[Dynamic Programming] Maximum Sum of 3 Non-Overlapping Subarrays - Hard

The problem of finding the biggest sum of three non-overlapping subarrays can be solved well with dynamic programming. Our goal is to find three different subarrays in a given array. We want their sum to be the highest. Each subarray must not overlap with the others. We need to think carefully about the indices and the possible sums of the subarrays. This way, we can find the best solution.

In this article, we will look at the details of the maximum sum of three non-overlapping subarrays problem. We will explain the problem clearly. We will show a dynamic programming method to find the solution. Also, we will give examples in Java, Python, and C++. We will talk about ways to make our solution better, the complexity of the problem, common mistakes when we implement it, and we will answer some common questions about this problem.

  • Dynamic Programming Maximum Sum of 3 Non-Overlapping Subarrays Solution Overview
  • Understanding the Problem Statement for Maximum Sum of 3 Non-Overlapping Subarrays
  • Dynamic Programming Approach for Maximum Sum of 3 Non-Overlapping Subarrays
  • Java Implementation of Maximum Sum of 3 Non-Overlapping Subarrays
  • Python Implementation of Maximum Sum of 3 Non-Overlapping Subarrays
  • C++ Implementation of Maximum Sum of 3 Non-Overlapping Subarrays
  • Optimization Techniques for Dynamic Programming in Maximum Sum Problem
  • Complexity Analysis of Maximum Sum of 3 Non-Overlapping Subarrays
  • Common Mistakes in Implementing the Maximum Sum of 3 Non-Overlapping Subarrays
  • Frequently Asked Questions

If we want to learn more about dynamic programming, we can check these related articles: Dynamic Programming Fibonacci Number, Dynamic Programming Maximum Subarray - Kadane’s Algorithm, and Dynamic Programming Edit Distance.

Understanding the Problem Statement for Maximum Sum of 3 Non-Overlapping Subarrays

We have a problem. We need to find the maximum sum of three non-overlapping subarrays. Here is how we can understand it:

We get an array nums of integers. Our goal is to choose three parts of this array that are next to each other, but they must not overlap. We should follow these rules:

  1. Each part must not overlap with the others.
  2. We want the total sum of the three parts to be the biggest.
  3. Each part must be at least L in size.

Input and Output

  • Input:
    • An integer array nums with length n (where n >= 3 * L).
    • An integer L that shows the minimum size for each part.
  • Output:
    • We give an array of three integers. These integers are the starting points of the three parts that give the highest total sum.

Example

Let’s look at an example:

  • Input: nums = [1,2,1,2,6,7,5,1], L = 2
  • Output: [0, 3, 5]
  • Explanation: The best parts are [1, 2], [2, 6], and [7, 5]. They add up to 1 + 2 + 6 + 7 + 5 = 21.

Constraints

  • The size of the input array must be enough to have three parts that do not overlap and meet the minimum size.
  • The ending index of one part must be less than the starting index of the next part.

This problem is a good example of using dynamic programming. We can use this method to pick parts while following the rules we said. We need to think carefully about the limits and the sums of the parts when we do the calculations.

For more details about dynamic programming, you can check articles like Dynamic Programming - Maximum Sum of Two Non-Overlapping Subarrays.

Dynamic Programming Approach for Maximum Sum of 3 Non-Overlapping Subarrays

We can solve the problem of finding the maximum sum of three non-overlapping subarrays using dynamic programming. This method helps us break the problem into smaller parts. We will define arrays to track the maximum sums at different indices.

Approach

  1. Define the Problem: We have an array nums. Our goal is to pick three non-overlapping subarrays to maximize their total sum.

  2. Prefix Sum Array: First, we make a prefix sum array called prefix. The value prefix[i] will hold the sum of the first i elements. This will help us quickly calculate the sum of any subarray in O(1) time.

    prefix = [0] * (len(nums) + 1)
for i in range(len(nums)):
    prefix[i + 1] = prefix[i] + nums[i]

  3. Dynamic Programming Arrays: We create two arrays:

    • maxSumLeft[i]: This holds the maximum sum of a subarray that ends at or before index i.
    • maxSumRight[i]: This holds the maximum sum of a subarray that starts at or after index i.

    These arrays help us find the maximum sum of the three subarrays by checking different split points.

