What is the time complexity of Maximum Strength of K Disjoint Subarrays?

The optimal solution has a time complexity of O(n*k).

What data structures are used to solve Maximum Strength of K Disjoint Subarrays?

This problem uses Array, Dynamic Programming, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Strength of K Disjoint Subarrays?

Insight 1: The problem can be effectively solved using dynamic programming because we need to find the maximum strength among all possible combinations of `k` disjoint subarrays. Insight 2: We can optimize the space complexity by using rolling arrays (two arrays) instead of a full 2D DP table.

Which companies ask LeetCode 3077. Maximum Strength of K Disjoint Subarrays in interviews?

This problem is frequently asked at DE Shaw.

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Maximum Strength of K Disjoint Subarrays - LeetCode 3077 Solution

Q: How to solve Maximum Strength of K Disjoint Subarrays?

The solution uses dynamic programming to find the maximum strength. It maintains two arrays, `dp` and `new_dp`, to store the maximum strength achieved by selecting `j` subarrays up to index `i`. The outer loop iterates through the number of subarrays `j` from 1 to `k`. The inner loop iterates through the prefixes of the input array `nums`. A variable `f` is used to store the maximum strength of `j` subarrays ending at index `i-1`. The dynamic programming recurrence relations are crucial: `f = max(f, dp[i-1]) + coeff * nums[i-1]` and `new_dp[i] = max(new_dp[i-1], f)`. The coefficient changes its sign in each iteration. At the end of each outer loop, `dp` is updated to `new_dp` for the next iteration.

Maximum Strength of K Disjoint Subarrays - Complete Solution Guide

Maximum Strength of K Disjoint Subarrays is LeetCode problem 3077, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array of integers nums with length n , and a positive odd integer k . Select exactly k disjoint subarrays sub 1 , sub 2 , ..., sub k from nums such that the last element of sub i appears before the first element of sub {i+1} for all 1 <= i <= k-1 . The goal is to maximize their combined strength. The strength of the selected subarrays is defined as: strength = k * sum(sub 1 )- (k - 1) * sum(sub 2 ) + (k - 2) * sum(sub 3 ) - ... - 2 * sum(sub {k-1} ) + sum(sub k ) where sum(sub i

Detailed Explanation

The problem requires you to select exactly `k` disjoint subarrays from a given array `nums` of length `n`, where `k` is an odd positive integer. The goal is to maximize the 'strength' of these subarrays. The 'strength' is calculated using a specific formula: `strength = k * sum(sub1) - (k - 1) * sum(sub2) + (k - 2) * sum(sub3) - ... - 2 * sum(sub{k-1}) + sum(subk)`. The order of the subarrays matters, as the formula depends on the index of the subarray. Disjoint means the selected subarrays cannot overlap. You are not required to cover the entire array `nums`.

Solution Approach

The solution uses dynamic programming to find the maximum strength. It maintains two arrays, `dp` and `new_dp`, to store the maximum strength achieved by selecting `j` subarrays up to index `i`. The outer loop iterates through the number of subarrays `j` from 1 to `k`. The inner loop iterates through the prefixes of the input array `nums`. A variable `f` is used to store the maximum strength of `j` subarrays ending at index `i-1`. The dynamic programming recurrence relations are crucial: `f = max(f, dp[i-1]) + coeff * nums[i-1]` and `new_dp[i] = max(new_dp[i-1], f)`. The coefficient changes its sign in each iteration. At the end of each outer loop, `dp` is updated to `new_dp` for the next iteration.

Step-by-Step Algorithm

Step 1: Initialize a `dp` array of size `n+1` with all elements set to 0. This array will store the maximum strength using a prefix of the input array and a certain number of subarrays.
Step 2: Iterate from `j = 1` to `k` (inclusive), where `j` represents the number of subarrays selected so far.
Step 3: Calculate the coefficient for the current subarray `j` based on the given formula. The coefficient alternates between positive and negative values depending on whether `j` is odd or even.
Step 4: Initialize a variable `f` to negative infinity. This variable stores the maximum strength of selecting `j` subarrays ending at index `i-1` of the input array.
Step 5: Initialize a `new_dp` array of size `n+1` with all elements set to negative infinity. This array will store the maximum strength for each prefix when considering j subarrays.
Step 6: Iterate from `i = 1` to `n` (inclusive), where `i` represents the current index of the input array prefix.
Step 7: Update `f` using the recurrence relation: `f = max(f, dp[i - 1]) + coeff * nums[i - 1]`. This calculates the maximum strength when the j-th subarray ends at index i-1.
Step 8: Update `new_dp[i]` using the recurrence relation: `new_dp[i] = max(new_dp[i - 1], f)`. This calculates the maximum strength using j subarrays up to index i, either by including nums[i-1] or excluding it.
Step 9: After the inner loop completes, update `dp` with `new_dp` to prepare for the next iteration of the outer loop.
Step 10: After the outer loop completes, return `dp[n]`, which represents the maximum strength achieved by selecting `k` subarrays from the entire input array.

Key Insights

Insight 1: The problem can be effectively solved using dynamic programming because we need to find the maximum strength among all possible combinations of `k` disjoint subarrays.
Insight 2: We can optimize the space complexity by using rolling arrays (two arrays) instead of a full 2D DP table.
Insight 3: The alternating signs in the strength formula require careful handling, which is achieved by tracking the coefficient based on whether the subarray index is odd or even.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Prefix Sum.

