What is the time complexity of Best Time to Buy and Sell Stock IV?

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

What data structures are used to solve Best Time to Buy and Sell Stock IV?

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

What are the key insights for solving Best Time to Buy and Sell Stock IV?

Insight 1: Dynamic programming is suitable for this problem because the maximum profit can be built up incrementally by considering the maximum profit achievable with fewer transactions or on previous days. Insight 2: The state can be defined as the maximum profit achievable with `j` transactions until day `i`. We need to track both buying and selling states.

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Best Time to Buy and Sell Stock IV - LeetCode 188 Solution

Q: Which companies ask LeetCode 188. Best Time to Buy and Sell Stock IV in interviews?

This problem is frequently asked at Arcesium, Citadel, Goldman Sachs, Google, Jump Trading and 3 more companies.

Best Time to Buy and Sell Stock IV - Complete Solution Guide

Best Time to Buy and Sell Stock IV is LeetCode problem 188, 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 integer array prices where prices[i] is the price of a given stock on the i th day, and an integer k . Find the maximum profit you can achieve. You may complete at most k transactions: i.e. you may buy at most k times and sell at most k times. Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again). Example 1: Input: k = 2, prices = [2,4,1] Output: 2 Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4),

Detailed Explanation

The problem asks us to find the maximum profit we can achieve by buying and selling a stock, given an array of prices where `prices[i]` is the price of the stock on day `i`. We are allowed to complete at most `k` transactions, meaning we can buy at most `k` times and sell at most `k` times. Crucially, we cannot engage in multiple transactions simultaneously; we must sell the stock before buying again.

Solution Approach

The solution uses dynamic programming to keep track of the maximum profit achievable with up to `k` transactions at each day. It uses two arrays, `buy` and `sell`, where `buy[j]` represents the maximum profit achievable after buying the stock for the `j`-th transaction, and `sell[j]` represents the maximum profit achievable after selling the stock for the `j`-th transaction. The algorithm iterates through the prices, updating the `buy` and `sell` arrays based on the current price and the maximum profit achievable with the previous transactions. If `k` is large enough (k >= n/2), it simplifies to finding all increasing price intervals and summing their differences.

Step-by-Step Algorithm

Step 1: Handle base cases: If the array `prices` has length 0 or 1, return 0, as no profit can be made.
Step 2: Handle the case where `k` is large: If `k` is greater than or equal to `n / 2`, it means we can make as many transactions as there are profitable price differences. Iterate through the `prices` array and sum up the differences where `prices[i] > prices[i-1]`.
Step 3: Initialize `buy` and `sell` arrays: Create two arrays, `buy` and `sell`, of size `k + 1`. Initialize `buy[j]` to negative infinity for all `j` and `sell[j]` to 0 for all `j`. Negative infinity indicates that no buy has happened yet for that transaction.
Step 4: Iterate through the `prices` array: For each `price` in `prices`, update `buy` and `sell` arrays. The inner loop iterates from 1 to `k` (inclusive).
Step 5: Update `buy[j]`: `buy[j] = max(buy[j], sell[j-1] - price)`. This means the maximum profit after buying on the `j`-th transaction is either the previous best profit after buying on the `j`-th transaction, or the profit obtained from selling on the `(j-1)`-th transaction and then buying at the current price.
Step 6: Update `sell[j]`: `sell[j] = max(sell[j], buy[j] + price)`. This means the maximum profit after selling on the `j`-th transaction is either the previous best profit after selling on the `j`-th transaction, or the profit obtained from buying on the `j`-th transaction and then selling at the current price.
Step 7: After iterating through all prices, return `sell[k]`, which represents the maximum profit achievable with at most `k` transactions.

Key Insights

Insight 1: Dynamic programming is suitable for this problem because the maximum profit can be built up incrementally by considering the maximum profit achievable with fewer transactions or on previous days.
Insight 2: The state can be defined as the maximum profit achievable with `j` transactions until day `i`. We need to track both buying and selling states.
Insight 3: A special case exists when `k` is large. In this scenario, it becomes optimal to perform as many transactions as possible, effectively buying whenever the price is lower than the next day and selling then.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(k)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Arcesium, Citadel, Goldman Sachs, Google, Jump Trading, Nielsen, Nvidia, PhonePe.

