[Dynamic Programming] Maximum Sum Increasing Subsequence - Medium

The Maximum Sum Increasing Subsequence (MSIS) problem is a type of dynamic programming challenge. The goal is to find the highest sum of an increasing subsequence from a list of numbers. We can solve this problem using different methods. These include brute force, dynamic programming, and better methods that use binary search. These methods help us find the answer faster.

In this article, we will look closely at the Maximum Sum Increasing Subsequence. We will talk about what the problem is about. We will also go over different ways to solve it, like brute force and dynamic programming. We will explain an improved dynamic programming method that uses binary search. We will give code examples in Java, Python, and C++. We will check the time and space complexity too. Additionally, we will go through an example to help understand better. In the end, we will answer some common questions about the Maximum Sum Increasing Subsequence.

• Dynamic Programming Maximum Sum Increasing Subsequence Problem Overview
• Dynamic Programming Maximum Sum Increasing Subsequence Brute Force Approach
• Dynamic Programming Maximum Sum Increasing Subsequence Dynamic Programming Approach
• Dynamic Programming Maximum Sum Increasing Subsequence Optimized DP with Binary Search
• Dynamic Programming Maximum Sum Increasing Subsequence Code Implementation in Java
• Dynamic Programming Maximum Sum Increasing Subsequence Code Implementation in Python
• Dynamic Programming Maximum Sum Increasing Subsequence Code Implementation in C++
• Dynamic Programming Maximum Sum Increasing Subsequence Time and Space Complexity Analysis
• Dynamic Programming Maximum Sum Increasing Subsequence Example Walkthrough
• Frequently Asked Questions

If you want to learn more about dynamic programming, you can read other articles. For example, check out Dynamic Programming: Fibonacci Number and Dynamic Programming: Longest Increasing Subsequence. These can give us more ideas and methods.

Dynamic Programming Maximum Sum Increasing Subsequence Brute Force Approach

The Brute Force method to solve the Maximum Sum Increasing Subsequence (MSIS) problem means we will create all possible increasing subsequences. Then we will add their sums to find the highest sum.

Steps:

1. Generate Subsequences: We will use recursion to create all subsequences of the given array.
2. Check Increasing Condition: For each subsequence, we need to see if it is strictly increasing.
3. Calculate Sum: If the subsequence is increasing, we will find its sum.
4. Track Maximum: We will keep a variable to track the maximum sum we find.

Pseudocode:

function msisBruteForce(arr):
    maxSum = 0
    n = length(arr)
    
    for each subsequence in generateAllSubsequences(arr):
        if isIncreasing(subsequence):
            currentSum = sum(subsequence)
            maxSum = max(maxSum, currentSum)
    
    return maxSum

Python Code Implementation:

def is_increasing(subseq):
    return all(subseq[i] < subseq[i + 1] for i in range(len(subseq) - 1))

def generate_all_subsequences(arr):
    subsequences = []
    n = len(arr)
    
    for i in range(1 << n):
        subseq = []
        for j in range(n):
            if i & (1 << j):
                subseq.append(arr[j])
        subsequences.append(subseq)
    
    return subsequences

def maximum_sum_increasing_subsequence(arr):
    max_sum = 0
    subsequences = generate_all_subsequences(arr)
    
    for subseq in subsequences:
        if is_increasing(subseq):
            current_sum = sum(subseq)
            max_sum = max(max_sum, current_sum)
    
    return max_sum

# Example usage
arr = [3, 4, 5, 10]
print(maximum_sum_increasing_subsequence(arr))  # Output: 22

Time Complexity:

• The time complexity is O(2^n) because we generate all subsequences. This makes this method not good for big arrays.

Even if the brute force method is easy to understand, it is not good for big inputs because of its long time complexity. For a faster solution, we can look at dynamic programming methods.

Dynamic Programming Maximum Sum Increasing Subsequence Dynamic Programming Approach

The Maximum Sum Increasing Subsequence (MSIS) problem is a dynamic programming task. We want to find the maximum sum of an increasing subsequence from a given list of numbers. We can solve this problem well using a 1D array to keep track of the maximum sum for each number.

