]));
    }

    public static void main(String[] args) {
        int[] nums = {1, -2, 0, 3};
        System.out.println("Maximum Sum with At Most Two Deletions: " + maxSumWithTwoDeletions(nums));
    }
}

Explanation of the Code

  • Input: We take an array of integers called nums.
  • DP Array: We create a 2D array dp to store the maximum sums for each deletion case.
  • Loop: We go through the input array. We update the DP table using the rules we defined.
  • Output: We print the maximum sum we can get with at most two deletions.

This dynamic programming method finds the best result. It helps us choose the best subarray while allowing for some deletions.

Dynamic Programming Approach for Maximum Sum of Subarray with Two Deletions in Python

We want to find the maximum sum of a subarray with at most two deletions. We can use dynamic programming for this. We will create a table to store our results. The main idea is to keep track of the maximum sum at each index while considering deletions.

Problem Breakdown

  1. Define State: We use dp[i][j] to show the maximum sum of a subarray up to index i with j deletions. Here j can be 0, 1, or 2.

  2. Transition:

    • When we do not delete any element, we can add the current element to the previous sum:

      dp[i][0] = max(dp[i-1][0] + arr[i], arr[i])

    • If we delete one element, we can delete the current one or keep the previous sum with one deletion:

      dp[i][1] = max(dp[i-1][1] + arr[i], dp[i-1][0])

    • If we delete two elements, we can choose to delete the current one or use the previous state with one deletion:

      dp[i][2] = max(dp[i-1][2] + arr[i], dp[i-1][1])

  3. Initialization:

    • For dp[0][0], we just need the first element:

      dp[0][0] = arr[0]
dp[0][1] = float('-inf')  # Impossible state
dp[0][2] = float('-inf')  # Impossible state

  4. Final Result: The result is the maximum value in dp[n-1][0], dp[n-1][1], and dp[n-1][2]. Here n is the length of the array.

Python Implementation

We can implement this in Python like this:

def max_sum_with_two_deletions(arr):
    n = len(arr)
    if n == 0:
        return 0
    if n == 1:
        return arr[0]

    dp = [[0] * 3 for _ in range(n)]
    dp[0][0] = arr[0]
    dp[0][1] = float('-inf')
    dp[0][2] = float('-inf')

    for i in range(1, n):
        dp[i][0] = max(dp[i-1][0] + arr[i], arr[i])
        dp[i][1] = max(dp[i-1][1] + arr[i], dp[i-1][0])
        dp[i][2] = max(dp[i-1][2] + arr[i], dp[i-1][1])

    return max(dp[n-1][0], dp[n-1][1], dp[n-1][2])

# Example usage
arr = [1, -2, 0, 3]
result = max_sum_with_two_deletions(arr)
print("Maximum sum of subarray with at most two deletions:", result)

Explanation of the Code

  • We first create the dp list to keep our results.
  • Then, we go through the array. We calculate the maximum sums based on our conditions.
  • In the end, we return the maximum of the three possible states at the last index.

This dynamic programming way helps us find the maximum sum of a subarray with at most two deletions in O(n) time and O(n) space.

For more about dynamic programming, you can check Dynamic Programming - Maximum Sum of Subarray with One Deletion.

Dynamic Programming Approach for Maximum Sum of Subarray with Two Deletions in C++

To find the maximum sum of a subarray with at most two deletions using dynamic programming in C++, we can follow a clear way.

Problem Breakdown

We make a dynamic programming array dp. Here, dp[i][j] shows the maximum sum of the subarray that ends at index i with j deletions. The value of j can be 0, 1, or 2.

