[Dynamic Programming] Maximum Sum of Subsequence with One Swap - Medium

The problem of finding the maximum sum of a subsequence with one swap is about finding a subsequence in an array. We can swap two elements to get the highest sum. This is a dynamic programming challenge. We need to think carefully about how the swap affects the total sum. This helps us find a good solution that works well in both time and space.

In this article, we will look into the details of the maximum sum of subsequence with one swap problem. We will begin with an overview of the problem. Then, we will provide a simple breakdown of the dynamic programming approach in Java, Python, and C++. We will also talk about ways to improve our solution and compare different methods. At the end, we will show a clear code walkthrough for each programming language and answer some common questions.

• [Dynamic Programming] Maximum Sum of Subsequence with One Swap Solution Overview
• Understanding the Problem Statement for Maximum Sum of Subsequence with One Swap
• Dynamic Programming Approach for Maximum Sum of Subsequence with One Swap in Java
• Dynamic Programming Approach for Maximum Sum of Subsequence with One Swap in Python
• Dynamic Programming Approach for Maximum Sum of Subsequence with One Swap in C++
• Optimizing the Dynamic Programming Solution for Maximum Sum of Subsequence with One Swap
• Comparative Analysis of Different Approaches to Maximum Sum of Subsequence with One Swap
• Code Walkthrough for Maximum Sum of Subsequence with One Swap in Java
• Code Walkthrough for Maximum Sum of Subsequence with One Swap in Python
• Frequently Asked Questions

For those who want to learn more about dynamic programming, we can suggest articles on related topics. You may find the Dynamic Programming Fibonacci Number and Dynamic Programming Minimum Path Sum in a Grid helpful.

Understanding the Problem Statement for Maximum Sum of Subsequence with One Swap

We need to find the maximum sum of a subsequence by swapping two elements in an array of integers. The subsequence can skip some numbers but still keeps the order. Our goal is to do this with just one swap.

Problem Breakdown

1. Input: We have an array of integers, arr. Each number can be either positive or negative.
2. Output: We want the highest possible sum of a subsequence after we swap one time.
3. Subsequence: This means we can take some numbers from the array without changing their order. We can leave out some numbers if we want.
4. Swap: We pick two positions i and j in the array. We swap the values at these places. After that, we calculate the highest sum of the new subsequence.

Constraints

• The size of the array, n, can be different. So, we have to think about how to make it fast.
• We should look at all pairs of positions to swap. This way, we can find the best sum.

Example

• Input: arr = [1, 2, 3, -1, 4]
• If we swap 1 and 4, we get the array [4, 2, 3, -1, 1].
• The highest subsequence sum after the swap is 4 + 2 + 3 = 9.

We can use a dynamic programming method to solve this problem. It helps us find the best sum after checking all possible swaps.

Dynamic Programming Approach for Maximum Sum of Subsequence with One Swap in Java

To solve the problem of finding the maximum sum of a subsequence with one swap using dynamic programming in Java, we can follow these steps:

1. Initialize Variables: We create an array to keep track of the maximum sum up to each index. We also need a variable to store the overall maximum sum.

2. Iterate through the Array: For each element, we calculate the new sum if we swap it with a previous element. We keep track of the maximum sum we find during these steps.

3. Dynamic Programming Table: We use a DP table to store the maximum sums we have calculated.

Implementation

Here is a Java implementation of the described approach:

public class MaxSumSubsequenceOneSwap {
    public static int maxSumWithOneSwap(int[] nums) {
        int n = nums.length;
        if (n <= 1) return 0;

        // Maximum sum of subsequence without swaps
        int[] dp = new int[n];
        System.arraycopy(nums, 0, dp, 0, n);
        int maxSum = dp[0];

        // Calculate max sum without any swaps
        for (int i = 1; i < n; i++) {
            dp[i] = Math.max(dp[i], dp[i - 1] + nums[i]);
            maxSum = Math.max(maxSum, dp[i]);
        }

        // Check for maximum sum with one swap
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                // Swap nums[i] and nums[j]
                int newSum = 0;
                for (int k = 0; k < n; k++) {
                    if (k == i) newSum += nums[j];
                    else if (k == j) newSum += nums[i];
                    else newSum += nums[k];
                }
                maxSum = Math.max(maxSum, newSum);
            }
        }

        return maxSum;
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 3, 4, 5};
        System.out.println("Maximum Sum of Subsequence with One Swap: " + maxSumWithOneSwap(nums));
    }
}

Explanation of Code

• Array Initialization: We start by initializing the dp array with the values from nums. This helps us keep track of maximum subsequence sums.
• Maximum Sum Calculation: The first loop calculates the maximum possible sum without any swaps.
• Double Loop for Swapping: The nested loop goes through each pair of indices. It calculates the new sum after swapping and updates the maximum sum if the new sum is bigger.