  4. Calculate Maximum Left Sums:

    maxSumLeft = [0] * len(nums)
maxSum = float('-inf')
for i in range(len(nums)):
    maxSum = max(maxSum, prefix[i + 1] - prefix[i + 1 - length])  # Calculate max for every length
    maxSumLeft[i] = maxSum

  5. Calculate Maximum Right Sums:

    maxSumRight = [0] * len(nums)
maxSum = float('-inf')
for i in range(len(nums) - 1, -1, -1):
    maxSum = max(maxSum, prefix[i + length + 1] - prefix[i + 1])  # Calculate max for every length
    maxSumRight[i] = maxSum

  6. Combine Results: Finally, we loop through the array to combine results from the left and right sums:

    maxSum = float('-inf')
for j in range(length * 2, len(nums) - length):
    maxSum = max(maxSum, maxSumLeft[j - length] + (prefix[j + length + 1] - prefix[j + 1]) + maxSumRight[j + 1])

Code Implementation

Here is the complete Python code:

def maxSumOfThreeSubarrays(nums, length):
    n = len(nums)
    prefix = [0] * (n + 1)
    
    for i in range(n):
        prefix[i + 1] = prefix[i] + nums[i]
    
    maxSumLeft = [0] * n
    maxSum = float('-inf')
    
    for i in range(length - 1, n):
        currentSum = prefix[i + 1] - prefix[i + 1 - length]
        if i == length - 1:
            maxSumLeft[i] = currentSum
        else:
            maxSumLeft[i] = max(maxSumLeft[i - 1], currentSum)
    
    maxSumRight = [0] * n
    maxSum = float('-inf')
    
    for i in range(n - 1, n - length - 1, -1):
        currentSum = prefix[i + 1] - prefix[i + 1 - length]
        if i == n - length:
            maxSumRight[i] = currentSum
        else:
            maxSumRight[i] = max(maxSumRight[i + 1], currentSum)

    maxSum = float('-inf')
    
    for j in range(length * 2 - 1, n - length):
        maxSum = max(maxSum, maxSumLeft[j - length] + (prefix[j + length + 1] - prefix[j + 1]) + maxSumRight[j + 1])
    
    return maxSum

This dynamic programming way helps us reduce time complexity to O(n). It works well for larger arrays. If you want to learn more about dynamic programming, you can check articles about topics like Dynamic Programming: Maximum Subarray - Kadane’s Algorithm.

Java Implementation of Maximum Sum of 3 Non-Overlapping Subarrays

To find the maximum sum of three non-overlapping subarrays, we can use a simple dynamic programming method. Below is a clear Java example that shows how we do this.

Java Code Implementation

public class MaximumSumOfThreeSubarrays {
    public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
        int n = nums.length;
        int[] result = new int[3];
        int[] sum = new int[n + 1];

        // Calculate prefix sums
        for (int i = 0; i < n; i++) {
            sum[i + 1] = sum[i] + nums[i];
        }

        // maxSum stores the maximum sum of a subarray of size k ending at index i
        int[] maxSum = new int[n];
        int maxIndex = 0;

        // Calculate maximum sum of one subarray
        for (int i = k - 1; i < n; i++) {
            int currentSum = sum[i + 1] - sum[i + 1 - k];
            if (i == k - 1 || currentSum > sum[maxIndex + 1] - sum[maxIndex + 1 - k]) {
                maxIndex = i - k + 1; // Update index of max sum subarray
            }
            maxSum[i] = maxIndex;
        }

        // Store the maximum sums for the second and third subarrays
        int maxTotal = 0;

        for (int j = 2 * k - 1; j < n; j++) {
            int firstIndex = maxSum[j - k];
            int secondSum = sum[j + 1] - sum[j + 1 - k] + sum[firstIndex + k] - sum[firstIndex];
            if (secondSum > maxTotal) {
                maxTotal = secondSum;
                result[0] = firstIndex;
                result[1] = maxSum[j - k];
                result[2] = j - k + 1;
            }
        }

        return result;
    }

    public static void main(String[] args) {
        MaximumSumOfThreeSubarrays solution = new MaximumSumOfThreeSubarrays();
        int[] nums = {1, 2, 1, 2, 6, 7, 5, 1};
        int k = 2;
        int[] result = solution.maxSumOfThreeSubarrays(nums, k);
        System.out.println("Indices of the three non-overlapping subarrays are: " + 
                           result[0] + ", " + result[1] + ", " + result[2]);
    }
}

Explanation of the Implementation

  • Prefix Sum Calculation: We first find the prefix sums. This helps us to quickly calculate the sums of subarrays.
  • Dynamic Programming Arrays: We use two arrays to track the maximum sums of the subarrays.
  • Iterate for Maximum Sums: We go through the array. We find the best indices for the three non-overlapping subarrays.
  • Output Result: At the end, we return the indices of the three maximum sum subarrays.

