[Dynamic Programming] Maximum Sum of a Subsequence with One Deletion and One Addition - Hard

The problem of finding the maximum sum of a subsequence with one deletion and one addition is a hard task in dynamic programming. Our goal is to find the maximum sum we can get by picking a subsequence from a given array. We will allow the removal of one element and the addition of another to make the total sum as high as possible. This way often leads to complicated calculations and typical properties of dynamic programming.

In this article, we will look at different parts of this challenge in dynamic programming. We will show optimized solutions in Java, Python, and C++. We will explain the problem in detail. Then we will break down the dynamic programming approach step by step. We will also discuss ways to make space complexity better and compare different methods for solving the problem. At the end, we will give a clear explanation of the code in Java and Python. We will also include a FAQ section to answer common questions.

[Dynamic Programming] Maximum Sum of a Subsequence with One Deletion and One Addition Optimized Solution in Java
Dynamic Programming Approach for Maximum Sum of a Subsequence with One Deletion and One Addition in Python
C++ Implementation of Maximum Sum of a Subsequence with One Deletion and One Addition
Understanding the Problem Statement for Maximum Sum of a Subsequence with One Deletion and One Addition
Step by Step Breakdown of the Dynamic Programming Approach
Optimized Space Complexity Techniques for Maximum Sum of a Subsequence
Comparative Analysis of Different Approaches for Maximum Sum of a Subsequence
Code Walkthrough of the Java Solution for Maximum Sum with One Deletion and One Addition
Python Code Explanation for Maximum Sum of a Subsequence with One Deletion and One Addition
Frequently Asked Questions

If we want to improve our dynamic programming skills, we can also check out related topics like the Fibonacci number problem here and here.

Dynamic Programming Approach for Maximum Sum of a Subsequence with One Deletion and One Addition in Python

To solve the problem of getting the maximum sum of a subsequence with one deletion and one addition using dynamic programming in Python, we can use an array to keep track of results. We will keep two arrays, dp and max_sum. Here dp[i] shows the maximum sum of the subsequence for the first i elements. The max_sum[i] array tells us the best sum we can get by deleting one element up to index i.

Here is the Python code for this method:

def max_sum_with_one_deletion_and_addition(arr):
    n = len(arr)
    if n < 2:
        return 0

    dp = [0] * n
    max_sum = [0] * n

    dp[0] = arr[0]
    for i in range(1, n):
        dp[i] = max(dp[i-1] + arr[i], arr[i])  # Max sum without deletion

    max_sum[0] = dp[0]
    for i in range(1, n):
        max_sum[i] = max(max_sum[i-1], dp[i])  # Max sum with one deletion

    result = float('-inf')
    for i in range(1, n):
        result = max(result, max_sum[i-1] + arr[i])  # Add current element after deletion

    return result

# Example usage
arr = [1, -2, 0, 3]
print("Maximum Sum of Subsequence with One Deletion and One Addition:", max_sum_with_one_deletion_and_addition(arr))

Explanation:

Initialization: We first check if the input array has less than two elements. If yes, we return 0 because we cannot do both operations.
Dynamic Programming Arrays:
- dp[i]: This keeps the maximum sum for the first i elements.
- max_sum[i]: This keeps the best sum of subsequences we can get by deleting one element.
Filling the Arrays:
- We go through the array to fill dp and max_sum.
- For every element, we choose to add it to the previous sum or start a new subsequence.
Final Calculation: We find the maximum possible sum by thinking about adding the current element after removing an earlier one.

This method runs in O(n) time and uses O(n) space. It is good for bigger datasets. If you want to learn more about similar dynamic programming methods, you can check out the Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

C++ Implementation of Maximum Sum of a Subsequence with One Deletion and One Addition

We can solve the problem of finding the maximum sum of a subsequence with one deletion and one addition using dynamic programming. The main idea is to keep two arrays that help us track the maximum sums based on our rules.

C++ Code Implementation

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

using namespace std;

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

    vector<int> dp(n);
    dp[0] = arr[0];
    int maxSum = arr[0];

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

    int finalMax = maxSum; // We track the maximum sum before deletion
    for (int i = 1; i < n; ++i) {
        // We think about deleting arr[i-1] and adding arr[i]
        if (i > 1) {
            finalMax = max(finalMax, dp[i - 2] + arr[i]);
        }
    }

    return finalMax;
}

int main() {
    vector<int> arr = {3, 2, 5, 10, 7};
    cout << "Maximum Sum of Subsequence with One Deletion and One Addition: " 
         << maxSumWithOneDeletionAndAddition(arr) << endl;
    return 0;
}

Explanation of the Code

Input Handling: The function maxSumWithOneDeletionAndAddition takes a vector of integers as input.
Dynamic Programming Array: We use the dp array to store maximum sums at each index without deletions.
Iterating Through the Array: For each element, we find the maximum sum by including the current element or adding it to the previous maximum.
Deletion and Addition Logic: After filling the dp array, we find the final maximum sum by considering the allowed deletion and addition.
Output: We print the result in the main function.

