Skip to content

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

Difficulty Topics

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 <= 104
  • ranges.length == n + 1
  • 0 <= ranges[i] <= 100

Solution

minimum-number-of-taps-to-open-to-water-a-garden.py
class Solution:
    def minTaps(self, n: int, ranges: List[int]) -> int:
        A = [0] * (n + 1)

        for i, x in enumerate(ranges):
            if x == 0: continue

            left = max(i - x, 0)
            A[left] = max(A[left], i + x)

        jumps = currStep = currFarthest = i = 0

        while i < len(A) and currStep < n:
            jumps += 1

            while i < len(A) and i <= currStep:
                currFarthest = max(currFarthest, A[i])
                i += 1

            if currStep == currFarthest: return -1

            currStep = currFarthest

        return jumps