How do you solve Best Time to Buy and Sell Stock V on LeetCode?

Best Time to Buy and Sell Stock V (LeetCode #3573) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(nk) time complexity and O(nk) space complexity. Key insight: Dynamic programming can optimize recursive solutions.

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

The optimal solution for Best Time to Buy and Sell Stock V runs in O(nk) time and uses O(nk) space. The time complexity is O(nk) due to iterating through n days for k transactions, while space complexity is O(nk) for the DP table.

What is the brute force solution for Best Time to Buy and Sell Stock V?

The brute force approach for Best Time to Buy and Sell Stock V is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible transactions by recursively trying to buy and sell stocks on different days. This approach checks every combination of transactions, leading to high time complexity. The optimal Optimal Solution approach improves this to O(nk).

What algorithm or data structure is used to solve Best Time to Buy and Sell Stock V?

Best Time to Buy and Sell Stock V uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Best Time to Buy and Sell Stock V?

Explain your thought process clearly. Discuss edge cases upfront. Practice coding without an IDE to simulate interview conditions.

What are common mistakes when solving Best Time to Buy and Sell Stock V?

Forgetting to handle edge cases like empty prices. Not considering short selling in profit calculations.

#3573

Best Time to Buy and Sell Stock V

Medium

ArrayDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(nk)
Space	O(1)	O(nk)

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Intuition

Time O(nk)Space O(nk)

Using dynamic programming, we can build a table to store the maximum profit for each transaction limit and day, allowing us to efficiently compute profits without redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Create a DP table where dp[t][d] represents max profit with t transactions by day d.
2Step 2: Iterate through each transaction limit and each day, calculating profits based on previous days.
3Step 3: Return the maximum profit from the last transaction on the last day.

solution.py11 lines

1def maxProfit(prices, k):
2    if not prices: return 0
3    n = len(prices)
4    if k >= n // 2: return sum(max(prices[i+1] - prices[i], 0) for i in range(n-1))
5    dp = [[0] * n for _ in range(k + 1)]
6    for t in range(1, k + 1):
7        max_diff = -prices[0]
8        for d in range(1, n):
9            dp[t][d] = max(dp[t][d-1], prices[d] + max_diff)
10            max_diff = max(max_diff, dp[t-1][d] - prices[d])
11    return dp[k][n-1]

ℹ

Complexity note: The time complexity is O(nk) due to iterating through n days for k transactions, while space complexity is O(nk) for the DP table.

1Dynamic programming can optimize recursive solutions.
2Understanding transaction types helps in formulating the DP approach.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.