This Java code gives us a clear and fast way to find the maximum sum of three non-overlapping subarrays. It uses the power of dynamic programming. For more details on dynamic programming, we can check this link on Dynamic Programming - Maximum Subarray.

Python Implementation of Maximum Sum of 3 Non-Overlapping Subarrays

To find the maximum sum of three non-overlapping subarrays, we can use a dynamic programming method. Our aim is to get the highest sum from three adjacent subarrays while making sure they do not overlap.

Implementation Steps

  1. Calculate Maximum Subarray Sums: First, we need to find maximum sums for all subarrays using Kadane’s algorithm. This helps us get the highest sum for each subarray of length k.

  2. Dynamic Programming Array: We will use a dynamic programming array to keep the best sums for the first two subarrays.

  3. Iterate for the Third Subarray: For each starting point of the third subarray, we compute the best total sum by adding the best results from the first two subarrays with the current one.

Python Code

Here is the code to implement this in Python:

def maxSumOfThreeSubarrays(nums, k):
    n = len(nums)
    if n < 3 * k:
        return []

    # Step 1: Calculate the sum of subarrays of length k
    sum_k = [0] * (n - k + 1)
    current_sum = sum(nums[:k])
    sum_k[0] = current_sum
    
    for i in range(1, n - k + 1):
        current_sum += nums[i + k - 1] - nums[i - 1]
        sum_k[i] = current_sum

    # Step 2: Prepare dp arrays for max sum of first and second subarrays
    left_max = [0] * (n - k + 1)
    right_max = [0] * (n - k + 1)

    # Fill left_max
    for i in range(n - k + 1):
        if i == 0:
            left_max[i] = sum_k[i]
        else:
            left_max[i] = max(left_max[i - 1], sum_k[i])

    # Fill right_max
    for i in range(n - k, -1, -1):
        if i == n - k:
            right_max[i] = sum_k[i]
        else:
            right_max[i] = max(right_max[i + 1], sum_k[i])

    # Step 3: Find the maximum sum of three non-overlapping subarrays
    max_sum = 0
    result_indices = []

    for j in range(k, n - 2 * k + 1):
        total = left_max[j - k] + sum_k[j] + right_max[j + k]
        if total > max_sum:
            max_sum = total
            result_indices = [left_max[j - k], j, right_max[j + k]]

    return result_indices  # Return the starting indices of the subarrays

Explanation of the Code

  • maxSumOfThreeSubarrays(nums, k): This function finds the starting indices of the three subarrays.
  • Kadane’s Algorithm: We use this to calculate the sum of subarrays that have fixed length k.
  • Dynamic Arrays: left_max and right_max store the maximum sums of the left and right parts of subarrays.
  • Final Calculation: The loop checks for the highest sum of three non-overlapping subarrays and gives the starting indices.

This method calculates the answer quickly in linear time. It works well even for large inputs. If you want to learn more about dynamic programming, check out this link: Dynamic Programming - Maximum Subarray (Kadane’s Algorithm).

C++ Implementation of Maximum Sum of 3 Non-Overlapping Subarrays

We can solve the problem of finding the maximum sum of three non-overlapping subarrays using dynamic programming. Here we have a C++ code that shows how to do this.

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

using namespace std;

class Solution {
public:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> sum(n + 1, 0);
        for (int i = 1; i <= n; ++i) {
            sum[i] = sum[i - 1] + nums[i - 1];
        }

        vector<int> first(n), second(n), third(n);
        int maxSum = 0;

        for (int i = k - 1; i < n; ++i) {
            int curSum = sum[i + 1] - sum[i + 1 - k];
            if (i == k - 1) {
                first[i] = curSum;
            } else {
                first[i] = max(first[i - 1], curSum);
            }
        }

        for (int i = 2 * k - 1; i < n; ++i) {
            int curSum = sum[i + 1] - sum[i + 1 - k];
            if (i == 2 * k - 1) {
                second[i] = first[i - k] + curSum;
            } else {
                second[i] = max(second[i - 1], first[i - k] + curSum);
            }
        }

        vector<int> result(3);
        for (int i = 3 * k - 1; i < n; ++i) {
            int curSum = sum[i + 1] - sum[i + 1 - k];
            if (i == 3 * k - 1) {
                maxSum = second[i - k] + curSum;
                result = {0, k - 1, 2 * k - 1};
            } else {
                if (second[i - k] + curSum > maxSum) {
                    maxSum = second[i - k] + curSum;
                    result[2] = i - k + 1;
                }
                result[1] = second[i - k];
            }
        }