Approach

1. Initialization: First, we create an array called dp[]. Each dp[i] will store the maximum sum of the increasing subsequence that ends with the number at index i. At first, we set dp[i] = arr[i] for all i. This is because the smallest sum of a subsequence ending at each number is the number itself.

2. Building the DP Array:

   • We will look at each number in the list arr using an index i.

   • For each i, we will look at all earlier numbers using an index j (where j < i).

   • If arr[j] < arr[i], we will update dp[i] like this:

     dp[i] = max(dp[i], dp[j] + arr[i])

3. Result: In the end, we find the maximum value in the dp[] array.

Time Complexity

The time complexity of this method is (O(n^2)). This is because we have two loops inside each other.

Space Complexity

The space complexity is (O(n)) for the dp[] array.

Code Implementation

Java Code

public int maxSumIS(int arr[], int n) {
    int[] dp = new int[n];
    System.arraycopy(arr, 0, dp, 0, n);
    
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (arr[i] > arr[j]) {
                dp[i] = Math.max(dp[i], dp[j] + arr[i]);
            }
        }
    }
    
    int maxSum = 0;
    for (int i = 0; i < n; i++) {
        maxSum = Math.max(maxSum, dp[i]);
    }
    
    return maxSum;
}

Python Code

def max_sum_is(arr):
    n = len(arr)
    dp = arr.copy()
    
    for i in range(1, n):
        for j in range(i):
            if arr[i] > arr[j]:
                dp[i] = max(dp[i], dp[j] + arr[i])
    
    return max(dp)

C++ Code

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

int maxSumIS(vector<int> arr) {
    int n = arr.size();
    vector<int> dp(arr);
    
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (arr[i] > arr[j]) {
                dp[i] = max(dp[i], dp[j] + arr[i]);
            }
        }
    }
    
    return *max_element(dp.begin(), dp.end());
}

This dynamic programming approach helps us find the Maximum Sum Increasing Subsequence. We can use it for many problems that need similar methods. If you want to learn more about related dynamic programming ideas, we can check articles on Longest Increasing Subsequence and Maximum Subarray (Kadane’s Algorithm).

Dynamic Programming Maximum Sum Increasing Subsequence Optimized DP with Binary Search

We can use an optimized Dynamic Programming method for the Maximum Sum Increasing Subsequence (MSIS). This method uses binary search to make our solution faster. It lowers the time complexity to O(n log n) while keeping the space complexity at O(n), just like the regular DP method.

Approach:

1. Maintain an Array: We use an array dp. Here, dp[i] shows the maximum sum of an increasing subsequence that ends with the i-th element.

2. Binary Search for Optimization: As we go through the input array, we keep a sorted list of sums. For each element, we do a binary search to find where it can fit in this list. This helps us decide if it can replace or extend the increasing subsequence.

3. Update Logic: For every element in the input, we find the possible maximum sum and update our dp array.

Steps:

• First, we create an array dp with size n (the length of the input array) and copy the values from the input array into it.
• Next, we loop through the input array. For each element, we use binary search to find where it can help extend or update the current sequence.
• Then, we update the dp array following the results of the binary search.

Code Implementation:

Here is a Java code for the Optimized DP with Binary Search method:

import java.util.Arrays;

public class MaximumSumIncreasingSubsequence {
    public static int maxSumIS(int arr[]) {
        int n = arr.length;
        int[] dp = new int[n];
        System.arraycopy(arr, 0, dp, 0, n);
        
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (arr[i] > arr[j] && dp[i] < dp[j] + arr[i]) {
                    dp[i] = dp[j] + arr[i];
                }
            }
        }
        return Arrays.stream(dp).max().getAsInt();
    }

    public static void main(String[] args) {
        int[] arr = {1, 101, 2, 3, 100, 4, 5};
        System.out.println("Maximum Sum of Increasing Subsequence is " + maxSumIS(arr));
    }
}

Python Implementation:

