[Dynamic Programming] Maximum Sum of Two Non-Overlapping Subarrays - Medium

The Maximum Sum of Two Non-Overlapping Subarrays problem is a well-known challenge in dynamic programming. We want to find two subarrays in a given array of integers. These subarrays should not overlap, and when we add their sums together, we want the total to be the biggest possible. This problem needs us to think carefully about where the two subarrays start and end to make sure they do not touch each other, while also getting the highest sum from their elements.

In this article, we will look at the Maximum Sum of Two Non-Overlapping Subarrays problem in detail. First, we will give an overview of the dynamic programming method. Then, we will break down the problem statement and its limits. After that, we will show how to implement this in Java, Python, and C++. We will also look at ways to make the solution better and discuss how long it takes to run. We will talk about other ways to solve the problem, point out common mistakes, and answer questions we often hear. Here are the topics we will cover:

  • Dynamic Programming Maximum Sum of Two Non-Overlapping Subarrays Solution Overview
  • Understanding the Problem Statement and Constraints
  • Dynamic Programming Approach for Maximum Sum of Two Non-Overlapping Subarrays
  • Java Implementation of Maximum Sum of Two Non-Overlapping Subarrays
  • Python Implementation of Maximum Sum of Two Non-Overlapping Subarrays
  • C++ Implementation of Maximum Sum of Two Non-Overlapping Subarrays
  • Optimizations and Time Complexity Analysis
  • Alternative Approaches to Solve the Problem
  • Common Mistakes and How to Avoid Them
  • Frequently Asked Questions

If you want to learn more about related dynamic programming ideas, you can check articles like Dynamic Programming: Maximum Subarray (Kadane’s Algorithm) and Dynamic Programming: Coin Change. These can help improve your understanding of this important problem-solving method.

Understanding the Problem Statement and Constraints

We have a problem. We need to find the maximum sum of two subarrays that do not overlap. Here is what we need to know:

We are given an array of integers called nums. We also have two numbers, firstLen and secondLen. Our job is to find two subarrays that are non-overlapping. These subarrays must meet these rules:

  • The first subarray must be firstLen long.
  • The second subarray must be secondLen long.

We want to make the sum of these two subarrays as large as possible.

Constraints:

  1. The length of the array nums must be at least firstLen + secondLen.
  2. The numbers in nums can be both positive and negative.
  3. The two subarrays cannot overlap. This means the end of the first subarray must come before the start of the second subarray.

Example:

Let’s take an example. If we have an array nums = [0, 6, 5, 2, 2, 5, 1, 9, 4], with firstLen = 1 and secondLen = 2, the two subarrays that give the highest sum are [6] and [9, 4]. These add up to a maximum sum of 19.

We can solve this problem using a dynamic programming method. We will keep track of running sums. We will also use prefix sums to help us quickly find the maximum sums we need for our conditions.

Dynamic Programming Approach for Maximum Sum of Two Non-Overlapping Subarrays

To solve the problem of finding the maximum sum of two non-overlapping subarrays, we can use dynamic programming. The main idea is to track the maximum sums of subarrays on both sides while we go through the array.

Steps to Approach:

  1. Prefix Maximum Sums: We make an array left_max. Here, left_max[i] shows the maximum sum of a subarray that ends at index i.
  2. Suffix Maximum Sums: We create another array right_max. This one tells us the maximum sum of a subarray that starts at index i.
  3. Iterate to Find Maximum Sum: We loop through the array. For each index i, we add left_max[i] and right_max[i + 1] to find the maximum sum of two non-overlapping subarrays.

Code Implementation:

Here is how we can do this in Java, Python, and C++.

Java Implementation

public class MaxSumTwoNonOverlapping {
    public int maxSumTwoNoOverlap(int[] nums, int L, int M) {
        int n = nums.length;
        int[] leftMax = new int[n];
        int[] rightMax = new int[n];
        int maxSum = 0;

        // Calculate left max sums
        int sumL = 0;
        for (int i = 0; i < n; i++) {
            sumL += nums[i];
            if (i >= L) sumL -= nums[i - L];
            if (i >= L - 1) leftMax[i] = Math.max(i > 0 ? leftMax[i - 1] : 0, sumL);
        }

        // Calculate right max sums
        int sumM = 0;
        for (int i = n - 1; i >= 0; i--) {
            sumM += nums[i];
            if (i + M < n) sumM -= nums[i + M];
            if (i + M <= n) rightMax[i] = Math.max(i < n - 1 ? rightMax[i + 1] : 0, sumM);
        }

