[Dynamic Programming] Maximum Subarray Sum with One Deletion - Medium

The problem of finding the biggest sum of a subarray with one deletion is a well-known challenge in dynamic programming. Our goal is to find the highest sum of a continuous subarray from a list of integers. We can delete one element to try to get a bigger sum. This method helps us to think about cases where removing one element gives us a better total than just looking for the biggest sum of a subarray without deletions.

In this article, we will look closely at the maximum subarray sum with one deletion. We will start by explaining the problem. Then we will explore a dynamic programming way to solve it in a smart way. We will also show how to implement this in Java, Python, and C++. We will talk about how to save space and check out different methods. Finally, we will look at common edge cases and answer some questions that people often ask about this topic.

  • [Dynamic Programming] Maximum Subarray Sum with One Deletion - Medium Solution Overview
  • Understanding the Problem Statement for Maximum Subarray Sum with One Deletion
  • Dynamic Programming Approach for Maximum Subarray Sum with One Deletion
  • Java Implementation of Maximum Subarray Sum with One Deletion
  • Python Code Solution for Maximum Subarray Sum with One Deletion
  • C++ Implementation for Maximum Subarray Sum with One Deletion
  • Optimizing Space Complexity in Maximum Subarray Sum with One Deletion
  • Comparative Analysis of Different Approaches for Maximum Subarray Sum with One Deletion
  • Testing and Edge Cases for Maximum Subarray Sum with One Deletion
  • Frequently Asked Questions

Understanding the Problem Statement for Maximum Subarray Sum with One Deletion

The problem of finding the Maximum Subarray Sum with One Deletion is a twist on the classic maximum subarray problem. We can use Kadane’s algorithm to solve it. In this case, we can delete one element from the array to make the sum of the other elements bigger.

Problem Definition

We have an integer array called arr. Our job is to find the biggest possible sum of a group of numbers in the array after we delete at most one element.

Constraints

  • The array can have negative numbers.
  • The length of the array can be from 1 to (10^5).

Example

Let’s look at the input array:

arr = [1, -2, 0, 3]
  • The maximum subarray sum without deletion is 4 (from the subarray [1, -2, 0, 3]).
  • If we delete -2, the maximum sum can be 4 (from the subarray [1, 0, 3]).

Output

The function should give back the maximum sum after we think about the possible deletion of one element.

Key Points

  • Kadane’s Algorithm: This is a common way to find the maximum subarray sum quickly.
  • Deletion Impact: We need to think about how deleting one element can help increase the subarray sum in some cases.

We need to understand these parts well to make a good solution using dynamic programming techniques.

Dynamic Programming Approach for Maximum Subarray Sum with One Deletion

We can solve the problem of finding the maximum subarray sum with one deletion using a dynamic programming method. The main idea is to keep track of two arrays or variables. One tracks the maximum subarrays without deletion. The other tracks the maximum subarrays with one deletion.

Problem Breakdown

  1. State Definition:
    • We say max_no_deletion[i] is the maximum subarray sum ending at index i without deletion.
    • We say max_with_deletion[i] is the maximum subarray sum ending at index i with one deletion.
  2. Recurrence Relations:

    • For max_no_deletion[i], we use this formula:

      max_no_deletion[i] = max(nums[i], max_no_deletion[i-1] + nums[i])

    • For max_with_deletion[i], we use this formula:

      max_with_deletion[i] = max(max_with_deletion[i-1] + nums[i], max_no_deletion[i-1])
  3. Base Cases:

    • We start with:

      max_no_deletion[0] = nums[0]
max_with_deletion[0] = -inf (or a very small number)
  4. Final Result:
    • The final answer is the maximum value between max_no_deletion[n-1] and max_with_deletion[n-1].

Implementation

Here is a simple implementation in Python:

def maximumSum(arr):
    n = len(arr)
    if n == 0:
        return 0
    
    max_no_deletion = arr[0]
    max_with_deletion = float('-inf')
    max_sum = arr[0]

    for i in range(1, n):
        max_no_deletion = max(arr[i], max_no_deletion + arr[i])
        max_with_deletion = max(max_with_deletion + arr[i], max_no_deletion - arr[i])
        max_sum = max(max_sum, max_no_deletion, max_with_deletion)

    return max_sum

Explanation of the Code

  • The function maximumSum goes through the array and updates the maximum sums based on our states.
  • It checks both maximum sums with and without deletion at each step.
  • The overall maximum updates as needed.