Steps:

  1. Initialize Variables:
    • n: Length of the input array.
    • dp: A 2D array that starts with zero. Here, dp[i][j] means the maximum sum of the subarray that ends at index i with j deletions.
    • maxSum: This variable keeps track of the overall maximum sum.
  2. Iterate through the Array:
    • For each index i, we find the maximum sum for 0 deletions, 1 deletion, and 2 deletions.
  3. Update the DP Table:
    • For dp[i][0]: We take the maximum of the previous value or the current value plus the previous maximum sum.
    • For dp[i][1]: We take the maximum of the previous deletion case or the sum of dp[i-1][0] plus the current value.
    • For dp[i][2]: It is similar to dp[i][1], but we look at the case of two deletions.

C++ Implementation:

#include <vector>
#include <algorithm>
#include <iostream>

using namespace std;

int maxSumAfterDeletions(vector<int>& arr) {
    int n = arr.size();
    if (n == 0) return 0;

    vector<vector<int>> dp(n, vector<int>(3, 0));
    dp[0][0] = arr[0]; // No deletions
    dp[0][1] = INT_MIN; // No valid value with one deletion at index 0
    dp[0][2] = INT_MIN; // No valid value with two deletions at index 0

    int maxSum = arr[0];

    for (int i = 1; i < n; i++) {
        dp[i][0] = max(dp[i-1][0] + arr[i], arr[i]); // Max sum with 0 deletions
        dp[i][1] = max(dp[i-1][1] + arr[i], dp[i-1][0]); // Max sum with 1 deletion
        dp[i][2] = max(dp[i-1][2] + arr[i], dp[i-1][1]); // Max sum with 2 deletions

        maxSum = max({maxSum, dp[i][0], dp[i][1], dp[i][2]});
    }

    return maxSum;
}

int main() {
    vector<int> arr = {1, -2, 0, 3};
    cout << "Maximum Sum of Subarray with Two Deletions: " << maxSumAfterDeletions(arr) << endl;
    return 0;
}

Explanation of Code:

  • Input: The function takes a vector arr as input.
  • Output: It gives back the maximum sum of a subarray with at most two deletions.
  • Main Logic: The loop goes through the array and updates the dp values for each deletion case. We update the maximum sum after each step.

This way gives us a good solution with a time complexity of O(n) and space complexity of O(n). This makes it work well for bigger input sizes. By using dynamic programming, we break the problem down and make our calculations better.

For more reading on similar dynamic programming methods, you can check Dynamic Programming: Maximum Sum of Subarray with One Deletion.

Optimizing Space Complexity in Maximum Sum of Subarray with Two Deletions

We want to make space better when we find the maximum sum of a subarray with at most two deletions. We can use a simple dynamic programming method. This method helps us avoid using too many extra arrays. We only keep the important previous states.

Approach

  1. State Definition:
    • We define dp[i][j] as the maximum sum of a subarray using the first i elements with j deletions. Here, j can be 0, 1, or 2.
  2. Transition Relation:
    • If we do not delete any element at index i, we can either add the current element to our subarray or start a new subarray with it:
      • dp[i][0] = max(dp[i-1][0] + nums[i], nums[i]) (no deletions)
      • dp[i][1] = max(dp[i-1][1] + nums[i], dp[i-1][0]) (one deletion)
      • dp[i][2] = max(dp[i-1][2] + nums[i], dp[i-1][1]) (two deletions)
  3. Space Optimization:
    • Instead of using a 2D array, we can use a 1D array of size 3. Each index in this array shows the number of deletions. We only need values from the last round to find the current values.

Implementation

Here is a Java code for the optimized way:

public class MaximumSumSubarray {
    public static int maxSumWithTwoDeletions(int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;

        int[] dp = new int[3]; // dp[0], dp[1], dp[2]
        int[] temp = new int[3];

        for (int i = 0; i < n; i++) {
            temp[0] = Math.max(dp[0] + nums[i], nums[i]);
            temp[1] = Math.max(dp[1] + nums[i], dp[0]);
            temp[2] = Math.max(dp[2] + nums[i], dp[1]);
            System.arraycopy(temp, 0, dp, 0, 3);
        }
        return Math.max(dp[0], Math.max(dp[1], dp[2]));
    }
}