Complexity Analysis

Time Complexity: O(n), where n is the number of elements in the array.
Space Complexity: O(n) for the dp array. We can make it O(1) if we only track the last two states.

This C++ code gives us a good way to find the maximum sum of a subsequence with one deletion and one addition using dynamic programming. If we want to learn more about dynamic programming, we can check related topics like Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

Understanding the Problem Statement for Maximum Sum of a Subsequence with One Deletion and One Addition

The problem is about finding the maximum sum of a subsequence. We can add one number and delete one number from a sequence of integers. The goal is to make the sum of the new subsequence as big as possible after doing these two actions.

We have an array arr with length n. Our task is to find the maximum sum of a subsequence where:

We can delete one number from the subsequence.
We can add one number to the subsequence. This can be any integer, not just from the original array.

Problem Breakdown:

Input: An array of integers arr.
Output: The maximum sum of a subsequence after we do one deletion and one addition.

Example:

Let’s look at an input array arr = [1, -2, 0, 3]:

If we delete -2 and add 4, the new subsequence is [1, 0, 3, 4]. The sum is 8.
If we delete 0 and add 5, the new subsequence is [1, -2, 3, 5]. The sum is 7.

Constraints:

The array can have positive, negative, and zero values.
We can choose the added number freely. It can be any integer and may change the total sum a lot.

Approach:

To solve this problem well, we can use dynamic programming. This way, we can keep track of the maximum sums while going through the array.

Iterate through each element and keep two arrays:
- One array tracks the maximum sum so far without any deletion or addition.
- Another array keeps the maximum sum with deletions and additions considered.
Calculate the maximum sums by checking possible deletions and additions in a smart way.

This dynamic programming method helps us check all options while keeping it efficient.

To see examples in Java, Python, and C++, please go to the sections for each language to find optimized solutions.

Step by Step Breakdown of the Dynamic Programming Approach

We want to solve the problem of finding the maximum sum of a subsequence. This subsequence allows for one deletion and one addition. We can do this using dynamic programming.

Problem Definition

We have an array of integers. Our goal is to find the maximum sum of a subsequence. We can remove one element from the subsequence. At the same time, we can add another element from the array.

Dynamic Programming State

We will use dp[i] to show the maximum sum of a subsequence up to the i-th index.
We also keep another array called max_sum. This tracks the maximum sum of subsequences when we consider deletions and additions.

Transition Relation

Base Case:
- We set dp[0] to array[0]. This is because the maximum sum ending at the first element is just that element.
Recurrence Relation:
- For each element i from 1 to n-1:
  - We calculate the maximum sum that includes the current element:
```
dp[i] = max(dp[i-1] + array[i], array[i])
```
  - To manage one deletion:
```
max_sum[i] = max(max_sum[i-1], dp[i-1] - array[j]) for all j < i
```
  - So, the final value at dp[i] will be:
```
dp[i] = max(dp[i], max_sum[i-1] + array[i])
```

Implementation

Here is how we can implement this in Python:

def maxSum(array):
    n = len(array)
    if n == 0:
        return 0

    dp = [0] * n
    max_sum = [0] * n
    dp[0] = array[0]
    max_sum[0] = 0  # No valid sum with a deletion at the first element

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

    return max(dp)

# Example usage
array = [3, 2, 5, 10, 7]
result = maxSum(array)
print(result)  # Output: 15 (subsequence could be [3, 5, 10])

Complexity Analysis

Time Complexity: O(n^2). This is because we have nested loops to find the maximum subsequence sum with one deletion.
Space Complexity: O(n). This is for storing dp and max_sum.

This breakdown helps us understand how to use the dynamic programming approach to find the maximum sum of a subsequence with one deletion and one addition.

Optimized Space Complexity Techniques for Maximum Sum of a Subsequence

When we want to find the maximum sum of a subsequence with one deletion and one addition, we can make the space use better. We can change the space complexity from (O(n)) to (O(1)). We can do this by using some simple techniques. Here is how we can do it:

In-Place Computation: We do not need a big DP array. We can use a few variables to track what we need. This way, we use much less space.
Two-Pass Approach: First, we calculate the maximum sum without any deletions. Then, in the second pass, we check the possible maximum sums by thinking about deleting one element and adding another.
Rolling Variables: We can use rolling variables to save the last two results we need. This means we do not need full arrays.