This dynamic programming method runs in O(n) time and O(1) space. It is good for large datasets.

If you want to learn more about dynamic programming, you can check the article on Dynamic Programming: Maximum Subarray (Kadane’s Algorithm).

Java Implementation of Maximum Subarray Sum with One Deletion

We will solve the problem of finding the maximum subarray sum with one deletion using Java. We can use a method called dynamic programming. The main idea is to have two arrays. One array will keep track of the maximum subarray sum ending at each index without any deletion. The other array will keep track of the maximum subarray sum ending at each index with one deletion.

Java Code

public class MaximumSubarraySumWithOneDeletion {
    public static int maximumSum(int[] arr) {
        int n = arr.length;
        if (n == 0) return 0;

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

        dp[0] = arr[0];
        dpDelete[0] = 0; // No deletion at the start

        int maxSum = arr[0];

        for (int i = 1; i < n; i++) {
            dp[i] = Math.max(arr[i], dp[i - 1] + arr[i]); // Maximum sum ending at i without deletion
            dpDelete[i] = Math.max(dp[i - 1], dpDelete[i - 1] + arr[i]); // Maximum sum ending at i with one deletion
            
            maxSum = Math.max(maxSum, Math.max(dp[i], dpDelete[i]));
        }

        return maxSum;
    }

    public static void main(String[] args) {
        int[] arr = {1, -2, 0, 3};
        System.out.println("Maximum Subarray Sum with One Deletion: " + maximumSum(arr)); // Output: 4
    }
}

Explanation of the Code

  • Initialization: We make two arrays dp and dpDelete to save the maximum sums.
  • Dynamic Programming Transition:
    • We update dp[i] to be the bigger value between the current element arr[i] or the sum of the last maximum dp[i - 1] + arr[i].
    • We update dpDelete[i] to take the bigger value between the last maximum without deletion dp[i - 1] or the sum of the last maximum with deletion dpDelete[i - 1] + arr[i].
  • Final Result: We find the maximum sum by taking the biggest value between dp[i] and dpDelete[i] for all indices.

This code runs in O(n) time. It also uses O(n) space. To make it better with space, we can do some more improvements.

Python Code Solution for Maximum Subarray Sum with One Deletion

We can solve the problem of finding the maximum subarray sum with one deletion using Python. We use a simple approach called dynamic programming. The main idea is to keep track of two things as we go through the array:

  1. max_with_deletion: This is the highest sum we can get by removing one element.
  2. max_without_deletion: This is the highest sum we can get without removing any elements.

We define these two states with the following formulas:

  • max_without_deletion[i] = max(max_without_deletion[i-1] + arr[i], arr[i])
  • max_with_deletion[i] = max(max_with_deletion[i-1] + arr[i], max_without_deletion[i-1])

At the end of our process, we will take the highest value from max_with_deletion and max_without_deletion for the whole array.

Here is the Python code for this solution:

def maximumSum(arr):
    if not arr:
        return 0
    
    max_without_deletion = arr[0]
    max_with_deletion = 0
    max_result = arr[0]
    
    for i in range(1, len(arr)):
        max_without_deletion = max(max_without_deletion + arr[i], arr[i])
        max_with_deletion = max(max_with_deletion + arr[i], max_without_deletion)
        max_result = max(max_result, max_without_deletion, max_with_deletion)
    
    return max_result

# Example usage
arr = [1, -2, 0, 3]
print(maximumSum(arr))  # Output: 4

In this code, we go through the array one time. We keep track of the two states which gives us a time complexity of (O(n)) and a space complexity of (O(1)). This means we only use a little extra space.

This dynamic programming solution helps us find the maximum subarray sum with one deletion in a good way. If you want to learn more about similar problems in dynamic programming, you can check Dynamic Programming: Maximum Subarray - Kadane’s Algorithm.

C++ Implementation for Maximum Subarray Sum with One Deletion

To find the maximum subarray sum with one deletion in C++, we can use a simple dynamic programming method. We will keep two arrays. One array will hold the maximum subarray sum that ends at the current index without deletion. The other array will hold the maximum subarray sum with one deletion.

