[Dynamic Programming] Maximum Sum of Subarray with At Most k Deletions - Medium

Dynamic Programming is a strong method we use to solve optimization problems. We break these problems into simpler parts. One such problem is finding the maximum sum of a subarray with at most k deletions. This problem asks us to think about different ways to delete elements from the array. Our goal is to maximize the sum of what remains. By using a dynamic programming approach, we can find the best solution quickly. We do this without having to redo calculations for parts that overlap. This makes our work better and faster.

In this article, we will look closely at the Maximum Sum of Subarray with At Most k Deletions problem. We will explain the problem statement and its optimal substructure. We will show dynamic programming solutions in Java, Python, and C++. We will also talk about ways to save space, analyze time complexity, and point out common mistakes we should avoid when we solve dynamic programming problems. Here is a list of the topics we will talk about:

  • Dynamic Programming Approach to Maximum Sum of Subarray with At Most k Deletions - Medium
  • Understanding the Problem Statement for Maximum Sum of Subarray
  • Optimal Substructure and Overlapping Subproblems in Dynamic Programming
  • Dynamic Programming Solution in Java for Maximum Sum of Subarray
  • Dynamic Programming Solution in Python for Maximum Sum of Subarray
  • Dynamic Programming Solution in C++ for Maximum Sum of Subarray
  • Space Optimization Techniques in Dynamic Programming
  • Analyzing Time Complexity of the Solution
  • Common Mistakes to Avoid in Dynamic Programming Problems
  • Frequently Asked Questions

Understanding the Problem Statement for Maximum Sum of Subarray

We have a problem to find the maximum sum of a subarray. We can delete at most k elements to help us get the best sum. This problem is an extension of the classic maximum subarray problem. We usually solve that with Kadane’s algorithm. But here, we also get the option to delete up to k items from the array.

Problem Definition:

We have an integer array nums and an integer k. Our goal is to find the maximum sum of a subarray after we delete at most k elements.

Constraints:

  • The length of nums can be up to n (1 ≤ n ≤ 10^5).
  • Each number in nums can be from -10^4 to 10^4.
  • We can delete up to k elements and k is not negative.

Example:

For nums = [1, -2, 0, 3] and k = 1: - The maximum sum of a subarray after deleting one element is 4 (we delete -2, so the sum is 1 + 0 + 3).

Approach:

We can use dynamic programming to solve this problem in a good way. We keep a DP array where dp[i][j] means the maximum sum of subarrays that end at index i with exactly j deletions. We can move between states by thinking about including or excluding elements based on how many deletions we can make.

The transition can be explained like this: - If we do not delete the current element: dp[i][j] = max(dp[i-1][j] + nums[i], nums[i]) - If we delete the current element: dp[i][j] = dp[i-1][j-1] (only if j > 0)

In the end, we will find the maximum value in the last row of the DP table. We check all possible deletions from 0 to k.

This problem helps us understand dynamic programming better. It also tests how fast the solution can be due to all the limits we have.

For more on similar dynamic programming methods, we can look at Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

Optimal Substructure and Overlapping Subproblems in Dynamic Programming

In dynamic programming, we need two main ideas to solve problems well: optimal substructure and overlapping subproblems. Knowing these ideas helps us when we want to find the maximum sum of a subarray with at most k deletions.

Optimal Substructure

A problem shows optimal substructure when we can make a good solution from good solutions of smaller problems. For our problem about the maximum sum of a subarray with at most k deletions, this is how it works:

  • If dp[i][j] means the maximum sum of subarrays that end at index i with at most j deletions, we can write the relation like this:

    [ dp[i][j] = (dp[i-1][j], dp[i-1][j-1] + nums[i]) ]

    Here, dp[i-1][j] looks at the case where we do not delete the element at index i. And dp[i-1][j-1] + nums[i] is for the case where we include it after deleting one element.

Overlapping Subproblems

Overlapping subproblems happen when we solve the same smaller problems many times while solving bigger problems. In our case:

  • We can calculate dp[i][j] using results we already found, like dp[i-1][j] and dp[i-1][j-1]. So each dp[i][j] depends on earlier results. This leads to us calculating the same things again unless we keep them stored.

