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

Best Time to Buy and Sell Stock II (LeetCode #122) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Buying low and selling high maximizes profit.

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

The optimal solution for Best Time to Buy and Sell Stock II runs in O(n) time and uses O(1) space. This complexity is linear because we only make a single pass through the prices array, checking adjacent days.

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

The brute force approach for Best Time to Buy and Sell Stock II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of buy and sell days to calculate the profit. This method is straightforward but inefficient as it examines all combinations. The optimal Optimal Solution approach improves this to O(n).

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

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

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

Explain your thought process clearly. Consider edge cases, like constant prices or decreasing prices. Discuss time and space complexity during your explanation.

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

Not considering multiple transactions. Confusing this problem with the single transaction version.

#122

Best Time to Buy and Sell Stock II

Medium

ArrayDynamic ProgrammingGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution takes advantage of the fact that we can buy and sell on the same day. Instead of checking every pair, we simply add up all the profits from each upward price movement.

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Algorithm

3 steps

1Step 1: Initialize a variable to store total profit.
2Step 2: Iterate through the prices array from the first to the last day.
3Step 3: If the price on the next day is higher than the current day, add the difference to the total profit.

solution.py6 lines

1def maxProfit(prices):
2    total_profit = 0
3    for i in range(1, len(prices)):
4        if prices[i] > prices[i - 1]:
5            total_profit += prices[i] - prices[i - 1]
6    return total_profit

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Complexity note: This complexity is linear because we only make a single pass through the prices array, checking adjacent days.

1Buying low and selling high maximizes profit.
2Multiple transactions can be beneficial in a fluctuating market.

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