Here is the C++ code for this:

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

using namespace std;

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

    vector<int> dp(n);
    vector<int> dpDelete(n);

    dp[0] = arr[0];
    dpDelete[0] = 0; // No deletion at the first index

    int maxSum = arr[0];

    for (int i = 1; i < n; i++) {
        dp[i] = max(arr[i], dp[i - 1] + arr[i]); // Max sum without deletion
        dpDelete[i] = max(dpDelete[i - 1] + arr[i], dp[i - 1]); // Max sum with one deletion
        maxSum = max(maxSum, max(dp[i], dpDelete[i])); // Update overall max sum
    }

    return maxSum;
}

int main() {
    vector<int> arr = {1, -2, 0, 3};
    cout << "Maximum Subarray Sum with One Deletion: " << maximumSum(arr) << endl;
    return 0;
}

Explanation of the Code:

  • Input: We have a vector of integers called arr which is our array.
  • Dynamic Programming Arrays:
    • dp[i]: This stores the maximum subarray sum ending at index i without deleting any element.
    • dpDelete[i]: This stores the maximum subarray sum ending at index i with one deletion.
  • Logic:
    • For every element, we calculate the maximum sum with and without deletion.
    • Then, we update the overall maximum sum by looking at both cases.
  • Output: We print the maximum subarray sum with one deletion.

This code runs fast and gives the maximum sum with a time of (O(n)) and uses (O(n)) space. We can make it even better by using just a few variables instead of arrays, so it uses (O(1)) space.

If you want to learn more about dynamic programming in C++, you can check the article on Dynamic Programming: Maximum Subarray (Kadane’s Algorithm).

Optimizing Space Complexity in Maximum Subarray Sum with One Deletion

In the problem of finding the maximum subarray sum with one deletion, we can make our space use better from O(n) to O(1). We do this by using a few simple variables instead of a whole DP array. The main idea is to keep track of the current maximum sums while we go through the array. We do not need to store all previous sums.

Approach

  1. Variables Used:
    • max_sum: This keeps the maximum subarray sum without deletion.
    • max_sum_with_deletion: This keeps the maximum subarray sum with one deletion.
    • current_max: This tracks the sum of the current subarray as we check the array.
  2. Iterate Through the Array:
    • For each element, we update current_max. It can be either the current element or the sum of current_max and the current element.
    • We update max_sum with the bigger value between max_sum and current_max.
    • We update max_sum_with_deletion to be the maximum of itself or max_sum before the current loop.

Implementation

Here is a Java code that shows the space optimization:

public class MaximumSubarraySumWithOneDeletion {
    public int maximumSum(int[] arr) {
        int max_sum = arr[0];
        int max_sum_with_deletion = 0;
        int current_max = arr[0];

        for (int i = 1; i < arr.length; i++) {
            current_max = Math.max(arr[i], current_max + arr[i]);
            max_sum = Math.max(max_sum, current_max);
            max_sum_with_deletion = Math.max(max_sum_with_deletion + arr[i], max_sum);
        }

        return Math.max(max_sum, max_sum_with_deletion);
    }
}

Python Implementation

Here is how we can do the same in Python:

def maximum_sum(arr):
    max_sum = arr[0]
    max_sum_with_deletion = 0
    current_max = arr[0]

    for i in range(1, len(arr)):
        current_max = max(arr[i], current_max + arr[i])
        max_sum = max(max_sum, current_max)
        max_sum_with_deletion = max(max_sum_with_deletion + arr[i], max_sum)

    return max(max_sum, max_sum_with_deletion)

# Example usage
arr = [1, -2, 0, 3]
print(maximum_sum(arr))  # Output: 4

C++ Implementation

Here is a C++ version of the same method:

class Solution {
public:
    int maximumSum(vector<int>& arr) {
        int max_sum = arr[0];
        int max_sum_with_deletion = 0;
        int current_max = arr[0];

        for (int i = 1; i < arr.size(); i++) {
            current_max = max(arr[i], current_max + arr[i]);
            max_sum = max(max_sum, current_max);
            max_sum_with_deletion = max(max_sum_with_deletion + arr[i], max_sum);
        }

        return max(max_sum, max_sum_with_deletion);
    }
};

This space-optimized way runs in O(n) time and O(1) space. This makes it good for big input arrays.