Example

Let’s look at a simple example to understand better:

Given the array nums = [1, -2, 0, 3] and k = 1, we can do these calculations:

  • For dp[0][0], the value is 1 (there is only one element).
  • For dp[1][0], the maximum sum without any deletion is 1.
  • For dp[1][1], we can either delete -2 or keep the previous maximum, which gives us 1.

The dynamic programming way uses optimal substructure and overlapping subproblems to build the solution step by step.

This method is very important in problems like Dynamic Programming - Maximum Subarray where we use similar ideas to get the best results.

Dynamic Programming Solution in Java for Maximum Sum of Subarray

We want to find the maximum sum of a subarray with at most k deletions using dynamic programming in Java. Here is how we can do it:

  1. Define the DP Table: We will use a 2D array called dp. The value dp[i][j] shows the maximum sum we can get from the first i elements of the array with j deletions allowed.

  2. Base Cases:

    • When j = 0, dp[i][0] is just the maximum subarray sum from the first i elements without any deletions. We can find this using Kadane’s algorithm.
    • When i = 0, dp[0][j] = 0 for all j because there are no elements to look at.

  3. Transition Formula: For each element nums[i-1], we can choose to:

    • Include it in the subarray. We add it to the previous maximum subarray sum without going over the deletion limit.
    • Delete it. We then check the maximum sum with one less deletion.

We can write the transition like this:

dp[i][j] = Math.max(dp[i-1][j] + nums[i-1], dp[i-1][j-1]);
  1. Final Result: The answer is the maximum value in the last row of the DP table for any j from 0 to k.

Here is the Java code for this:

public class MaximumSumSubarray {
    public int maxSumAfterKDeletions(int[] nums, int k) {
        int n = nums.length;
        int[][] dp = new int[n + 1][k + 1];

        // Base case initialization
        for (int j = 0; j <= k; j++) {
            dp[0][j] = 0; // No elements to consider
        }

        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= k; j++) {
                // Include nums[i-1]
                dp[i][j] = dp[i - 1][j] + nums[i - 1];

                // If we can delete an element
                if (j > 0) {
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - 1]);
                }
            }
        }

        // Find the maximum sum with at most k deletions
        int maxSum = Integer.MIN_VALUE;
        for (int j = 0; j <= k; j++) {
            maxSum = Math.max(maxSum, dp[n][j]);
        }

        return maxSum;
    }

    public static void main(String[] args) {
        MaximumSumSubarray solution = new MaximumSumSubarray();
        int[] nums = {1, -2, 0, 3};
        int k = 1;
        System.out.println("Maximum sum of subarray with at most " + k + " deletions: " + solution.maxSumAfterKDeletions(nums, k)); // Output: 4
    }
}

This Java solution helps us find the maximum sum of a subarray with at most k deletions. It uses dynamic programming ideas. We take advantage of optimal substructure and overlapping problems that come up in these types of tasks.

Dynamic Programming Solution in Python for Maximum Sum of Subarray

To find the maximum sum of a subarray with at most k deletions, we use dynamic programming. We can make a 2D DP array. Here, dp[i][j] shows the maximum sum of a subarray ending at index i with j deletions.

Python Implementation

def maximum_sum(nums, k):
    n = len(nums)
    if n == 0:
        return 0

    # Initialize DP array
    dp = [[0] * (k + 1) for _ in range(n)]
    dp[0][0] = nums[0]

    for j in range(1, k + 1):
        dp[0][j] = max(dp[0][j - 1], nums[0])

    for i in range(1, n):
        dp[i][0] = max(dp[i - 1][0] + nums[i], nums[i])
        for j in range(1, k + 1):
            dp[i][j] = max(dp[i - 1][j] + nums[i], dp[i - 1][j - 1])

    # Get the maximum value from the last row
    return max(dp[n - 1])

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

Explanation of the Code

  • Initialization: We create a DP table dp. It holds the maximum sum of subarrays ending at index i with j deletions.
  • Base Case: The first element sets dp[0][0] to nums[0]. For j > 0, it tracks the maximum sum with deletions.
  • DP Transition:
    • If we do not make deletions, we can take the current number or add it to the previous maximum (no deletions).
    • If deletions are allowed, we check both cases. One where we delete the current element and one where we do not.
  • Result Extraction: The maximum value in the last row of the DP table gives the maximum sum we want.