        return result;
    }
};

int main() {
    Solution sol;
    vector<int> nums = {1, 2, 1, 2, 6, 7, 5, 1};
    int k = 2;
    vector<int> result = sol.maxSumOfThreeSubarrays(nums, k);
    
    cout << "Indices of the three non-overlapping subarrays: ";
    for (int index : result) {
        cout << index << " ";
    }
    cout << endl;

    return 0;
}

Explanation of the Code

  1. Prefix Sum Calculation: First we compute a prefix sum array. This helps us to find the sum of any subarray quickly.
  2. Dynamic Programming Arrays:
    • first[i]: This is the maximum sum of one subarray that ends at or before index i.
    • second[i]: This is the maximum sum of two subarrays where the second one ends at or before index i.
  3. Final Calculation: For the third subarray, we loop through the array again. We compute the maximum sum of three subarrays using the two arrays we computed before.

Complexity Analysis

  • Time Complexity: O(n). Here n is the length of the input array.
  • Space Complexity: O(n). We need space to store the prefix sums and dynamic programming arrays.

This method helps us find the maximum sum of three non-overlapping subarrays in a quick way. This is very important for large input sizes.

For more reading about dynamic programming, we can look at articles like Dynamic Programming: Maximum Sum of Non-Adjacent Elements.

Optimization Techniques for Dynamic Programming in Maximum Sum Problem

Dynamic Programming (DP) is a strong way to solve optimization problems. One example is the Maximum Sum of 3 Non-Overlapping Subarrays problem. We can use some techniques to make DP solutions better and faster in this case.

  1. Prefix Sum Array:
    • We can use a prefix sum array to quickly get sums of subarrays. This helps us find the sum of any subarray really fast.
    int[] prefixSum = new int[n + 1];
for (int i = 0; i < n; i++) {
    prefixSum[i + 1] = prefixSum[i] + nums[i];
}
  2. Sliding Window Technique:
    • Instead of finding sums for every subarray again, we keep a window of the maximum sum as we go through the array. This makes our time from O(n^3) to O(n) when we look for subarrays.
  3. Dynamic Programming with States:
    • We keep states that show the maximum sum of subarrays until a certain index. For three non-overlapping subarrays, we track the maximum sums of the first, second, and third subarrays.
    int[] maxSum1 = new int[n]; // Maximum sum of first subarray ending at each index
int[] maxSum2 = new int[n]; // Maximum sum of first two subarrays
int[] maxSum3 = new int[n]; // Maximum sum of three subarrays
  4. Iterative Updates:
    • Instead of using loops inside loops, we update the maximum sums while going through the array. This cuts down on repeated calculations and keeps the best results as we go.
  5. Memory Optimization:
    • Instead of using big arrays to save results, we can use a rolling array or just a few variables to keep the needed values. This makes our use of space smaller.

These techniques help a lot to make the performance and efficiency of dynamic programming better in problems like the Maximum Sum of 3 Non-Overlapping Subarrays. Using these methods can help us run faster and use less resources. This makes it easier to work with larger datasets.

For more ideas on dynamic programming techniques, we can read the Dynamic Programming: Maximum Subarray (Kadane’s Algorithm) article. It gives good tips on how to optimize DP solutions using known methods.

Complexity Analysis of Maximum Sum of 3 Non-Overlapping Subarrays

In this analysis, we look at the “Maximum Sum of 3 Non-Overlapping Subarrays” problem. We will check the time and space complexities of the dynamic programming method we use to solve it.

Time Complexity

  1. Precomputation of Sums:
    First, we need to calculate the prefix sums for the array. This step takes O(n) time. Here, n is the length of the input array.

  2. Dynamic Programming Calculation:
    Next, we use dynamic programming to find the maximum sums for three non-overlapping subarrays. We loop through the array. For each starting point of the third subarray, we look at all valid positions for the first and second subarrays.
    If we say maxSum(i) is the maximum sum we can get with subarrays that end before index i, our total check usually leads to O(n^2) complexity for these sums.

  3. Overall Time Complexity:
    So, the overall time complexity is O(n^2). This happens because we have nested loops to find the best partitions.

Space Complexity

  1. Auxiliary Space:
    We need space to store the prefix sums. This takes O(n) space.
    We also keep some extra arrays to track maximum sums and indices. This also needs O(n) space.