Python Implementation

Here is a Python version of the optimized solution:

def max_sum_with_two_deletions(nums):
    n = len(nums)
    if n == 0:
        return 0

    dp = [0] * 3  # dp[0], dp[1], dp[2]

    for i in range(n):
        temp = [0] * 3
        temp[0] = max(dp[0] + nums[i], nums[i])
        temp[1] = max(dp[1] + nums[i], dp[0])
        temp[2] = max(dp[2] + nums[i], dp[1])
        dp = temp

    return max(dp)

# Example usage
print(max_sum_with_two_deletions([-1, 2, -2, 3, -4, 5]))  # Output the maximum sum

C++ Implementation

Below is the C++ version of the optimized way:

#include <vector>
#include <algorithm>
using namespace std;

int maxSumWithTwoDeletions(vector<int>& nums) {
    int n = nums.size();
    if (n == 0) return 0;

    vector<int> dp(3, 0);

    for (int i = 0; i < n; i++) {
        vector<int> temp(3, 0);
        temp[0] = max(dp[0] + nums[i], nums[i]);
        temp[1] = max(dp[1] + nums[i], dp[0]);
        temp[2] = max(dp[2] + nums[i], dp[1]);
        dp = temp;
    }
    return max({dp[0], dp[1], dp[2]});
}

Conclusion

By using these methods, we can lower the space needed for the dynamic programming solution for the Maximum Sum of Subarray with At Most Two Deletions problem. This way, we keep a linear time complexity and use only a small space. This makes it good for larger inputs. For more reading, you can check similar dynamic programming methods like Maximum Sum of Subarray with One Deletion or Maximum Sum of Subarray with K Deletions.

Alternative Sliding Window Technique for Maximum Sum of Subarray with Two Deletions

We can use the alternative sliding window technique to find the maximum sum of a subarray with at most two deletions. This method keeps a window that changes based on the number of deletions we can make and the sum inside that window.

Approach

  1. Initial Setup: We need to create variables to track the maximum sum, current sum, and the number of deletions we used.

  2. Iterate through the Array: We use two pointers to grow the window by adding elements to the current sum. We keep doing this until we have more than two deletions. If we do, we shrink the window from the left side until we have the allowed number of deletions.

  3. Update Maximum Sum: In each step, we update the maximum sum we find.

Pseudocode

def maxSumTwoDeletions(arr):
    n = len(arr)
    if n == 0:
        return 0
    
    max_sum = float('-inf')
    current_sum = 0
    left = 0
    deletions = 0
    
    for right in range(n):
        current_sum += arr[right]
        
        while deletions > 2:
            current_sum -= arr[left]
            left += 1
        
        max_sum = max(max_sum, current_sum)
        
        if right - left + 1 >= 3:
            deletions += 1
    
    return max_sum

Implementation in Different Languages

Java

public int maxSumTwoDeletions(int[] arr) {
    int n = arr.length;
    if (n == 0) return 0;

    int maxSum = Integer.MIN_VALUE;
    int currentSum = 0;
    int left = 0;
    int deletions = 0;

    for (int right = 0; right < n; right++) {
        currentSum += arr[right];

        while (deletions > 2) {
            currentSum -= arr[left];
            left++;
        }

        maxSum = Math.max(maxSum, currentSum);

        if (right - left + 1 >= 3) {
            deletions++;
        }
    }

    return maxSum;
}

C++

int maxSumTwoDeletions(vector<int>& arr) {
    int n = arr.size();
    if (n == 0) return 0;

    int max_sum = INT_MIN;
    int current_sum = 0;
    int left = 0;
    int deletions = 0;

    for (int right = 0; right < n; right++) {
        current_sum += arr[right];

        while (deletions > 2) {
            current_sum -= arr[left];
            left++;
        }

        max_sum = max(max_sum, current_sum);

        if (right - left + 1 >= 3) {
            deletions++;
        }
    }

    return max_sum;
}

Considerations

  • Time Complexity: O(n). Here n is the length of the array. Each element is checked at most two times.
  • Space Complexity: O(1). We only use a fixed amount of space for variables.
  • Edge Cases: We should check cases where the array size is less than three. If there are not enough elements, we cannot make two deletions.