This way, we explore the benefits of one swap to maximize the subsequence sum. We use dynamic programming principles for better efficiency.

For more understanding of dynamic programming, we can check related articles like Dynamic Programming: Fibonacci Number or Dynamic Programming: Maximum Subarray (Kadane’s Algorithm).

Dynamic Programming Approach for Maximum Sum of Subsequence with One Swap in Python

To find the maximum sum of a subsequence with one swap using dynamic programming in Python, we can do several steps:

1. Initialization: We create an array dp to keep the maximum sum up to each index without swaps. We also need another variable to track the maximum sum with one swap.

2. Dynamic Programming Transition:

   • We go through the array and find the maximum sum for each index without swaps.
   • For each element, we think about how swapping it with a previous element can change the maximum sum.

3. Implementation: Here is how we can write the code:

def maxSumWithOneSwap(arr):
    n = len(arr)
    dp = [0] * n
    max_sum = 0
    
    # Calculate maximum sum without any swap
    dp[0] = arr[0]
    for i in range(1, n):
        dp[i] = max(dp[i - 1] + arr[i], arr[i])
        max_sum = max(max_sum, dp[i])
    
    # Calculate maximum sum with one swap
    for i in range(n):
        for j in range(i):
            if arr[i] != arr[j]:  # Only swap if they are different
                swapped_sum = sum(arr) - arr[i] - arr[j] + arr[j] + arr[i]
                max_sum = max(max_sum, swapped_sum)
    
    return max_sum

# Example usage
arr = [1, 2, 3, 4, 5]
print(maxSumWithOneSwap(arr))  # Output will be the maximum sum after one swap

4. Complexity Analysis:
   • Time Complexity: O(n^2) because of the nested loop for swaps.
   • Space Complexity: O(n) for the dp array.

This approach helps us find the maximum sum of a subsequence with one swap. It uses dynamic programming to keep track of sums and possible swaps. If we want to learn more about dynamic programming concepts, we can check this Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

Dynamic Programming Approach for Maximum Sum of Subsequence with One Swap in C++

To solve the problem of finding the maximum sum of a subsequence with one swap using dynamic programming in C++, we can split the solution into easy steps. We need to keep two arrays. One array is for maximum sums without swapping. The other is for maximum sums with one swap.

Approach

1. Initialize Arrays:
   • We create an array dp to store the maximum sum of the subsequence ending at each index without a swap.
   • We also create another array swap_dp to keep the maximum sum with one swap.
2. Iterate through the Array:
   • For each element, we calculate the maximum sum for dp by checking all previous elements.
   • For swap_dp, we look at the previous sums and find new sums if we do a swap.
3. Update Maximum Values:
   • We keep a running maximum to help with calculating swap_dp.

Implementation

Here is a C++ code for this approach:

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

int maxSumWithOneSwap(vector<int>& nums) {
    int n = nums.size();
    if (n <= 1) return n ? nums[0] : 0;

    vector<int> dp(n, 0);
    vector<int> swap_dp(n, 0);

    dp[0] = nums[0];
    int max_so_far = nums[0];
    int overall_max = nums[0];

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

    for (int i = 0; i < n; i++) {
        swap_dp[i] = dp[i];
        if (i > 0) {
            swap_dp[i] = max(swap_dp[i], max_so_far + nums[i]);
        }
        max_so_far = max(max_so_far, dp[i]);
    }

    return *max_element(swap_dp.begin(), swap_dp.end());
}

int main() {
    vector<int> nums = {1, -2, 0, 3};
    cout << "Maximum Sum of Subsequence with One Swap: " << maxSumWithOneSwap(nums) << endl;
    return 0;
}

Explanation of the Code

• Input: The function maxSumWithOneSwap takes a vector of integers as input.
• Dynamic Programming Arrays:
  • dp[i] keeps track of the maximum sum of the subsequence ending at index i.
  • swap_dp[i] calculates the maximum sum if we swap with any previous element.
• Calculations:
  • The for loops calculate values for dp and swap_dp.
• Output: The maximum sum is printed in the main function.