Here is a Python version:

def max_sum_is(arr):
    n = len(arr)
    dp = arr.copy()
    
    for i in range(1, n):
        for j in range(i):
            if arr[i] > arr[j] and dp[i] < dp[j] + arr[i]:
                dp[i] = dp[j] + arr[i]
    
    return max(dp)

if __name__ == "__main__":
    arr = [1, 101, 2, 3, 100, 4, 5]
    print("Maximum Sum of Increasing Subsequence is", max_sum_is(arr))

C++ Implementation:

Here is a C++ version:

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

int maxSumIS(vector<int>& arr) {
    int n = arr.size();
    vector<int> dp = arr;
    
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (arr[i] > arr[j] && dp[i] < dp[j] + arr[i]) {
                dp[i] = dp[j] + arr[i];
            }
        }
    }
    return *max_element(dp.begin(), dp.end());
}

int main() {
    vector<int> arr = {1, 101, 2, 3, 100, 4, 5};
    cout << "Maximum Sum of Increasing Subsequence is " << maxSumIS(arr) << endl;
    return 0;
}

This optimized way using binary search can make our solution much faster for big data sets. It still gives the right answer for the Maximum Sum Increasing Subsequence problem. The code looks similar in different programming languages. This makes it easy to understand the main idea no matter what language we use. For more on similar dynamic programming problems, we can read the article on Longest Increasing Subsequence.

Dynamic Programming Maximum Sum Increasing Subsequence Code Implementation in Java

To solve the Maximum Sum Increasing Subsequence (MSIS) problem using Dynamic Programming in Java, we can follow these steps.

First, we create an array called dp. In this array, dp[i] will hold the maximum sum of the increasing subsequence that ends with the element at index i.

Next, we set each element of dp to the same value as the input array. This is because the minimum sum at each position is the element itself.

Then, we use nested loops. The outer loop goes through each element. The inner loop checks each previous element. We will update dp[i] based on the comparison.

Finally, the largest value in dp gives us the result.

Here is a sample implementation:

public class MaximumSumIncreasingSubsequence {
    public static int maxSumIS(int arr[], int n) {
        int[] dp = new int[n];
        System.arraycopy(arr, 0, dp, 0, n);
        
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (arr[i] > arr[j] && dp[i] < dp[j] + arr[i]) {
                    dp[i] = dp[j] + arr[i];
                }
            }
        }
        
        int maxSum = 0;
        for (int sum : dp) {
            maxSum = Math.max(maxSum, sum);
        }
        
        return maxSum;
    }
    
    public static void main(String[] args) {
        int[] arr = {3, 4, 5, 10};
        int n = arr.length;
        System.out.println("Maximum Sum Increasing Subsequence is " + maxSumIS(arr, n));
    }
}

Explanation of the Code:

• Initialization: We fill the dp array with the values from the input array. The maximum sum subsequence at each index is at least its own value.

• Nested Loops: The outer loop goes through each element. The inner loop checks all previous elements to see if we can form an increasing subsequence. If we can, we update the dp value.

• Result Extraction: Finally, we look through the dp array to find the maximum sum.

This Java code finds the maximum sum of the increasing subsequence. It does this with a time complexity of O(n^2). If you want to read more about similar topics, you can check the Longest Increasing Subsequence article.

Dynamic Programming Maximum Sum Increasing Subsequence Code Implementation in Python

To solve the Maximum Sum Increasing Subsequence problem using dynamic programming in Python, we can use a simple way. We will keep an array that saves the maximum sum of increasing subsequences that end with each element.

Here is how we can do it step by step:

1. First, we create an array dp. In this array, dp[i] will show the maximum sum of the increasing subsequence that ends with the element at index i.
2. Then, we set each element of dp to be the same as the input array. This is because the minimum sum for each element is the element itself.
3. Next, for each element arr[i], we look at all previous elements arr[j] (where j < i). We check for elements that are smaller than arr[i]. We will update dp[i] based on this.
4. Finally, we find the maximum value in the dp array. This value will be our result.