This sliding window method gives a better solution than regular dynamic programming methods. It is especially good because it uses less space while still keeping the time complexity low for this problem.

Handling Edge Cases in Maximum Sum of Subarray with Two Deletions

When we implement the solution for “Maximum Sum of Subarray with At Most Two Deletions,” we need to think about different edge cases. This helps make the algorithm strong. Here are some common edge cases and how we can handle them:

  1. Empty Array:
    • If the input array is empty, we set the result to 0. There are no elements to add up.
    if (nums.length == 0) return 0;
  2. Array with One Element:
    • If the array has only one element, the maximum sum is that element. It does not matter if we delete it or not.
    if (nums.length == 1) return nums[0];
  3. All Negative Numbers:
    • If all numbers are negative, the maximum sum can be a single number or a mix of numbers that gives the least negative value. Deleting some numbers may not help.
    int maxSum = Integer.MIN_VALUE;
for (int num : nums) {
    maxSum = Math.max(maxSum, num);
}
  4. Less than Two Deletions:
    • If the number of deletions we can do is greater than or equal to the length of the array, the result should be 0. We can delete all elements.
    if (k >= nums.length) return 0;
  5. Handling Maximum Subarray with Deletions:
    • When we find maximum sums with deletions, we must keep a running sum. We also need to think about how each deletion affects the maximum sum.
    • We should use dynamic programming to track the maximum sum at each index while allowing deletions.
  6. Multiple Identical Maximums:
    • If the maximum sum shows up in many subarrays because of the same values, we should return the first one or the correct maximum sum based on what the problem asks.
  7. Deletions Near Array Edges:
    • We must make sure our algorithm finds sums correctly when we delete from the start or end of the array. This can change the sum a lot.

Here is an example code snippet in Java that handles these edge cases:

public int maxSumWithTwoDeletions(int[] nums) {
    if (nums.length == 0) return 0;
    if (nums.length == 1) return nums[0];
    
    int n = nums.length;
    int[][] dp = new int[n][3]; // dp[i][j] keeps the max sum using j deletions up to index i
    
    for (int j = 0; j < 3; j++) {
        dp[0][j] = nums[0]; // Start with first element
    }
    
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < 3; j++) {
            dp[i][j] = Math.max(dp[i - 1][j] + nums[i], dp[i - 1][j - 1]);
            if (j > 0) {
                dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - 1]); // Think about deletion
            }
        }
    }
    
    return Math.max(dp[n - 1][0], Math.max(dp[n - 1][1], dp[n - 1][2]));
}

This code handles edge cases well. It makes sure our solution is strong for different input situations while finding the maximum sum of a subarray with at most two deletions.

Performance Analysis of Different Approaches for Maximum Sum of Subarray with Two Deletions

When we look at how different methods perform for finding the maximum sum of a subarray with at most two deletions, we must think about time and space complexity.

Dynamic Programming Approach

The dynamic programming method usually gives us a time complexity of (O(n)) and a space complexity of (O(n)), where (n) is the length of the input array. In this method, we keep a DP array to store maximum sums calculated for each index while we think about the allowed deletions.

