MEDIUM NC#113 Dynamic Programming
309. Best Time to Buy and Sell Stock with Cooldown
📖 Problem
You are given an array prices where prices[i] is the price of a given stock on the ith day. Find the maximum profit you can achieve. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times) with the following restrictions: - After you sell your stock, you cannot buy stock on the next day (i.e., cooldown 1 day). - You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •Array DP table updates
- •State transition thinking
- •Base case initialization
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Dynamic Programming reasoning pattern
💡 Approach
- → Use DP to track max profit at each state
- → Three states: hold (bought), sold (just sold), cooldown
- → hold[i] = max(hold[i-1], cooldown[i-1] - prices[i])
- → sold[i] = max(sold[i-1], hold[i-1] + prices[i])
- → cooldown[i] = max(cooldown[i-1], sold[i-1])
- → Time: O(n), Space: O(1)
🛠️ Hints & Pitfalls
Hints
- •Use DP to track max profit at each state
- •Three states: hold (bought), sold (just sold), cooldown
- •hold[i] = max(hold[i-1], cooldown[i-1] - prices[i])
Common Pitfalls
- •sold[i] = max(sold[i-1], hold[i-1] + prices[i])
- •cooldown[i] = max(cooldown[i-1], sold[i-1])
- •Time: O(n), Space: O(1)
🧪 Test Cases
Hidden tests on submit: 1
Test Case 1
Not run Input:
maxProfitWithCooldown([1,2,3,0,2]); Expected:
2 Test Case 2
Not run Input:
maxProfitWithCooldown([1]); Expected:
0 Test Case 3
Not run Input:
maxProfitWithCooldown([]); Expected:
0