This dynamic programming method finds the maximum sum of a subarray with at most k deletions. It works in O(n*k) time and uses O(n*k) space.

For more about dynamic programming methods, you can look at Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

Dynamic Programming Solution in C++ for Maximum Sum of Subarray

To find the maximum sum of a subarray with at most k deletions using dynamic programming in C++, we define a table dp[i][j]. Here:

  • i is the index in the array.
  • j is the number of deletions we use.

The value of dp[i][j] keeps the maximum sum we can get using the first i elements of the array and j deletions.

Algorithm Steps:

  1. Initialization:
    • We create a 2D array dp of size n x (k + 1). We fill it with 0. Here, n is the length of the input array.
    • We set dp[0][0] = arr[0]. This is because we can only take the first element without deleting anything.
  2. Filling the DP Table:
    • We loop through each element i from 0 to n-1.
    • For each element, we loop through each possible number of deletions j from 0 to k.
    • We update dp[i][j] by looking at:
      • Not deleting the current element: dp[i][j] = max(dp[i][j], dp[i-1][j] + arr[i])
      • Deleting the current element (only if j > 0): dp[i][j] = max(dp[i][j], dp[i-1][j-1])
  3. Result Extraction:
    • We find the maximum value in the last row of the dp table. This considers all possible deletions from 0 to k.

C++ Implementation:

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

int maxSumAfterKDeletions(vector<int>& arr, int k) {
    int n = arr.size();
    vector<vector<int>> dp(n, vector<int>(k + 1, 0));

    dp[0][0] = arr[0]; // Initialize the first element

    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= k; j++) {
            if (i > 0) {
                // Not deleting the current element
                dp[i][j] = max(dp[i][j], dp[i - 1][j] + arr[i]);
            }
            if (j > 0 && i > 0) {
                // Deleting the current element
                dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
            }
        }
    }

    // Get the maximum sum possible with at most k deletions
    int maxSum = 0;
    for (int j = 0; j <= k; j++) {
        maxSum = max(maxSum, dp[n - 1][j]);
    }

    return maxSum;
}

int main() {
    vector<int> arr = {1, -2, 0, 3};
    int k = 1;
    cout << "Maximum Sum of Subarray with at most " << k << " deletions: " << maxSumAfterKDeletions(arr, k) << endl;
    return 0;
}

Explanation of Code:

  • The function maxSumAfterKDeletions does the dynamic programming solution.
  • It starts the DP table and goes through the input array. It checks both options of deleting and not deleting.
  • In the end, it finds the maximum sum by looking at all possible values in the last row of the DP table.

This C++ code shows a simple way to solve the problem of finding the maximum sum of a subarray with at most k deletions using dynamic programming. If you want to learn more about similar problems, check out Maximum Subarray and Minimum Path Sum.

Space Optimization Techniques in Dynamic Programming

In dynamic programming, space optimization techniques are very important. They help us reduce the memory used by algorithms while keeping them efficient. Here are some good ways to optimize space in dynamic programming problems. We focus on the maximum sum of subarray with at most k deletions.

  1. Use Iterative Approaches Over Recursive:
    • We should prefer iterative methods. These use a bottom-up DP table. This way, we avoid the extra memory used by recursive stack space.
  2. Reduce the Size of the DP Table:
    • Instead of keeping a full table for all subproblems, we can only store the previous states that we really need.
    • For example, when we calculate the maximum sum of subarrays, we often only need the last two rows of the DP table.
  3. Use One-Dimensional Arrays:
    • If the state only depends on the last state, we can use a one-dimensional array.
    • Example: In the maximum sum of subarray problem, we can find the current state using just the last state. This change reduces space complexity from O(n^2) to O(n).
  4. In-Place Modifications:
    • When we can, we should change the input array to save space. We need to make sure that this does not change the final results.