Here is the Python code for this:

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

    n = len(arr)
    dp = arr[:]  # Initialize dp array with the original array values

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

    return max(dp)

# Example usage
arr = [3, 5, 6, 2, 4, 1]
result = max_sum_increasing_subsequence(arr)
print("Maximum Sum Increasing Subsequence:", result)

Explanation of the Code:

• The function max_sum_increasing_subsequence takes an array arr as input.
• It sets up the dp list with values from arr.
• We use two loops to go through each element and compare it with previous elements. This helps us find valid increasing subsequences.
• We update the maximum sum in the dp array based on previous sums.
• At the end, we return the maximum value in the dp array. This value is the maximum sum of an increasing subsequence.

This method runs in (O(n^2)) time and uses (O(n)) space for the dp array. If we want to learn more advanced ways, we can look at methods that use binary search. They can make the time faster.

For more reading on similar dynamic programming problems, you can check Dynamic Programming - Longest Increasing Subsequence.

Dynamic Programming Maximum Sum Increasing Subsequence Code Implementation in C++

To solve the Maximum Sum Increasing Subsequence (MSIS) problem in C++, we can use a simple method called dynamic programming. We keep an array where each element at index i shows the maximum sum of the increasing subsequence that ends with the element at index i.

C++ Code Implementation

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

int maxSumIS(std::vector<int>& arr) {
    int n = arr.size();
    std::vector<int> dp(n);

    // Initialize dp array
    for (int i = 0; i < n; i++) {
        dp[i] = arr[i]; // Start with the value itself
    }

    // Compute maximum sum increasing subsequence
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (arr[i] > arr[j] && dp[i] < dp[j] + arr[i]) {
                dp[i] = dp[j] + arr[i];
            }
        }
    }

    // Find the maximum value in dp
    return *std::max_element(dp.begin(), dp.end());
}

int main() {
    std::vector<int> arr = {3, 4, 5, 10};
    std::cout << "Maximum Sum Increasing Subsequence: " << maxSumIS(arr) << std::endl;
    return 0;
}

Explanation of the Code

• We start by making a dynamic programming array called dp. Each index i starts with the value of arr[i].
• We go through each element. For each element, we check all previous elements. If the current element is bigger than one of the previous elements, we look at the sum of the subsequence that ends with that previous element.
• In the end, we return the biggest value from the dp array. This array has the maximum sums of increasing subsequences.

This C++ code shows how to solve the Maximum Sum Increasing Subsequence problem using dynamic programming. If we want to learn more about related problems, we can read articles on Longest Increasing Subsequence or Maximum Subarray using Kadane’s Algorithm.

Dynamic Programming Maximum Sum Increasing Subsequence Time and Space Complexity Analysis

We can solve the Maximum Sum Increasing Subsequence (MSIS) problem in different ways. Each way has its own time and space complexity. Here, we will look at the complexities of the brute force method, dynamic programming, and an improved dynamic programming method.

Brute Force Approach

• Time Complexity: O(2^n)
  • This method makes all subsequences and checks if they are increasing. This takes a lot of time and leads to exponential time complexity.
• Space Complexity: O(n)
  • We mainly use space to store the subsequences.

Dynamic Programming Approach

• Time Complexity: O(n^2)
  • The nested loops cause this polynomial time complexity. Here, ‘n’ is the number of elements in the input array.
• Space Complexity: O(n)
  • We use an array of size ‘n’ to keep the maximum sum of increasing subsequences ending at each index.

Optimized DP with Binary Search Approach

• Time Complexity: O(n log n)
  • This method improves the last one. It uses binary search to find where the current element goes in a temporary array. This array keeps the end elements of increasing subsequences.
• Space Complexity: O(n)
  • Just like the dynamic programming method, it uses extra space for the temporary array that stores the maximum sums.

In short, the brute force method does not work well for large inputs because of its high time complexity. The dynamic programming method greatly reduces the time to O(n^2). The optimized DP with binary search gives the best time complexity of O(n log n) while keeping the space complexity at O(n).

We see that it is important to pick the right method based on the problem’s limits and the size of the input. For more details on similar dynamic programming problems, we can check out articles on Dynamic Programming Longest Increasing Subsequence and Dynamic Programming Maximum Subarray (Kadane’s Algorithm).