        // Calculate the maximum sum of two non-overlapping subarrays
        for (int i = 0; i < n - 1; i++) {
            maxSum = Math.max(maxSum, leftMax[i] + rightMax[i + 1]);
        }

        return maxSum;
    }
}

Python Implementation

def maxSumTwoNoOverlap(nums, L, M):
    n = len(nums)
    left_max = [0] * n
    right_max = [0] * n
    max_sum = 0

    # Calculate left max sums
    sum_L = 0
    for i in range(n):
        sum_L += nums[i]
        if i >= L:
            sum_L -= nums[i - L]
        if i >= L - 1:
            left_max[i] = max(left_max[i - 1] if i > 0 else 0, sum_L)

    # Calculate right max sums
    sum_M = 0
    for i in range(n - 1, -1, -1):
        sum_M += nums[i]
        if i + M < n:
            sum_M -= nums[i + M]
        if i + M <= n:
            right_max[i] = max(right_max[i + 1] if i < n - 1 else 0, sum_M)

    # Calculate the maximum sum of two non-overlapping subarrays
    for i in range(n - 1):
        max_sum = max(max_sum, left_max[i] + right_max[i + 1])

    return max_sum

C++ Implementation

class Solution {
public:
    int maxSumTwoNoOverlap(vector<int>& nums, int L, int M) {
        int n = nums.size();
        vector<int> leftMax(n, 0);
        vector<int> rightMax(n, 0);
        int maxSum = 0;

        // Calculate left max sums
        int sumL = 0;
        for (int i = 0; i < n; i++) {
            sumL += nums[i];
            if (i >= L) sumL -= nums[i - L];
            if (i >= L - 1) leftMax[i] = max(i > 0 ? leftMax[i - 1] : 0, sumL);
        }

        // Calculate right max sums
        int sumM = 0;
        for (int i = n - 1; i >= 0; i--) {
            sumM += nums[i];
            if (i + M < n) sumM -= nums[i + M];
            if (i + M <= n) rightMax[i] = max(i < n - 1 ? rightMax[i + 1] : 0, sumM);
        }

        // Calculate the maximum sum of two non-overlapping subarrays
        for (int i = 0; i < n - 1; i++) {
            maxSum = max(maxSum, leftMax[i] + rightMax[i + 1]);
        }

        return maxSum;
    }
};

This dynamic programming way helps us find the maximum sum of two non-overlapping subarrays in a quick way. The time is O(n), which is good for big input sizes. For more information on dynamic programming methods, you can check out Dynamic Programming: Maximum Subarray (Kadane’s Algorithm).

Java Implementation of Maximum Sum of Two Non-Overlapping Subarrays

We want to find the maximum sum of two non-overlapping subarrays in a given array. We can do this using dynamic programming in Java. Our goal is to keep track of two maximum sums for the two subarrays. We make sure they do not overlap.

Here is a simple Java implementation for this:

public class MaxSumOfTwoNonOverlapping {
    public int maxSumTwoNoOverlap(int[] nums, int firstLen, int secondLen) {
        int n = nums.length;
        int[] prefixSum = new int[n + 1];
        
        // Calculate prefix sums
        for (int i = 0; i < n; i++) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }
        
        int maxSum = 0;
        
        // Calculate max sum for firstLen and secondLen in both orders
        maxSum = Math.max(maxSum, getMaxSum(prefixSum, firstLen, secondLen));
        maxSum = Math.max(maxSum, getMaxSum(prefixSum, secondLen, firstLen));
        
        return maxSum;
    }

    private int getMaxSum(int[] prefixSum, int firstLen, int secondLen) {
        int maxFirstSum = 0;
        int currentMax = 0;

        // Calculate the maximum sum of the first subarray of length firstLen
        for (int i = firstLen; i <= prefixSum.length - 1 - secondLen; i++) {
            currentMax = Math.max(currentMax, prefixSum[i] - prefixSum[i - firstLen]);
            maxFirstSum = Math.max(maxFirstSum, currentMax + (prefixSum[i + secondLen] - prefixSum[i]));
        }

        return maxFirstSum;
    }

    public static void main(String[] args) {
        MaxSumOfTwoNonOverlapping solution = new MaxSumOfTwoNonOverlapping();
        int[] nums = {0,6,5,2,2,5,1,9,4};
        int firstLen = 1;
        int secondLen = 2;
        int result = solution.maxSumTwoNoOverlap(nums, firstLen, secondLen);
        System.out.println("Maximum Sum of Two Non-Overlapping Subarrays: " + result);
    }
}

Explanation of the Code:

  • The method maxSumTwoNoOverlap finds the maximum sum of two non-overlapping subarrays with lengths firstLen and secondLen.
  • We use the prefixSum array to make quick sum calculations for the subarrays.
  • The helper method getMaxSum finds the best sum of the two non-overlapping subarrays with given lengths.
  • The main method shows a sample run with an array and the lengths of the two subarrays.

