2026-06-01 16:12:21 +08:00
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#+PROPERTY: STUDY_DECK_02
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2026-06-01 17:12:10 +08:00
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* TODO 0121. Best Time to Buy And Sell Stock :easy:
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2026-06-01 02:33:30 +08:00
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:PROPERTIES:
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2026-06-01 17:22:07 +08:00
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:NEETCODE: [[file:../../roadmap.org::*0121. Best Time to Buy And Sell Stock][0121. Best Time to Buy And Sell Stock]]
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2026-06-01 02:33:30 +08:00
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:END:
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2026-06-01 17:22:07 +08:00
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You are given an array ~prices~ where ~prices[i]~ is the price of a given stock on the ~i^{th}~ day.
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You want to maximize your profit by choosing a *single day* to buy one stock and choosing a *different day in the future* to sell that stock.
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Return /the maximum profit you can achieve from this transaction/. If you cannot achieve any profit, return ~0~.
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*Example 1:*
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#+begin_src
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Input: prices = [7,1,5,3,6,4]
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Output: 5
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Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
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Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
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#+end_src
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*Example 2:*
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#+begin_src
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Input: prices = [7,6,4,3,1]
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Output: 0
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Explanation: In this case, no transactions are done and the max profit = 0.
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#+end_src
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*Constraints:*
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- ~1 <= prices.length <= 10^{5}~
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- ~0 <= prices[i] <= 10^{4}~
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2026-06-01 02:39:53 +08:00
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** TODO Approach
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Write your approach here.
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** TODO Python
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#+begin_src python
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2026-06-01 17:22:07 +08:00
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class Solution:
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def maxProfit(self, prices: List[int]) -> int:
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2026-06-01 02:39:53 +08:00
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#+end_src
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** TODO C++
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2026-06-01 02:33:30 +08:00
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#+begin_src cpp
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2026-06-01 17:22:07 +08:00
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class Solution {
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public:
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int maxProfit(vector<int>& prices) {
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}
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};
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2026-06-01 02:33:30 +08:00
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#+end_src
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