Best Time to Buy and Sell Stock IV | LeetCode 188

Hard

Array Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int maxProfit(int k, int[] prices) {
        int n = prices.length;
        if (n <= 1) {
            return 0;
        }

        if (k >= n / 2) {
            int profit = 0;
            for (int i = 1; i < n; i++) {
                if (prices[i] > prices[i - 1]) {
                    profit += prices[i] - prices[i - 1];
                }
            }
            return profit;
        }

        int[] buy = new int[k + 1];
        int[] sell = new int[k + 1];

        for (int i = 0; i <= k; i++) {
            buy[i] = Integer.MIN_VALUE;
            sell[i] = 0;
        }

        for (int price : prices) {
            for (int j = 1; j <= k; j++) {
                buy[j] = Math.max(buy[j], sell[j - 1] - price);
                sell[j] = Math.max(sell[j], buy[j] + price);
            }
        }

        return sell[k];
    }
}

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Problem Info

Description

Problem 188: Best Time to Buy and Sell Stock IV - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

ArcesiumCitadelGoldman SachsGoogleJump TradingNielsenNvidiaPhonePe

Stats

Difficulty:Hard

Languages:4

Companies:8

Problem #:188

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

7 steps

1
Step 1: Handle base cases: If the array `prices` has length 0 or 1, return 0, as no profit can be made.
2
Step 2: Handle the case where `k` is large: If `k` is greater than or equal to `n / 2`, it means we can make as many transactions as there are profitable price differences. Iterate through the `prices` array and sum up the differences where `prices[i] > prices[i-1]`.
3
Step 3: Initialize `buy` and `sell` arrays: Create two arrays, `buy` and `sell`, of size `k + 1`. Initialize `buy[j]` to negative infinity for all `j` and `sell[j]` to 0 for all `j`. Negative infinity indicates that no buy has happened yet for that transaction.
4
Step 4: Iterate through the `prices` array: For each `price` in `prices`, update `buy` and `sell` arrays. The inner loop iterates from 1 to `k` (inclusive).
5
Step 5: Update `buy[j]`: `buy[j] = max(buy[j], sell[j-1] - price)`. This means the maximum profit after buying on the `j`-th transaction is either the previous best profit after buying on the `j`-th transaction, or the profit obtained from selling on the `(j-1)`-th transaction and then buying at the current price.
6
Step 6: Update `sell[j]`: `sell[j] = max(sell[j], buy[j] + price)`. This means the maximum profit after selling on the `j`-th transaction is either the previous best profit after selling on the `j`-th transaction, or the profit obtained from buying on the `j`-th transaction and then selling at the current price.
7
Step 7: After iterating through all prices, return `sell[k]`, which represents the maximum profit achievable with at most `k` transactions.

💡

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(k)

📝

Why?

The outer loop iterates `n` times (prices array), and the inner loop iterates `k` times. The space used is determined by the `buy` and `sell` arrays, each having a size of `k+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

Divide and Conquer

The problem can theoretically be approached using divide and conquer, dividing the problem into subproblems based on different ranges of days. However, this typically doesn't improve the complexity and often increases overhead.

Time: O(n*k*log(n)) likelySpace: O(n) likely

Trade-offs: Divide and conquer might have higher overhead than the DP approach and a higher time complexity. It is not recommended for this problem.

✨

Key Insights

1Insight 1: Dynamic programming is suitable for this problem because the maximum profit can be built up incrementally by considering the maximum profit achievable with fewer transactions or on previous days.
2Insight 2: The state can be defined as the maximum profit achievable with `j` transactions until day `i`. We need to track both buying and selling states.
3Insight 3: A special case exists when `k` is large. In this scenario, it becomes optimal to perform as many transactions as possible, effectively buying whenever the price is lower than the next day and selling then.

📚

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: `prices` array is empty or has only one element. The code handles this by returning 0, as no profit can be made.
Edge case 2: `k` is very large. If k is greater than or equal to `n / 2`, where n is the number of days (length of `prices`), the problem simplifies to finding all increasing price differences and summing them up. This optimization significantly reduces the runtime when `k` is large.
Edge case 3: `k = 0`. In this case, the loops do not execute, and `sell[0]` which is initialized as 0 is returned. This handles the case with zero transactions

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge case where `k` is large. This can lead to Time Limit Exceeded errors for large input sizes.

✗Common mistake 2: Incorrect initialization of the `buy` and `sell` arrays. `buy` array should be initialized to negative infinity to indicate that no stock has been bought yet for each transaction, preventing incorrect profit calculations.

✗Common mistake 3: Not understanding the order of updates for `buy` and `sell` arrays. The `buy` array needs to be updated based on the `sell` array from the previous transaction, and the `sell` array needs to be updated based on the `buy` array of the current transaction.

✅

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: Algorithmic trading strategies often involve optimizing buy and sell decisions to maximize profit, subject to constraints like the number of allowed trades. This problem is a simplified version of these strategies.

🏦Real-world scenario 2: Resource allocation problems can be modeled similarly. Imagine allocating a limited number of resources (like transactions) to different investment opportunities (like prices) to maximize the overall return (profit).

💡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 9, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Best Time to Buy and Sell Stock IV"?

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.

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