Example Implementation in Java

public int maxSumAfterKDeletions(int[] arr, int k) {
    int n = arr.length;
    if (n == 0) return 0;
    
    int[] dp = new int[n + 1];
    for (int i = 1; i <= n; i++) {
        dp[i] = dp[i - 1] + arr[i - 1]; // Sum so far
    }

    for (int del = 1; del <= k; del++) {
        for (int i = n; i >= del; i--) {
            dp[i] = Math.max(dp[i], dp[i - del]); // Consider deletion
        }
    }

    return dp[n];
}

Example Implementation in Python

def max_sum_after_k_deletions(arr, k):
    n = len(arr)
    if n == 0:
        return 0

    dp = [0] * (n + 1)
    for i in range(1, n + 1):
        dp[i] = dp[i - 1] + arr[i - 1]  # Sum so far

    for del_count in range(1, k + 1):
        for i in range(n, del_count - 1, -1):
            dp[i] = max(dp[i], dp[i - del_count])  # Consider deletion

    return dp[n]

Example Implementation in C++

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

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

    vector<int> dp(n + 1, 0);
    for (int i = 1; i <= n; i++) {
        dp[i] = dp[i - 1] + arr[i - 1]; // Sum so far
    }

    for (int del = 1; del <= k; del++) {
        for (int i = n; i >= del; i--) {
            dp[i] = max(dp[i], dp[i - del]); // Consider deletion
        }
    }

    return dp[n];
}
  1. Iterative Updates:
    • Instead of keeping the whole DP table, we can update the values in the same array. We only keep the results we need for the calculations.
  2. Memoization:
    • We can use memoization to store results of subproblems only when we need them. This can really help reduce the space we use.

When we use these space optimization techniques in dynamic programming, we can improve performance. This makes algorithms work better with memory. For more information about dynamic programming principles, we can check Dynamic Programming: Maximum Subarray - Kadane’s Algorithm.

Analyzing Time Complexity of the Solution

When we solve the problem of Maximum Sum of Subarray with At Most k Deletions using dynamic programming, we need to look at the time complexity. This helps us see how good our solution is.

Dynamic Programming Approach

We have an array arr with size n and an integer k. In the dynamic programming approach, we create a DP table. In this table, dp[i][j] shows the maximum sum of the subarray that ends at index i with at most j deletions.

We can write the recurrence relation like this:

  • If we do not delete the current element: [ dp[i][j] = dp[i - 1][j] + arr[i] ]

  • If we delete the current element: [ dp[i][j] = dp[i - 1][j - 1] ]

We decide to take the current element or delete it. This helps us build our solution.

Time Complexity Analysis

  1. DP Table Construction: The DP table has dimensions of (n+1) x (k+1). So, filling this table will take: [ O(n k) ]

  2. Final Result Calculation: After we build the DP table, we need to find the maximum value from the last row. This takes: [ O(k) ]

Overall Time Complexity

When we combine these steps, the overall time complexity of the solution is: [ O(n k) ]

Space Complexity Consideration

The space complexity of this solution is (O(n k)) because of the DP table. But we can make it better to (O(k)). We just need to store the last row of the DP table. The current state only depends on the previous state.

Example Implementation in Python

Here is a simple implementation that shows the time complexity ideas:

def maxSumAfterKDeletions(arr, k):
    n = len(arr)
    dp = [[0] * (k + 1) for _ in range(n + 1)]

    for i in range(1, n + 1):
        for j in range(min(i, k + 1)):
            dp[i][j] = max(dp[i - 1][j] + arr[i - 1], dp[i - 1][j - 1] if j > 0 else 0)

    return max(dp[n][j] for j in range(k + 1))

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

This Python function shows how the solution works and follows the time complexity of (O(n k)). This makes it good for medium sizes of (n) and (k).