Companies

Asked at: DE Shaw.

Maximum Strength of K Disjoint Subarrays | LeetCode 3077

Hard

Array Dynamic Programming Prefix Sum

Time: O(n*k)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long maximumStrength(int[] nums, int k) {
        int n = nums.length;
        
        // dp[i] represents g[j-1][i]: max strength using prefix nums[0...i-1] for j-1 subarrays.
        // Initially for j=1, we need g[0], which is 0 for 0 subarrays.
        long[] dp = new long[n + 1];
        
        for (int j = 1; j <= k; j++) {
            // coeff is the multiplier for the j-th subarray's sum.
            int coeff = (k - j + 1) * (j % 2 == 1 ? 1 : -1);
            
            // f[j][i]: max strength for j subarrays, j-th ends at i-1.
            // We only need to track the latest f value.
            long f = Long.MIN_VALUE;
            
            // new_dp[i] represents g[j][i]: max strength for j subarrays in prefix i.
            long[] new_dp = new long[n + 1];
            Arrays.fill(new_dp, Long.MIN_VALUE);
            
            for (int i = 1; i <= n; i++) {
                // Recurrence for f: f[j][i] = max(f[j][i-1], g[j-1][i-1]) + coeff * nums[i-1]
                // In our variables: f_new = max(f_old, dp[i-1]) + coeff * nums[i-1]
                f = Math.max(f, dp[i - 1]) + (long)coeff * nums[i - 1];
                
                // Recurrence for g: g[j][i] = max(g[j][i-1], f[j][i])
                // In our variables: new_dp[i] = max(new_dp[i-1], f_new)
                new_dp[i] = Math.max(new_dp[i - 1], f);
            }
            
            dp = new_dp;
        }
        
        return (long)dp[n];
    }
}

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Problem Info

Description

Problem 3077: Maximum Strength of K Disjoint Subarrays - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Prefix Sum

Companies

DE Shaw

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3077

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a `dp` array of size `n+1` with all elements set to 0. This array will store the maximum strength using a prefix of the input array and a certain number of subarrays.
2
Step 2: Iterate from `j = 1` to `k` (inclusive), where `j` represents the number of subarrays selected so far.
3
Step 3: Calculate the coefficient for the current subarray `j` based on the given formula. The coefficient alternates between positive and negative values depending on whether `j` is odd or even.
4
Step 4: Initialize a variable `f` to negative infinity. This variable stores the maximum strength of selecting `j` subarrays ending at index `i-1` of the input array.
5
Step 5: Initialize a `new_dp` array of size `n+1` with all elements set to negative infinity. This array will store the maximum strength for each prefix when considering j subarrays.
6
Step 6: Iterate from `i = 1` to `n` (inclusive), where `i` represents the current index of the input array prefix.
7
Step 7: Update `f` using the recurrence relation: `f = max(f, dp[i - 1]) + coeff * nums[i - 1]`. This calculates the maximum strength when the j-th subarray ends at index i-1.
8
Step 8: Update `new_dp[i]` using the recurrence relation: `new_dp[i] = max(new_dp[i - 1], f)`. This calculates the maximum strength using j subarrays up to index i, either by including nums[i-1] or excluding it.
9
Step 9: After the inner loop completes, update `dp` with `new_dp` to prepare for the next iteration of the outer loop.
10
Step 10: After the outer loop completes, return `dp[n]`, which represents the maximum strength achieved by selecting `k` subarrays from the entire input array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*k)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n*k) because there are two nested loops: the outer loop iterates `k` times, and the inner loop iterates `n` times. The space complexity is O(n) because we use two arrays, `dp` and `new_dp`, each of size `n+1`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

A recursive approach with memoization can also be used to solve this problem. The state would be (index, number of subarrays selected).

Time: O(n*k)Space: O(n*k)

Trade-offs: Recursion with memoization is generally less efficient than the iterative DP approach due to function call overhead and potentially higher space complexity from the recursion stack, particularly for larger inputs. But it may be more intuitive to formulate.

✨

Key Insights

1Insight 1: The problem can be effectively solved using dynamic programming because we need to find the maximum strength among all possible combinations of `k` disjoint subarrays.
2Insight 2: We can optimize the space complexity by using rolling arrays (two arrays) instead of a full 2D DP table.
3Insight 3: The alternating signs in the strength formula require careful handling, which is achieved by tracking the coefficient based on whether the subarray index is odd or even.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: When all numbers are negative and k=1, the code correctly returns the largest negative number.
Edge case 2: When all numbers are zero, the code correctly returns 0.
Edge case 3: When k=n and all elements must be selected, the code handles alternating signs correctly to calculate total strength.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not correctly handling the alternating signs in the strength formula, which can lead to incorrect results.

✗Common mistake 2: Using a full 2D DP table, leading to higher space complexity. Rolling arrays significantly reduce the space requirements.

✗Common mistake 3: Not initializing the `f` and `new_dp` arrays with negative infinity, which can lead to incorrect maximum values being calculated.

✗Common mistake 4: Integer overflow: remember to use `long` or `long long` to store the intermediate results.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Portfolio optimization: Selecting k investments (subarrays) with different weights (coefficients) to maximize the overall return (strength).

🏦Real-world scenario 2: Resource allocation: Assigning resources (subarrays) to different tasks with varying priorities (coefficients) to maximize the overall efficiency (strength).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 18, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Strength of K Disjoint Subarrays"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?