  2. Overall Space Complexity:
    In the end, the overall space complexity stays O(n).

To sum up, we can find the maximum sum of 3 non-overlapping subarrays with a time complexity of O(n^2) and a space complexity of O(n). This shows the limits of this method, especially when we deal with big input sizes.

Common Mistakes in Implementing the Maximum Sum of 3 Non-Overlapping Subarrays

When we try to find the maximum sum of 3 non-overlapping subarrays, we can make some common mistakes. These mistakes can give us wrong results or slow performance. Here are some things to watch out for:

  1. Incorrect Index Handling:
    • We need to make sure that the indices for the subarrays do not overlap. If we miscalculate the boundaries, the subarrays can overlap. This goes against the problem rules.
  2. Failure to Optimize for Edge Cases:
    • If we do not handle edge cases, like arrays with fewer than 3 elements, we can get index out-of-bounds errors. We should always check the array length before we do calculations.
  3. Not Using Prefix Sums:
    • If we skip using prefix sums, it can make our calculations slow. If we calculate sums without prefix sums, it can lead to O(n^3) complexity instead of O(n).
    // Example of prefix sum in Java
int[] prefix = new int[n + 1];
for (int i = 0; i < n; i++) {
    prefix[i + 1] = prefix[i] + nums[i];
}
  4. Improper Dynamic Programming Table Initialization:
    • We must ensure that our dynamic programming (DP) arrays or tables are set up right. If we do it wrong, we can get wrong results.
  5. Incorrect Transition Logic:
    • We must make sure the DP transition shows the maximum sum correctly. If we do not define the transitions well, the algorithm can miss the best solutions.
    # Example of DP transition in Python
dp = [[0] * (k + 1) for _ in range(n)]
for i in range(n):
    for j in range(k):
        dp[i][j] = max(dp[i - 1][j], current_sum + dp[i - 2][j - 1])
  6. Neglecting to Track Maximum Values:
    • We should always track the maximum values we find as we go. If we forget to update these values, we can end up with bad solutions.
  7. Ignoring Time Complexity:
    • If we do not think about the time complexity of our solution, we can have performance problems, especially with bigger inputs. We should aim for O(n) or O(n^2) complexity for our solutions to work well.
  8. Not Testing with Diverse Inputs:
    • If we do not test different types of inputs, like negative numbers or zeros, we can miss bugs. We should always check with many test cases.

By being careful about these common mistakes, we can better implement a solution for the maximum sum of 3 non-overlapping subarrays. This way, we can ensure our solution is correct and efficient. For more information about dynamic programming, we can check other articles like Dynamic Programming Fibonacci Number and Dynamic Programming Maximum Subarray - Kadane’s Algorithm.

Frequently Asked Questions

What is the maximum sum of 3 non-overlapping subarrays problem?

The maximum sum of 3 non-overlapping subarrays problem is about finding the best way to pick three connected subarrays from a list of numbers. We want the total of their numbers to be as high as possible. The chosen subarrays cannot touch each other. Each subarray must have at least one number. We can solve this problem well using dynamic programming methods, which we talk about in our article.

How do you implement the maximum sum of 3 non-overlapping subarrays in Java?

To implement the maximum sum of 3 non-overlapping subarrays in Java, we can use a dynamic programming method. This keeps track of total sums and highest values for different parts of the array. We go through the array and look for the best sums for the different sections. For detailed code examples, look at our Java implementation section in the article.

What dynamic programming techniques are used to solve this problem?

Dynamic programming techniques for the maximum sum of 3 non-overlapping subarrays problem usually need cumulative sums and prefix arrays. By saving the highest sums for possible sections, we can find the answer without doing extra calculations. This method makes the problem easier to solve within a good time frame.

Can you provide a Python solution for the maximum sum of 3 non-overlapping subarrays?

Sure! The Python solution for the maximum sum of 3 non-overlapping subarrays uses dynamic programming. We use lists to keep track of the highest sums for segments of the array. This simple method makes it easy to understand and change. To see the full code and explanation, please go to the Python implementation section of our article.

What are some common mistakes while implementing the maximum sum of 3 non-overlapping subarrays?

Common mistakes when we implement the maximum sum of 3 non-overlapping subarrays are not handling the edges of the subarrays right, not keeping the non-overlapping rule, and getting cumulative sums wrong. Also, not making the code efficient can lead to slow solutions that do not work well. Check our article for more tips on how to avoid these mistakes.

For more related dynamic programming topics, check out these articles on Fibonacci numbers and Kadane’s algorithm.