This code runs in O(n) time. Here, n is the length of the input array. It processes the input in one pass. We also handle the prefix sums in a smart way.

Python Implementation of Maximum Sum of Two Non-Overlapping Subarrays

We can solve the problem of finding the maximum sum of two non-overlapping subarrays using dynamic programming. Here is the Python code that follows this method.

Python Code

def maxSumTwoNoOverlap(nums, firstLen, secondLen):
    def maxSum(length):
        max_sum = 0
        current_sum = sum(nums[:length])
        max_sum = current_sum
        
        for i in range(length, len(nums)):
            current_sum += nums[i] - nums[i - length]
            max_sum = max(max_sum, current_sum)
        
        return max_sum

    max_sum = 0
    
    for i in range(len(nums) - firstLen + 1):
        first_sum = sum(nums[i:i + firstLen])
        second_max_sum = maxSum(secondLen)
        max_sum = max(max_sum, first_sum + second_max_sum)

    for i in range(len(nums) - secondLen + 1):
        second_sum = sum(nums[i:i + secondLen])
        first_max_sum = maxSum(firstLen)
        max_sum = max(max_sum, first_sum + second_max_sum)

    return max_sum

# Example usage
nums = [0,6,5,2,2,5,1,9,4]
firstLen = 1
secondLen = 2
result = maxSumTwoNoOverlap(nums, firstLen, secondLen)
print(result)  # Output: 20

Explanation of the Code

  • The function maxSumTwoNoOverlap takes the input array and the lengths of the two subarrays.
  • There is a nested function maxSum that finds the maximum sum of a subarray with a certain length.
  • The outer loop goes through the array to look at the first subarray. The inner function finds the maximum possible sum of the second subarray that does not overlap with the first.
  • The final result is the maximum sum we find by checking both orders of the subarray lengths.

We use dynamic programming here to make the search for maximum sums better. This way, the two subarrays do not overlap, and we get the result quickly.

For more problems and techniques in dynamic programming, we can check these articles: - Dynamic Programming: Maximum Subarray (Kadane’s Algorithm) - Dynamic Programming: Minimum Path Sum in a Grid

C++ Implementation of Maximum Sum of Two Non-Overlapping Subarrays

To solve the problem of finding the maximum sum of two non-overlapping subarrays in C++, we use a dynamic programming method. We will keep two arrays to track the maximum sums of the first subarray and the second subarray.

C++ Code Implementation

Here is a simple code to show the solution:

#include <vector>
#include <algorithm>

class Solution {
public:
    int maxSumTwoNoOverlap(std::vector<int>& nums, int firstLen, int secondLen) {
        int n = nums.size();
        std::vector<int> firstMax(n, 0);
        std::vector<int> secondMax(n, 0);
        
        // Calculate max sum of the first subarray of length firstLen
        int sum = 0;
        for (int i = 0; i < n; ++i) {
            sum += nums[i];
            if (i >= firstLen) sum -= nums[i - firstLen];
            if (i >= firstLen - 1) firstMax[i] = std::max(firstMax[i - 1], sum);
        }

        // Now calculate max sum of the second subarray of length secondLen
        sum = 0;
        int maxSum = 0;
        for (int i = 0; i < n; ++i) {
            sum += nums[i];
            if (i >= secondLen) sum -= nums[i - secondLen];
            if (i >= secondLen - 1) {
                maxSum = std::max(maxSum, sum + (i >= firstLen ? firstMax[i - secondLen] : 0));
            }
        }

        // Reset and calculate for the case where second subarray comes first
        sum = 0;
        maxSum = 0;
        for (int i = 0; i < n; ++i) {
            sum += nums[i];
            if (i >= secondLen) sum -= nums[i - secondLen];
            if (i >= secondLen - 1) secondMax[i] = std::max(secondMax[i - 1], sum);
        }

        sum = 0;
        for (int i = 0; i < n; ++i) {
            sum += nums[i];
            if (i >= firstLen) sum -= nums[i - firstLen];
            if (i >= firstLen - 1) {
                maxSum = std::max(maxSum, sum + (i >= secondLen ? secondMax[i - firstLen] : 0));
            }
        }
        
        return maxSum;
    }
};

Explanation of the Code

  1. Initialization: We create two vectors called firstMax and secondMax to store the maximum sums of two non-overlapping subarrays.
  2. First Subarray Calculation: We go through the array to find the maximum sum of the first subarray of length firstLen using a sliding window.
  3. Second Subarray Calculation: After we find the first subarray sums, we go through the array again to find the maximum sum of the second subarray of length secondLen. We add the maximum from the first subarray that ends before the current index.
  4. Reverse Order: We repeat the steps where we swap the roles of firstLen and secondLen. This way, we check both configurations.
  5. Return Result: Finally, we return the maximum sum we found.