Example Implementation in Python

Here is a simple code that shows these techniques:

def max_sum_with_one_deletion_and_addition(arr):
    if not arr:
        return 0

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

    # Initialize variables
    max_sum = arr[0]
    prev_max = arr[0]
    prev_prev_max = 0

    for i in range(1, n):
        current = arr[i]
        # Update max_sum considering deletion and addition
        max_sum = max(max_sum, prev_max + current - arr[i - 1])
        prev_prev_max = prev_max
        prev_max = max(prev_max + current, current)

    return max(max_sum, prev_max)

# Example usage
arr = [4, 3, 5, 1, 3]
print(max_sum_with_one_deletion_and_addition(arr))  # Output: 12

Explanation of the Code

Initialization: First, we check for empty or single-element arrays.
Iterate through the array: For each element, we update the maximum sum. We think about both the current element and the previous sums.
Rolling Variables: We use prev_max and prev_prev_max to keep the sums we need for the next step. This way, we do not need extra space for an array.

By using these techniques, we make the space complexity of the algorithm (O(1)). The time complexity stays at (O(n)), so it works well for big inputs. This method is very helpful in dynamic programming problems where we need to save space.

If you want to read more about similar dynamic programming methods, you can look at articles on Dynamic Programming: Maximum Subarray (Kadane’s Algorithm) or Dynamic Programming: Maximum Product Subarray.

Comparative Analysis of Different Approaches for Maximum Sum of a Subsequence

When we try to find the maximum sum of a subsequence with one deletion and one addition, we can use different methods. Each method has its own time and space complexity. Here are three main ways to solve the problem:

Dynamic Programming Approach:

Description: We use a dynamic programming table. This table helps us keep track of results as we go. It makes the calculation of the maximum sum faster.
Time Complexity: O(n^2)
Space Complexity: O(n)

Java Implementation:

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

    int[] dp = new int[n];
    dp[0] = arr[0];
    for (int i = 1; i < n; i++) {
        dp[i] = Math.max(arr[i], dp[i - 1] + arr[i]);
        if (i > 1) {
            dp[i] = Math.max(dp[i], dp[i - 2] + arr[i]);
        }
    }
    return dp[n - 1];
}

Optimized Space Complexity Approach:

Description: We only use a few variables instead of a full array. This way, we save space but still get the same result.
Time Complexity: O(n)
Space Complexity: O(1)

Python Implementation:

def maxSumWithOneDeletionAndAddition(arr):
    n = len(arr)
    if n == 0:
        return 0

    current_max = prev_max = 0
    for i in range(n):
        temp = current_max
        current_max = max(arr[i], current_max + arr[i], prev_max + arr[i])
        prev_max = temp
    return current_max

Brute Force Approach:

Description: In this method, we check all possible subsequences. We calculate their sums and make changes for deletion and addition.
Time Complexity: O(2^n) because we look at all subsequences.
Space Complexity: O(1), since we do not need extra space besides the input.

C++ Implementation:

int maxSumWithOneDeletionAndAddition(vector<int>& arr) {
    int n = arr.size();
    int max_sum = INT_MIN;

    for (int i = 0; i < n; ++i) {
        for (int j = i + 1; j < n; ++j) {
            int sum = 0;
            for (int k = 0; k < n; ++k) {
                if (k != i && k != j) {
                    sum += arr[k];
                }
            }
            max_sum = max(max_sum, sum + arr[j]);
        }
    }
    return max_sum;
}

Comparison: - The brute force approach helps us understand the problem better. But it takes too long for bigger inputs. - The dynamic programming approach is better for time and space. It works well for medium-sized inputs. - The optimized space approach is great for large inputs. It keeps the speed of dynamic programming and uses less memory.

For more information about dynamic programming, we can look at the Dynamic Programming Fibonacci Number and Dynamic Programming Maximum Subarray (Kadane’s Algorithm).

Code Walkthrough of the Java Solution for Maximum Sum with One Deletion and One Addition

We can solve the problem of finding the maximum sum of a subsequence with one deletion and one addition using dynamic programming in Java. Our goal is to get the highest sum of the subsequence while following the rules for deletion and addition.