This way, we can compute the maximum sum of a subsequence with one swap. Dynamic programming helps us manage time well.

For more learning about dynamic programming, you can check Dynamic Programming: Maximum Subarray (Kadane’s Algorithm) for basic ideas.

Optimizing the Dynamic Programming Solution for Maximum Sum of Subsequence with One Swap

We can make the Dynamic Programming solution for the “Maximum Sum of Subsequence with One Swap” faster. We will use prefix sums and careful indexing. This helps us reduce the time complexity from O(n^2) to O(n). The main idea is to keep track of the maximum sum of subsequences while going through the array. We also need to note any possible gains from swaps.

Approach

1. Initial Setup:
   • We create an array dp. The value dp[i] shows the maximum sum of the subsequence that ends at index i.
   • We set dp[0] to the first element of the array. The rest of the values will be zero.
2. Iterate Through the Array:
   • For each element at index i, we calculate dp[i]. This is based on the maximum sum of subsequences that can include the element at index i.
   • We also keep track of the maximum value of dp[j] for all j < i. This helps us see if a swap can give a higher sum.
3. Swap Logic:
   • For each element, we check if swapping it with the maximum element before it gives a better subsequence sum.
   • We store the highest possible sum by looking at both the original and swapped cases.

Code Implementation

Here is how we can write the optimized solution in Java:

public class MaxSumSubsequenceOneSwap {
    public static int maxSumWithOneSwap(int[] arr) {
        int n = arr.length;
        if (n <= 1) return n == 0 ? 0 : arr[0];

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

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

            // Calculate potential swap gain
            if (i > 1) {
                maxSum = Math.max(maxSum, dp[i - 1] + maxElement - arr[i]);
            }

            maxElement = Math.max(maxElement, arr[i]);
        }
        return maxSum;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4, 5};
        System.out.println("Maximum Sum of Subsequence with One Swap: " + maxSumWithOneSwap(arr));
    }
}

Python Implementation

Here is the same logic in Python:

def max_sum_with_one_swap(arr):
    n = len(arr)
    if n <= 1:
        return arr[0] if n == 1 else 0

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

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

        # Calculate potential swap gain
        if i > 1:
            max_sum = max(max_sum, dp[i - 1] + max_element - arr[i])

        max_element = max(max_element, arr[i])

    return max_sum

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5]
    print("Maximum Sum of Subsequence with One Swap:", max_sum_with_one_swap(arr))

C++ Implementation

Here is the C++ version of the same logic:

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

int maxSumWithOneSwap(vector<int> &arr) {
    int n = arr.size();
    if (n <= 1) return n == 0 ? 0 : arr[0];

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

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

        // Calculate potential swap gain
        if (i > 1) {
            maxSum = max(maxSum, dp[i - 1] + maxElement - arr[i]);
        }

        maxElement = max(maxElement, arr[i]);
    }
    return maxSum;
}

int main() {
    vector<int> arr = {1, 2, 3, 4, 5};
    cout << "Maximum Sum of Subsequence with One Swap: " << maxSumWithOneSwap(arr) << endl;
    return 0;
}

This approach helps us quickly find the maximum sum of subsequences with one swap. The time complexity is O(n). For more on dynamic programming, we can read articles like Dynamic Programming Fibonacci Number.

Comparative Analysis of Different Approaches to Maximum Sum of Subsequence with One Swap

When we solve the problem of finding the maximum sum of a subsequence with one swap, we can use different ways. Each way has its own pros and cons regarding how complex or efficient it is. Here are some common methods, their benefits, and drawbacks.

1. Brute Force Approach

• Description: We generate all possible subsequences and check the maximum sum after one swap.
• Time Complexity: O(2^n) because we generate all subsequences.
• Pros: It is easy to understand and use.
• Cons: It does not work well for larger arrays.