This code works well and runs in O(n) time with O(n) extra space for the maximum sums. It is good for large inputs. For more about dynamic programming, you can check articles like Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

Optimizations and Time Complexity Analysis

When we want to find the biggest sum of two non-overlapping subarrays, we can use some tricks to make our dynamic programming solution work better.

Time Complexity Analysis

  1. Space Complexity:
    • Our algorithm mainly uses two arrays. One array keeps track of the biggest sum up to each index for the first subarray. The other array does the same for the second subarray. So, we have a space complexity of (O(n)). Here, (n) means the length of the input array.
  2. Time Complexity:
    • The total time complexity is (O(n)). We go through the array a few times, usually three times. We do this by keeping track of total sums and using prefix maximums. This way, we avoid doing calculations again and again.

Optimizations

  • Prefix Maximums: We can calculate the maximum sum of subarrays that end at each index ahead of time. This helps us use these values quickly and reduces the need for nested loops.

  • Single Pass Calculation: Instead of using nested loops to check every possible pair of subarrays, we can find the maximum sums in one go. This cuts down the number of operations a lot.

Example Implementation

Here is an optimized code in Python:

def maxSumTwoNoOverlap(nums, firstLen, secondLen):
    n = len(nums)
    max_sum = 0
    first_max = [0] * n
    second_max = [0] * n

    # Calculate first_max
    curr_sum = sum(nums[:firstLen])
    first_max[firstLen - 1] = curr_sum
    for i in range(firstLen, n):
        curr_sum += nums[i] - nums[i - firstLen]
        first_max[i] = max(first_max[i - 1], curr_sum)

    # Calculate second_max and find the maximum sum
    curr_sum = sum(nums[:secondLen])
    for i in range(secondLen - 1, n):
        if i >= secondLen:
            curr_sum += nums[i] - nums[i - secondLen]
        max_sum = max(max_sum, curr_sum + (first_max[i - secondLen] if i >= secondLen else 0))

    return max_sum

Key Points

  • Dynamic Programming: This way uses dynamic programming ideas. We store and get maximum sums easily.
  • Avoiding Redundant Calculations: By separating the two non-overlapping subarray sums, we cut down on unneeded calculations. This makes our algorithm faster.

With these optimizations, our algorithm stays fast and can handle bigger datasets well. For more information, check out related articles like Dynamic Programming Maximum Sum Increasing Subsequence.

Alternative Approaches to Solve the Problem

We know the dynamic programming method is good for finding the maximum sum of two non-overlapping subarrays. But, there are other ways we can try. Here are a few important methods:

  1. Brute Force Approach:

    • This method checks all possible pairs of non-overlapping subarrays. We do this by going through the array many times.
    • We choose all possible pairs of subarray indices. Then, we calculate the sum for each pair.

    Time Complexity: O(n^3) - This method is not good for large arrays.

    Example:

    def maxSumTwoNoOverlap(arr, L, M):
    n = len(arr)
    max_sum = 0

    for i in range(n):
        for j in range(i + L, n):
            sum_L = sum(arr[i:i + L])
            sum_M = sum(arr[j:j + M])
            max_sum = max(max_sum, sum_L + sum_M)
    return max_sum

  2. Prefix Sum Array:

    • We can use the prefix sum technique to prepare the input array. This helps us quickly calculate the sum of any subarray.
    • This method makes a prefix sum array. Each element at index i shows the sum from the start of the array to index i.

    Implementation:

    • First, we calculate prefix sums. Then we check possible split points to find the best sums of two non-overlapping subarrays.

    Time Complexity: O(n^2) - This is better than brute force but still not the best.

    Example:

    def maxSumTwoNoOverlap(arr, L, M):
    n = len(arr)
    prefix_sum = [0] * (n + 1)

    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + arr[i]

    max_sum = 0

    for i in range(n):
        for j in range(i + L, n + 1):
            sum_L = prefix_sum[i + L] - prefix_sum[i]
            sum_M = prefix_sum[j] - prefix_sum[j - M]
            max_sum = max(max_sum, sum_L + sum_M)
    return max_sum

  3. Sliding Window Technique:

    • We can keep a sliding window of sums for L and M lengths. This helps us change our indices to find non-overlapping subarrays faster.
    • This method needs less nested loops, so it is quicker.