Common Mistakes to Avoid in Dynamic Programming Problems

Dynamic programming (DP) is a strong method for solving optimization problems. But we can easily make small mistakes. These mistakes can lead to wrong answers or slow algorithms. Here are some common mistakes to watch out for in dynamic programming problems:

  1. Not Identifying the Optimal Substructure:
    • Before we use DP, we need to check if we can break the problem into smaller parts. If we can build the best solution from the best solutions of its smaller parts, then we have an optimal substructure.
  2. Ignoring Overlapping Subproblems:
    • DP works best when we solve the same small problems many times. If we calculate the same small problem again and again, we should use memoization or tabulation to keep the results.
  3. Misdefining State Variables:
    • We must clearly define the state variables that will show the problem. If we define these variables wrong, we can get wrong answers.
  4. Incorrect Transition Formula:
    • The transition formula tells us how to make a solution from subproblems. We need to make sure it matches the problem’s needs. Mistakes in this formula can give us wrong values in the DP table.
  5. Forgetting to Handle Base Cases:
    • Base cases are very important for starting our DP solution. If we miss these cases or define them wrong, we can get errors when running or wrong final answers.
  6. Using Incorrect Data Types:
    • We need to check that the data types in our DP tables can handle the values we are calculating. For example, if we use an integer type for big sums, it can cause overflow.
  7. Not Considering All Conditions:
    • DP often needs to think about many conditions. For example, we might need to decide if we should include an element. We have to make sure we think about all situations in the transition states.
  8. Neglecting Space Optimization:
    • If the problem allows, we should save space by only keeping the necessary states. We don’t need to keep a full DP table if we can avoid it.
  9. Not Testing Edge Cases:
    • We should always test our solution with edge cases. This means testing things like empty inputs or inputs that can cause problems at the limits.
  10. Failure to Analyze Complexity:
    • After we create a DP solution, we should look at its time and space complexity. This helps us see how good and scalable our solution is.

By knowing these common mistakes, we can get better at dynamic programming. We can create stronger solutions. For more information about dynamic programming problems, we can read articles on topics like Dynamic Programming: Maximum Subarray (Kadane’s Algorithm).

Frequently Asked Questions

1. What is the maximum sum of subarray with at most k deletions problem?

The maximum sum of subarray with at most k deletions problem is about finding the biggest sum of a continuous subarray in a given array. We can delete up to k elements. We can solve this problem well by using dynamic programming. This method helps us take advantage of the best parts of the problem and the parts that repeat. By setting it up right, we can find the best sum while following the deletion rules.

2. How can dynamic programming be applied to solve the maximum sum of subarray with at most k deletions?

Dynamic programming works great for our problem. We can create a state that shows the maximum sum we can get up to a certain index with a certain number of deletions. Then, we can build a relation that helps us find the maximum sum step by step. We update our state based on what we calculated before and what the current element adds. This way, we get the best solution.

3. What is the time complexity of the dynamic programming solution for the maximum sum of subarray with at most k deletions?

The time complexity of the dynamic programming way to solve our problem is O(n * k). Here, n is the length of the array and k is the maximum number of deletions we can make. This happens because for every element in the array, we might have to check all the deletion options up to k. This makes a nested loop structure.

4. Can you provide a sample code for solving the maximum sum of subarray with at most k deletions in Python?

Sure! Here is a simple Python code to solve the maximum sum of subarray with at most k deletions:

def maxSumAfterKDeletions(arr, k):
    n = len(arr)
    dp = [[0] * (k + 1) for _ in range(n)]
    
    for j in range(k + 1):
        dp[0][j] = arr[0] if j == 0 else 0
    
    for i in range(1, n):
        for j in range(k + 1):
            dp[i][j] = max(dp[i - 1][j] + arr[i], dp[i - 1][j - 1] if j > 0 else 0)
    
    return max(dp[n - 1])

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

5. What common mistakes should be avoided when solving dynamic programming problems like this?

When we solve the maximum sum of subarray with at most k deletions, we should avoid some common mistakes. First, we need to initialize the dp array correctly. We also must not forget edge cases like when k is 0. It is important to define the state transitions well. Lastly, we should understand the optimal substructure idea, as it helps us make a good dynamic programming solution.

For more insights into dynamic programming, we can check out related articles like Dynamic Programming: Maximum Subarray (Kadane’s Algorithm) and Dynamic Programming: Minimum Path Sum in a Grid to understand similar ideas better.