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Minimum Number of Taps to Open to Water a Garden

Difficulty: Hard

Problem Description

There is a one-dimensional garden on the x-axis. The garden starts at the point 0 and ends at the point 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] 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.

Key Insights

  • Each tap can cover a specific range based on its position and its range value.
  • The problem can be transformed into an interval covering problem.
  • A greedy approach can be used to select the minimum number of taps needed to cover the entire garden.

Space and Time Complexity

Time Complexity: O(n log n)
Space Complexity: O(n)

Solution

To solve the problem, we can utilize a greedy algorithm that focuses on covering the garden with the minimum number of taps. The approach involves the following steps:

  1. Define Coverage Intervals: For each tap, determine the range it can cover, which is defined by the interval [i - ranges[i], i + ranges[i]].
  2. Sort Intervals: Sort these intervals based on their starting points. If two intervals have the same starting point, sort by their ending points in descending order.
  3. Iterate and Select Intervals: Starting from the leftmost point of the garden (0), iteratively choose the tap that extends the coverage the farthest until the entire garden (0 to n) is covered or it is determined that it cannot be covered.

Code Solutions

def minTaps(n, ranges):
    # Create an array to store the coverage intervals
    intervals = []
    
    for i in range(n + 1):
        left = max(0, i - ranges[i])
        right = min(n, i + ranges[i])
        intervals.append((left, right))
    
    # Sort intervals based on the starting point (and by ending point if needed)
    intervals.sort()
    
    taps = 0
    farthest = 0
    current_end = 0
    index = 0
    
    while current_end < n:
        # Find the farthest point we can reach with the current taps
        while index < len(intervals) and intervals[index][0] <= current_end:
            farthest = max(farthest, intervals[index][1])
            index += 1
        
        if current_end == farthest:
            return -1  # Can't reach any further
        
        taps += 1
        current_end = farthest  # Move to the furthest point reached
    
    return taps
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