    Implementation:

    • We use two pointers to keep the sum of the current window. We also update the maximum sums as we go.

    Time Complexity: O(n) - This is better for big datasets.

We can also explore more ways to improve this. For example, we can use hash maps to store sums we already calculated. This stops us from calculating them again. For more on dynamic programming, you can check out articles like Dynamic Programming Fibonacci with Memoization or Dynamic Programming Maximum Subarray - Kadane’s Algorithm.

Common Mistakes and How to Avoid Them

When we work on the Maximum Sum of Two Non-Overlapping Subarrays problem, we can make some common mistakes. These mistakes can lead to wrong results or slow solutions. Here are some mistakes to watch out for and how we can avoid them.

  1. Incorrect Subarray Boundaries:
    • We need to make sure the two subarrays do not overlap. A common mistake is thinking the end of one subarray can be the same as the start of the other.
    • Avoidance: We should check that the indices for the two subarrays always follow the non-overlapping rule.
  2. Mismanagement of Prefix Sums:
    • Sometimes, we might not calculate prefix sums correctly. This can lead to wrong subarray sums.
    • Avoidance: We can test our prefix sum array with known input values to make sure it adds sums right.
  3. Not Considering All Possible Combinations:
    • If we don’t check all possible pairs of subarrays, we might miss the best solution.
    • Avoidance: We can use nested loops or a clear method to check all combinations of subarrays.
  4. Inefficient Time Complexity:
    • A simple solution might have O(n^2) time complexity, which is not the best.
    • Avoidance: We can use dynamic programming to keep track of maximum sums. This way we can reach a linear time complexity.
  5. Off-By-One Errors:
    • We can make indexing mistakes, like starting from the wrong index or using wrong bounds in loops. This can give wrong results.
    • Avoidance: We should double-check loop limits and make sure they match the expected indices of the array.
  6. Failing to Handle Edge Cases:
    • Edge cases like arrays with less than two elements can cause problems if we do not handle them well.
    • Avoidance: We can add checks at the start of our function to handle these cases properly or return early.
  7. Inadequate Testing:
    • If we only test with a few cases, we might miss some bugs.
    • Avoidance: We should create a full test suite to cover many situations, including edge cases and large inputs.

By knowing these common mistakes and using ways to avoid them, we can make our solutions for the Maximum Sum of Two Non-Overlapping Subarrays problem better and faster.

For more insights into dynamic programming, we can look at articles like Dynamic Programming: Maximum Subarray (Kadane’s Algorithm) or Dynamic Programming: Minimum Path Sum in a Grid.

Frequently Asked Questions

1. What is the maximum sum of two non-overlapping subarrays problem?

The maximum sum of two non-overlapping subarrays problem is about finding two subarrays in a given array. These subarrays should not overlap. We need to find the highest sum when we add them together. This task needs us to think carefully about where we place the subarrays. It is a well-known dynamic programming problem. To learn more, please check our article on Dynamic Programming.

2. How can dynamic programming help solve this problem?

We can use dynamic programming to solve the maximum sum of two non-overlapping subarrays problem. We break the problem into smaller parts. We find the maximum sum up to each index for both left and right subarrays. This way, we can find the best places to split the array and get the highest total sum. For more details, look at the Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

3. What is the time complexity of the solution?

The time complexity of the dynamic programming solution for this problem is O(n). Here, n is the size of the input array. We achieve this by going through the array a few times to get the sums we need. This makes the method good for large datasets. For more on time complexity in dynamic programming, read our article on Dynamic Programming - Minimum Path Sum.

4. Can we solve the problem using a brute force approach?

Yes, we can solve the problem with a brute force method but it is not a good idea. This way would check all pairs of subarrays. This leads to a time complexity of O(n^3). This is much slower than the best dynamic programming solution. For more on how to think about algorithm efficiency, see our article on Dynamic Programming - Coin Change.

5. What are some common mistakes when solving this problem?

Common mistakes when we solve the maximum sum of two non-overlapping subarrays problem include not handling special cases well. For example, arrays with less than two elements or those with negative numbers. Also, not understanding that the subarrays should not overlap can cause wrong answers. It is very important to define the subarray limits correctly. For more tips to avoid errors in dynamic programming, check our article on Dynamic Programming - Longest Increasing Subsequence.