Comparative Analysis of Different Approaches for Maximum Subarray Sum with One Deletion

When we try to find the maximum subarray sum with one deletion, we can look at different ways to solve it. Each way has its good and not-so-good points. Let us look at these methods:

  1. Dynamic Programming Approach:

    • This method is the fastest one. It works in O(n) time and uses O(1) space.

    • We keep track of two arrays:

      • max_ending_here: This shows the maximum sum of subarrays ending at the current index.
      • max_ending_here_with_deletion: This shows the maximum sum of subarrays ending at the current index with one deletion.

    • The steps are:

      for i in range(n):
    max_ending_here = max(nums[i], max_ending_here + nums[i])
    max_ending_here_with_deletion = max(max_ending_here_with_deletion + nums[i], max_ending_here)
    max_sum = max(max_sum, max_ending_here, max_ending_here_with_deletion)
  2. Kadane’s Algorithm Variant:
    • We can change Kadane’s algorithm a bit. We treat deletion as a special case.
    • This runs in O(n) time, just like the normal Kadane’s algorithm for maximum subarray sum.
    • We use the same variables as in the dynamic programming approach.
    • This way is good when we do not want to use more space.
  3. Brute Force Approach:
    • Here, we check every possible subarray and calculate the sum after deleting one element.
    • The time complexity is O(n^2) because for each element, we may have to go through the array again to find the maximum sum.
    • This method is easy to understand and use. But it is not good for big arrays.
  4. Divide and Conquer:
    • This method is more complex. We divide the array into halves and calculate the maximum subarray sum with one deletion in each half.
    • This can give us O(n log n) time complexity. But it is usually not as good as the dynamic programming approach for this problem.
  5. Space Optimization:
    • We can improve the dynamic programming approach to use O(1) space. We do this by keeping track of only the needed variables. Here is a Java example:
    public int maxSumAfterOneDeletion(int[] arr) {
    int maxEndingHere = arr[0], maxEndingHereWithDeletion = 0, maxSum = arr[0];
    for (int i = 1; i < arr.length; i++) {
        maxEndingHere = Math.max(arr[i], maxEndingHere + arr[i]);
        maxEndingHereWithDeletion = Math.max(maxEndingHereWithDeletion + arr[i], maxEndingHere);
        maxSum = Math.max(maxSum, Math.max(maxEndingHere, maxEndingHereWithDeletion));
    }
    return maxSum;
}

Each method has its own use cases. But for finding the maximum subarray sum with one deletion, the dynamic programming solution is the best. It is the fastest and uses less space, especially for big datasets. If you want to read more about similar dynamic programming problems, you can check the articles on Dynamic Programming with Kadane’s Algorithm.

Testing and Edge Cases for Maximum Subarray Sum with One Deletion

When we implement the Maximum Subarray Sum with One Deletion solution, we need to think about different test cases. This helps us make sure the algorithm works in all situations. Here are some edge cases and what we expect from them:

  1. All Positive Numbers:
    • Input: [1, 2, 3, 4, 5]
    • Expected Output: 15
    • Explanation: If we delete any number, the sum will be smaller.
  2. All Negative Numbers:
    • Input: [-1, -2, -3, -4, -5]
    • Expected Output: -1
    • Explanation: The biggest sum we can get after deletion is the least negative number.
  3. Single Element Array:
    • Input: [10]
    • Expected Output: 0
    • Explanation: If we delete the only number, the sum is zero.
  4. Two Elements, One Positive and One Negative:
    • Input: [1, -2]
    • Expected Output: 1
    • Explanation: If we delete -2, we get the biggest sum.
  5. Zeroes in the Array:
    • Input: [0, -1, -2, 0, -3]
    • Expected Output: 0
    • Explanation: If we delete -3 or -1, we still get a max sum of zero.
  6. Large Array with Mixed Values:
    • Input: [-1, 2, -2, 3, -4, 4]
    • Expected Output: 6
    • Explanation: If we delete -4, the max sum becomes 2 + (-2) + 3 + 4.
  7. Array with Repeated Maximum Values:
    • Input: [1, 2, 3, 4, 5, 5]
    • Expected Output: 20
    • Explanation: Deleting one 5 gives the same max sum.
  8. Alternating Positive and Negative Values:
    • Input: [2, -1, 3, -2, 5, -3]
    • Expected Output: 9
    • Explanation: If we delete -3, we get the max subarray sum of 2 + (-1) + 3 + (-2) + 5.
  9. Edge Case with Large Numbers:
    • Input: [100000, -100000, 100000, -100000]
    • Expected Output: 100000
    • Explanation: Deleting the first or second -100000 keeps the max sum the same.

