[LeetCode] 0259. 3Sum Smaller

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內容目錄

Hard

You are given an array prices where prices[i] is the price of a given stock on the i^th day.

Find the maximum profit you can achieve. You may complete at most two transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

Input: prices = [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Constraints:

1 <= prices.length <= 10⁵
0 <= prices[i] <= 10⁵

Python

				
					# time complexity: O(n^2)
# space completity: O(1)
from typing import List


class Solution:
    def threeSumSmaller(self, nums: List[int], target: int) -> int:
        result = 0
        nums.sort()
        for i in range(len(nums) - 1):
            result += self.twoSumSmaller(nums, i + 1, target - nums[i])
        return result

    def twoSumSmaller(self, nums: List[int], startIdx: int, target: int) -> int:
        left = startIdx
        right = len(nums) - 1
        result = 0
        while left < right:
            if nums[left] + nums[right] < target:
                result += right - left
                left += 1
            else:
                right -= 1
        return result


nums = [-2, 0, 1, 3]
target = 2
print(Solution().threeSumSmaller(nums, target))

Hard

Python

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