2. Dynamic Programming Approach

• Description: We use dynamic programming to keep track of results and build the solution. We calculate the maximum sum of subsequences while thinking about the swap of elements.
• Time Complexity: O(n^2) where n is the length of the array.
• Pros: It works better than brute force. It balances complexity and performance.
• Cons: It needs extra space for dynamic programming arrays.

Java Implementation

public int maxSumWithOneSwap(int[] arr) {
    int n = arr.length;
    int[] dp = new int[n];
    dp[0] = arr[0];
    for (int i = 1; i < n; i++) {
        dp[i] = Math.max(dp[i - 1] + arr[i], arr[i]);
    }
    int maxSum = dp[n - 1];
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int swappedSum = dp[n - 1] - arr[i] + arr[j];
            maxSum = Math.max(maxSum, swappedSum);
        }
    }
    return maxSum;
}

Python Implementation

def max_sum_with_one_swap(arr):
    n = len(arr)
    dp = [0] * n
    dp[0] = arr[0]
    for i in range(1, n):
        dp[i] = max(dp[i - 1] + arr[i], arr[i])
    max_sum = dp[n - 1]
    for i in range(n):
        for j in range(i + 1, n):
            swapped_sum = dp[n - 1] - arr[i] + arr[j]
            max_sum = max(max_sum, swapped_sum)
    return max_sum

3. Greedy Approach

• Description: We find two elements to swap that can give the most increase in the subsequence sum.
• Time Complexity: O(n) to find the elements and O(1) for the swap.
• Pros: It is quick and efficient.
• Cons: It might not always give the best solution compared to dynamic programming in some cases.

4. Optimized Dynamic Programming

• Description: This builds on dynamic programming. We reduce the number of swaps and keep a running maximum.
• Time Complexity: O(n).
• Pros: It is efficient and keeps the structure of dynamic programming.
• Cons: It is more complex to implement.

Summary of Comparison

• Brute Force: Simple but not good for big inputs.
• Dynamic Programming: Effective and works for many cases.
• Greedy: Fast but can give bad results sometimes.
• Optimized DP: Good mix of efficiency and effectiveness but harder to implement.

In conclusion, the choice of approach depends on the problem’s needs. This includes the size of the input array and how fast we need the solution. For more complex dynamic programming problems, we can look at articles like Dynamic Programming - Maximum Subarray (Kadane’s Algorithm) for more information.

Code Walkthrough for Maximum Sum of Subsequence with One Swap in Java

The problem of finding the maximum sum of a subsequence with one swap can be solved well with dynamic programming. In this part, we will show how to do this in Java.

Java Implementation

The main idea is to use a dynamic programming array. This array will help us track the maximum sum we can get up to each index. We will also look for the maximum sum if we swap any two elements in the subsequence.

Here is the Java code for the solution:

public class MaxSumSubsequenceOneSwap {
    public static int maxSumWithOneSwap(int[] nums) {
        int n = nums.length;
        if (n < 2) return 0;

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

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

        // Now check for the maximum sum with one swap
        int totalSum = 0;
        for (int num : nums) {
            totalSum += num;
        }

        // Find the best two elements to swap
        int bestSwapGain = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int gain = Math.max(nums[i], nums[j]) - Math.min(nums[i], nums[j]);
                bestSwapGain = Math.max(bestSwapGain, gain);
            }
        }

        return Math.max(maxSum, totalSum + bestSwapGain);
    }

    public static void main(String[] args) {
        int[] nums = {3, 5, 1, 2, 9};
        System.out.println("Maximum Sum of Subsequence with One Swap: " + maxSumWithOneSwap(nums));
    }
}

Explanation of the Code

1. Initialization:
   • We start with a dynamic programming array dp. Here, dp[i] keeps the maximum sum of the subsequence that ends at index i.
   • We set the first element to nums[0].
2. Dynamic Programming Calculation:
   • For each element, we update dp[i]. We either add the current element to the maximum sum from the previous element or we start fresh from the current element.
3. Calculating Total Sum:
   • We go through the original array to find the total sum of all elements.
4. Best Swap Gain Calculation:
   • We look at all pairs of indices (i, j) to find the gain from swapping nums[i] and nums[j].
   • The gain is the difference between the bigger and smaller of the two numbers.
5. Final Result:
   • We return the maximum value between the best sum we found without a swap and the total sum changed by the best swap gain.