Java Implementation:

public int maxSumAfterDeletions(int[] nums) {
    int n = nums.length;
    int[] dp = new int[n + 1];
    int maxSum = Integer.MIN_VALUE;

    for (int i = 1; i <= n; i++) {
        dp[i] = dp[i - 1] + nums[i - 1];
        maxSum = Math.max(maxSum, dp[i]);
    }

    for (int i = 2; i <= n; i++) {
        dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i - 1]);
        maxSum = Math.max(maxSum, dp[i]);
    }

    return maxSum;
}

Python Implementation:

def max_sum_after_deletions(nums):
    n = len(nums)
    dp = [0] * (n + 1)
    max_sum = float('-inf')

    for i in range(1, n + 1):
        dp[i] = dp[i - 1] + nums[i - 1]
        max_sum = max(max_sum, dp[i])

    for i in range(2, n + 1):
        dp[i] = max(dp[i - 1], dp[i - 2] + nums[i - 1])
        max_sum = max(max_sum, dp[i])

    return max_sum

C++ Implementation:

int maxSumAfterDeletions(vector<int>& nums) {
    int n = nums.size();
    vector<int> dp(n + 1, 0);
    int maxSum = INT_MIN;

    for (int i = 1; i <= n; i++) {
        dp[i] = dp[i - 1] + nums[i - 1];
        maxSum = max(maxSum, dp[i]);
    }

    for (int i = 2; i <= n; i++) {
        dp[i] = max(dp[i - 1], dp[i - 2] + nums[i - 1]);
        maxSum = max(maxSum, dp[i]);
    }

    return maxSum;
}

Space Optimization

We can reduce the space complexity to (O(1)) by tracking only the needed variables instead of keeping the whole DP array. This change is very helpful for larger arrays. It makes the algorithm use less memory.

Sliding Window Technique

We also have another method that uses a sliding window. This method can also reach good performance. It works well when the array has a special structure. This technique keeps a window of valid elements while we calculate the maximum sum. It gives us a time complexity of (O(n)) and a space complexity of (O(1)).

Performance Summary

  • Dynamic Programming: (O(n)) time, (O(n)) space (can be improved to (O(1)).
  • Sliding Window: (O(n)) time, (O(1)) space.
  • The choice between these methods often depends on the details of the problem, like the size of the input array and the need for saving space.

For more reading on related dynamic programming problems, we can check articles on Dynamic Programming - Maximum Subarray (Kadane’s Algorithm) or Dynamic Programming - Maximum Sum of Subarray with One Deletion.

Frequently Asked Questions

1. What is the Maximum Sum of Subarray with at Most Two Deletions problem?

The Maximum Sum of Subarray with at Most Two Deletions problem is about finding the biggest sum of a contiguous subarray from a list of integers. You can delete up to two elements in this case. We can use dynamic programming to solve this. It helps us to use earlier calculations to quickly find the sums of possible subarrays while thinking about the deletions.

2. How can I implement the Maximum Sum of Subarray with Two Deletions in Java?

To implement the Maximum Sum of Subarray with Two Deletions in Java, we can use dynamic programming. We keep an array to track the maximum sums while we go through the input array. We calculate the maximum sums for zero, one, and two deletions separately. Then we update the maximum result as we go. This way, we can get the result in O(n) time.

3. Is there a Python solution for the Maximum Sum of Subarray with Two Deletions?

Yes, we can implement the Maximum Sum of Subarray with Two Deletions in Python using a similar dynamic programming approach. We keep arrays to store the maximum sums with and without deletions. This helps us to find the result while going through the array. This solution is clean and works well for Python developers.

4. What is the space complexity of the Maximum Sum of Subarray with Two Deletions solution?

The space complexity for the Maximum Sum of Subarray with Two Deletions can usually be improved to O(1). We can just keep a few variables to track the maximum sums instead of using more arrays. This way, we save memory and still get good results.

5. How does the sliding window technique apply to the Maximum Sum of Subarray with Two Deletions?

We can use the sliding window technique for the Maximum Sum of Subarray with Two Deletions. We adjust the window to include valid subarrays and make changes for deletions when needed. This method helps us find maximum sums by focusing on continuous parts of the array. It can also lower the total time we need for calculations.

For more reading on related dynamic programming ideas, check out Dynamic Programming: Maximum Subarray (Kadane’s Algorithm). This will help you understand better this advanced problem.