leetcode - 1326. Minimum Number of Taps to Open to Water a Garden-Toy模板网

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Description

There is a one-dimensional garden on the x-axis. The garden starts at the point 0 and ends at the point n. (i.e The length of the garden is n).

There are n + 1 taps located at points [0, 1, …, n] in the garden.

Given an integer n and an integer array ranges of length n + 1 where ranges[i] (0-indexed) means the i-th tap can water the area [i - ranges[i], i + ranges[i]] if it was open.

Return the minimum number of taps that should be open to water the whole garden, If the garden cannot be watered return -1.

Example 1:
leetcode - 1326. Minimum Number of Taps to Open to Water a Garden,OJ题目记录,leetcode,算法,职场和发展

Input: n = 5, ranges = [3,4,1,1,0,0]
Output: 1
Explanation: The tap at point 0 can cover the interval [-3,3]
The tap at point 1 can cover the interval [-3,5]
The tap at point 2 can cover the interval [1,3]
The tap at point 3 can cover the interval [2,4]
The tap at point 4 can cover the interval [4,4]
The tap at point 5 can cover the interval [5,5]
Opening Only the second tap will water the whole garden [0,5]

Example 2:

Input: n = 3, ranges = [0,0,0,0]
Output: -1
Explanation: Even if you activate all the four taps you cannot water the whole garden.

Constraints:

1 <= n <= 10^4
ranges.length == n + 1
0 <= ranges[i] <= 100

Solution

For example2, we need to water the interval between 1 and 2.

Draw the range that each water tap could cover, like example1 does, then this problem is converted into 45. 跳跃游戏 II.

Build jump list, jump[i - span] = i + span

Then use greedy to solve jump game.

Time complexity: $o (n)$
Space complexity: $o (1)$ 文章来源地址https://www.toymoban.com/news/detail-687052.html

Code

class Solution:
    def minTaps(self, n: int, ranges: List[int]) -> int:
        intervals = [i for i in range(n + 1)]
        for index, each_range in enumerate(ranges):
            left = max(0, index - each_range)
            right = min(n, index + each_range)
            intervals[left] = max(intervals[left], right)
        res = 0
        cur_max = -1
        next_max = intervals[0]
        for i, jump in enumerate(intervals):
            if i > next_max:
                return -1
            if i > cur_max:
                res += 1
                cur_max = next_max
            next_max = max(next_max, jump)
        return res