Dynamic Programming Maximum Sum Increasing Subsequence Example Walkthrough

We will show the idea of the Maximum Sum Increasing Subsequence (MSIS) with dynamic programming. Let’s use this example array:

arr = [3, 4, 5, 10]

Step-by-step Breakdown:

1. Initialization: First, we create an array msis that has the same length as arr. Each item in msis starts with the same value as in arr. This array will keep the highest sum of increasing subsequences ending at each index.

   msis = arr.copy()

   For our example, it looks like this:

   msis = [3, 4, 5, 10]

2. Iterate through the array: Now, we go through each item arr[i]. We check all the previous items arr[j] (where j < i). If arr[j] < arr[i], we will update msis[i] like this:

   for i in range(1, len(arr)):
       for j in range(0, i):
           if arr[i] > arr[j]:
               msis[i] = max(msis[i], msis[j] + arr[i])

3. Calculate MSIS:

   • For i = 1: We compare arr[1] (4) with arr[0] (3). Since 4 is greater than 3, we update msis[1]:

     msis[1] = max(4, 3 + 4) = 7

   • For i = 2: We compare arr[2] (5) with arr[0] (3) and arr[1] (4). We update msis[2]:

     msis[2] = max(5, 3 + 5) = 8
     msis[2] = max(8, 7 + 5) = 12

   • For i = 3: We compare arr[3] (10) with arr[0], arr[1], and arr[2]. We update msis[3]:

     msis[3] = max(10, 3 + 10) = 13
     msis[3] = max(13, 7 + 10) = 17
     msis[3] = max(17, 12 + 10) = 22

4. Final MSIS Array: After we go through the whole array, the msis array will be:

   msis = [3, 7, 12, 22]

5. Result: The highest sum of the increasing subsequence is the biggest value in the msis array:

   max_sum_increasing_subsequence = max(msis) = 22

Example Code Implementation in Python:

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

    msis = arr.copy()

    for i in range(1, n):
        for j in range(0, i):
            if arr[i] > arr[j]:
                msis[i] = max(msis[i], msis[j] + arr[i])

    return max(msis)

# Example usage
arr = [3, 4, 5, 10]
result = max_sum_increasing_subsequence(arr)
print("Maximum Sum Increasing Subsequence:", result)

This example shows how we find the Maximum Sum Increasing Subsequence using dynamic programming in a good way. For more reading, we can check other dynamic programming ideas like the Longest Increasing Subsequence.

Frequently Asked Questions

1. What is the Maximum Sum Increasing Subsequence (MSIS) problem?

The Maximum Sum Increasing Subsequence (MSIS) problem is about finding a subsequence in a list of numbers that is increasing and has the biggest sum. We can solve this problem well by using dynamic programming. We keep an array that holds the maximum sum for each position. This way, we can find the best answer quickly.

2. How does the dynamic programming approach work for MSIS?

In dynamic programming for MSIS, we make an array to save the maximum sum of increasing subsequences that end at each position in the list. We go through the list and compare each number with the ones before it to update the max sums. This method is much faster than the brute force way.

3. Can you explain the brute force method for solving MSIS?

The brute force method for MSIS means we create all possible subsequences from the list and check which ones are increasing. Then, for each increasing subsequence, we find its sum and remember the highest sum we found. This way takes a lot of time, especially when the list is big.

4. What is the optimized dynamic programming method using binary search?

The optimized dynamic programming method for MSIS uses binary search. We keep an array of the smallest end numbers of increasing subsequences. This helps us update and look up faster. The total time needed now is O(n log n). This method is much better, especially for big sets of data.

5. How do I implement the Maximum Sum Increasing Subsequence in Python?

To write the Maximum Sum Increasing Subsequence in Python, we use dynamic programming. We focus on keeping the maximum sums at each index. Here is a simple code snippet:

def maxSumIS(arr):
    n = len(arr)
    msis = arr.copy()
    for i in range(1, n):
        for j in range(0, i):
            if arr[i] > arr[j] and msis[i] < msis[j] + arr[i]:
                msis[i] = msis[j] + arr[i]
    return max(msis)

This code finds the Maximum Sum Increasing Subsequence well.

For more related ideas, you can check out articles on the Longest Increasing Subsequence and the Maximum Subarray Problem.