By testing these edge cases, we can check that our Maximum Subarray Sum with One Deletion algorithm works well in many situations. We should use dynamic programming to keep good performance while handling these cases.

Here is a sample implementation to test these cases:

def maxSumAfterOneDeletion(arr):
    n = len(arr)
    if n == 1:
        return 0
    max_sum = [0] * n
    max_sum[0] = arr[0]
    max_sum[1] = max(arr[1], arr[1] + arr[0])
    
    for i in range(2, n):
        max_sum[i] = max(max_sum[i - 1] + arr[i], arr[i], max_sum[i - 1])
    
    max_with_deletion = [0] * n
    max_with_deletion[1] = max_sum[0]
    
    for i in range(2, n):
        max_with_deletion[i] = max(max_with_deletion[i - 1], max_sum[i - 1])
    
    return max(max_with_deletion[n - 1], max_sum[n - 1])

# Test Cases
test_cases = [
    [1, 2, 3, 4, 5],
    [-1, -2, -3, -4, -5],
    [10],
    [1, -2],
    [0, -1, -2, 0, -3],
    [-1, 2, -2, 3, -4, 4],
    [1, 2, 3, 4, 5, 5],
    [2, -1, 3, -2, 5, -3],
    [100000, -100000, 100000, -100000]
]

for case in test_cases:
    print(f"Input: {case}, Output: {maxSumAfterOneDeletion(case)}")

This implementation and the test cases will help us check if the algorithm is correct for different edge cases.

Frequently Asked Questions

1. What is the Maximum Subarray Sum with One Deletion problem?

The Maximum Subarray Sum with One Deletion problem is about finding the highest sum of a continuous part of an integer array. You can delete one element if you want. This task needs us to know about dynamic programming. It helps us find the best solution without checking every possible subarray.

2. How can dynamic programming be applied to solve this problem?

We can use dynamic programming for the Maximum Subarray Sum with One Deletion problem. We keep track of two things: the highest sum without deletion and the highest sum with one deletion. As we go through the array, we update these two sums based on the current number. This way, we can find the final highest sum quickly.

3. What is the time complexity of the dynamic programming solution for this problem?

The time complexity for this dynamic programming solution is O(n). Here, n is the number of elements in the input array. This means we can solve the problem in one go through the array. We also use a small amount of extra space for the needed variables.

4. Can you provide a Java implementation for the Maximum Subarray Sum with One Deletion problem?

Sure! Here is a simple Java implementation for the Maximum Subarray Sum with One Deletion problem:

public class MaxSubarraySumOneDeletion {
    public int maximumSum(int[] arr) {
        int n = arr.length;
        int maxSum = arr[0];
        int maxEndHere = arr[0];
        int maxEndHereWithDeletion = 0;

        for (int i = 1; i < n; i++) {
            maxEndHereWithDeletion = Math.max(maxEndHere, maxEndHereWithDeletion + arr[i]);
            maxEndHere = Math.max(arr[i], maxEndHere + arr[i]);
            maxSum = Math.max(maxSum, Math.max(maxEndHere, maxEndHereWithDeletion));
        }
        
        return maxSum;
    }
}

This code calculates the maximum subarray sum while allowing one deletion.

5. What are common edge cases to consider when solving this problem?

When we solve the Maximum Subarray Sum with One Deletion problem, we should think about edge cases. These cases include arrays with only one element, arrays with all negative numbers, and arrays where the best option means deleting a high-value element. Testing these cases helps us make sure our solution works well in different situations. If you want to learn more about dynamic programming, you can check topics like Dynamic Programming: Maximum Subarray (Kadane’s Algorithm).