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

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:

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, and also jump[i] = i + span.

Then use greedy to solve jump game.

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

class Solution:
    def minTaps(self, n: int, ranges: List[int]) -> int:
        if not ranges:
            return -1
        range_array = list(range(len(ranges)))
        for i, span in enumerate(ranges):
            range_array[i] = max(range_array[i], i + span)
            update_index = max(0, i - span)
            range_array[update_index] = max(range_array[update_index], i + span)
        cur_most = -1
        next_most = range_array[0]
        res = 0
        for i, jump_step in enumerate(range_array):
            if i > next_most:
                res = -1
                break
            if i > cur_most:
                res += 1
                cur_most = next_most
            next_most = max(next_most, jump_step)
        return res