Java Implementation

Here is a simple Java implementation of the solution:

public class MaximumSumSubsequence {
    public static int maximumSum(int[] arr) {
        int n = arr.length;

        if (n == 0) return 0;

        int[] dp = new int[n];
        dp[0] = arr[0];
        int maxSum = dp[0];

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

        // Consider deletion and addition
        int[] maxWithDeletion = new int[n];
        maxWithDeletion[0] = arr[0];
        for (int i = 1; i < n; i++) {
            maxWithDeletion[i] = Math.max(maxWithDeletion[i - 1], dp[i - 1] + arr[i]);
            maxSum = Math.max(maxSum, maxWithDeletion[i]);
        }

        return maxSum;
    }

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

Explanation of Code

Initialization: We use the array dp to keep track of the maximum sum subsequence ending at each index. The variable maxSum holds the overall maximum sum we find.
Dynamic Programming Logic:
- For each element, we calculate the maximum sum we can get by either including or excluding that element.
- We update maxSum with the highest value found in dp.
Incorporating Deletion:
- The second time we go through the array, we think about deleting an element.
- The array maxWithDeletion saves the maximum sum we can get after deleting an element. This lets us add more elements after that.
Output: We print the final result in the main method.

This implementation helps us find the maximum sum of a subsequence with one deletion and one addition. It works well and is efficient. For more on dynamic programming, we can look at other ideas like Dynamic Programming: Maximum Subarray (Kadane’s Algorithm).

Python Code Explanation for Maximum Sum of a Subsequence with One Deletion and One Addition

We can solve the problem of finding the maximum sum of a subsequence with one deletion and one addition using dynamic programming. We will use two arrays. The first is max_ending_here and the second is max_ending_here_with_deletion. These arrays help us to track the maximum sums.

def max_sum_subsequence_with_deletion_and_addition(arr):
    n = len(arr)
    if n == 0:
        return 0

    max_ending_here = [0] * n
    max_ending_here_with_deletion = [0] * n

    max_ending_here[0] = arr[0]
    max_ending_here_with_deletion[0] = 0 

    for i in range(1, n):
        max_ending_here[i] = max(arr[i], max_ending_here[i - 1] + arr[i])
        max_ending_here_with_deletion[i] = max(max_ending_here_with_deletion[i - 1], max_ending_here[i - 1])

    return max(max(max_ending_here), max_ending_here_with_deletion)

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

Code Explanation:

Initialization:
- max_ending_here[i] keeps the maximum sum of the subsequence that ends at index i.
- max_ending_here_with_deletion[i] keeps the maximum sum of the subsequence up to index i with one deletion.
Iteration:
- We go through the array starting from index 1.
- We update max_ending_here[i] by deciding to start new with arr[i] or extend the previous sum by adding arr[i].
- We update max_ending_here_with_deletion[i] by taking the maximum of the last value or the value from max_ending_here at the previous index.
Final Result: The final answer is the highest value from max_ending_here or max_ending_here_with_deletion.

This way we can get the result in O(n) time and O(n) space. This makes it good for larger inputs.

For more about dynamic programming, we can look at other problems like Dynamic Programming - Fibonacci Number and Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

Frequently Asked Questions

1. What is the maximum sum of a subsequence with one deletion and one addition?

We need to find the maximum sum of a subsequence when we can delete one number and add one number. This means we will calculate the best sum we can get by changing the subsequence. We use dynamic programming to do this. It helps us to find the answer fast. We keep track of the best sums at each step while thinking about what to delete and what to add.

2. How can I implement the solution in Java for this dynamic programming problem?

To solve the maximum sum of a subsequence with one deletion and one addition in Java, we can use dynamic programming. We will make a state array to remember the maximum sums. You can check the example in the article. It shows an easy way to understand the logic and how to write the code.

3. What are the key differences between the Python and C++ implementations of this dynamic programming problem?

Python and C++ both solve the maximum sum of a subsequence with one deletion and one addition in a similar way. But they look different because of their syntax and data structures. Python is usually shorter, while C++ gives us more control over memory. You can look at the sections in the article to see the detailed code for both languages.

4. How does dynamic programming optimize the solution for the maximum sum of a subsequence with one deletion and one addition?

Dynamic programming helps us by breaking the problem into smaller parts that overlap. We save the results we get so we do not have to calculate the same things again. This makes the process faster. It helps us find the maximum sum without wasting time. For more details, check the step-by-step guide in the article.

5. Are there similar problems that I can practice to enhance my skills in dynamic programming?

Yes, there are many problems like this that we can practice to get better at dynamic programming. For example, we can try Dynamic Programming: Maximum Subarray (Kadane’s Algorithm) or Dynamic Programming: Minimum Cost Climbing Stairs. These problems are good practice for improving our dynamic programming skills.