This code efficiently calculates the maximum sum of a subsequence with one swap. It works in O(n^2) time. This is okay for medium input sizes.

Code Walkthrough for Maximum Sum of Subsequence with One Swap in Python

We will solve the problem of finding the maximum sum of a subsequence with one swap. We can use a simple dynamic programming method. The main idea is to go through the array. We will keep track of the maximum sums we can get with and without a swap.

Here is a basic Python code that shows how we can do this:

def maxSumWithOneSwap(arr):
    n = len(arr)
    if n <= 1:
        return sum(arr)
    
    # Step 1: Calculate the initial maximum sum without any swaps using Kadane's algorithm.
    max_sum_no_swap = float('-inf')
    current_sum = 0
    for num in arr:
        current_sum += num
        if current_sum > max_sum_no_swap:
            max_sum_no_swap = current_sum
        if current_sum < 0:
            current_sum = 0

    # Step 2: Iterate to find the maximum sum with one swap.
    max_sum_with_swap = max_sum_no_swap
    for i in range(n):
        for j in range(i + 1, n):
            # Swap arr[i] and arr[j] and calculate the potential new sum
            arr[i], arr[j] = arr[j], arr[i]

            # Calculate the maximum sum after the swap using Kadane's algorithm
            current_sum = 0
            for num in arr:
                current_sum += num
                if current_sum > max_sum_with_swap:
                    max_sum_with_swap = current_sum
                if current_sum < 0:
                    current_sum = 0
            
            # Swap back to original order
            arr[i], arr[j] = arr[j], arr[i]

    return max_sum_with_swap

# Example usage:
arr = [1, -2, 0, 3]
result = maxSumWithOneSwap(arr)
print(f"The maximum sum of the subsequence with one swap is: {result}")

Explanation of the Code:

• Initialization: First, we check the length of the array. If it is one or less, we just return the sum of the array.
• Kadane’s Algorithm: The first part of the code finds the maximum sum of the subsequence without any swaps.
• Swap Logic: We use nested loops to check each pair of indices. We will swap the values, calculate the new maximum subsequence sum using Kadane’s algorithm, and then swap back.
• Final Output: The function gives back the maximum sum we found, whether we did a swap or not.

This Python function shows how we can use the dynamic programming method to find the maximum sum of a subsequence with one swap. It keeps the time efficient while still being easy to understand. For more on dynamic programming, we can check out Dynamic Programming: Maximum Subarray (Kadane’s Algorithm).

Frequently Asked Questions

1. What is the Maximum Sum of Subsequence with One Swap problem?

The Maximum Sum of Subsequence with One Swap problem is about finding a subsequence in an array. We want to make the sum of its elements as high as possible. We can swap two elements just once. We can solve this problem well using dynamic programming. This helps us look at different subsequences and find the best swap to get the highest sum.

2. How can I implement the Maximum Sum of Subsequence with One Swap in Java?

To implement the Maximum Sum of Subsequence with One Swap in Java, we can use a dynamic programming method. First, we calculate the maximum sum of the original subsequence. Then, we go through the array to check possible swaps. We update the maximum sum based on the swaps we find to get the best answer. For a clear example, see the dynamic programming approach for Maximum Sum of Subsequence with One Swap in Java.

3. What is the time complexity of the Maximum Sum of Subsequence with One Swap solution?

The time complexity for the Maximum Sum of Subsequence with One Swap solution depends on how we use dynamic programming. Usually, the solution works in O(n^2) time. Here, n is the length of the array. This happens because we need to check possible swaps for each element in the subsequence. It is important to make the approach as efficient as we can.

4. Can I solve the Maximum Sum of Subsequence with One Swap problem using Python?

Yes, we can solve the Maximum Sum of Subsequence with One Swap problem well using Python. We can use dynamic programming techniques like those in Java. We can make a function that finds the maximum sum of the subsequence while thinking about one swap. For a hands-on example, look at the dynamic programming approach for Maximum Sum of Subsequence with One Swap in Python.

5. Are there other related dynamic programming problems I can explore?

Yes, there are many other dynamic programming problems that can help us learn more. For example, we can look at the maximum subarray sum using Kadane’s algorithm or the maximum sum of non-adjacent elements. Checking these problems will give us a good base in dynamic programming techniques.