Minimum Number of Taps to Open to Water a Garden

class Solution {
public:
    int minimumTaps(int gardenLength, vector<int>& radii) {
        // Prepare vector to store the furthest each point can be covered to
        vector<int> coverage(gardenLength + 1, 0);

        // Determine the max coverage for each garden point based on tap range
        for (int tap = 0; tap < radii.size(); tap++) {
            int leftBoundary = max(0, tap - radii[tap]);
            int rightBoundary = min(gardenLength, tap + radii[tap]);
            coverage[leftBoundary] = max(coverage[leftBoundary], rightBoundary);
        }

        int numTaps = 0; // Counter for minimum taps required
        int lastCover = 0; // Variable to track end of the current coverage
        int furthestCover = 0; // Variable to track the furthest possible end the next tap can cover

        // Iteratively explore each position in the garden
        for (int position = 0; position <= gardenLength; position++) {
            // Check if the current position is out of current range cover
            if (position > furthestCover) {
                return -1; // Impossible to cover this position
            }

            // Update and proceed to a new tap if needed
            if (position > lastCover) {
                numTaps++;
                lastCover = furthestCover; // Extend the covered range to the furthest known
            }

            // Update the furthest coverage from current position
            furthestCover = max(furthestCover, coverage[position]);
        }

        return numTaps; // Return the computed minimum number of taps required
    }
};

This solution implements a method to calculate the minimum number of taps required to water an entire garden of a specified length using taps placed at various points, each with a given radius range. This problem is tackled using a greedy approach, which involves determining the optimal placement and usage of each tap to maximize the area covered.

• Initialize a vector, coverage, where each index represents the maximum right boundary that can be covered starting from that index. This vector is populated by iterating over the radii of each tap and updating the coverage accordingly.
• Set two counters, numTaps and lastCover, to track the number of taps used and the endpoint of the last coverage range, and furthestCover to keep track of the farthest point that can potentially be covered.
• Iterate through each position in the garden using a for loop. If at any position, you find that the area up to that position cannot be covered by any of the taps used so far (position > furthestCover), it implies that the garden cannot be fully watered, and the function returns -1.
• If position exceeds lastCover, it indicates the need for an additional tap, hence numTaps is incremented and lastCover is updated to furthestCover.
• At each position, update furthestCover to ensure it reflects the maximum reach possible with the already considered taps.
• The loop continues until the garden's length is fully considered. The function finally returns numTaps, which represents the minimum number of taps required to cover the entire garden.

class Solution {
    public int minimumTaps(int gardenLength, int[] tapRanges) {
        int[] maxCover = new int[gardenLength + 1];

        for (int i = 0; i < tapRanges.length; i++) {
            int left = Math.max(0, i - tapRanges[i]);
            int right = Math.min(gardenLength, i + tapRanges[i]);
            maxCover[left] = Math.max(maxCover[left], right);
        }

        int usedTaps = 0;
        int currentReach = 0;
        int furthestReach = 0;

        for (int i = 0; i <= gardenLength; i++) {
            if (i > furthestReach) {
                return -1;
            }
            if (i > currentReach) {
                usedTaps++;
                currentReach = furthestReach;
            }
            furthestReach = Math.max(furthestReach, maxCover[i]);
        }

        return usedTaps;
    }
}

This solution addresses the problem of determining the minimum number of taps needed to water the entire length of a garden using Java. The approach involves creating a maxCover array to represent the maximum distance that can be covered when a tap at a particular position is turned on.

• Start by constructing the maxCover array, where each index indicates the furthest point to the right that the water can reach if a tap is opened at that index. This setup uses a combination of minimum and maximum calculations to avoid overshooting the garden's boundaries or under-utilizing a tap’s potential.
• Initialize usedTaps to keep count of how many taps need to be turned on, currentReach to store the current maximum distance covered, and furthestReach to track the furthest point reachable with all currently considered taps.
• Iteratively determine the number of taps required:
  • If you surpass the furthestReach and still have some garden left to cover, then it's not possible to water the whole garden with the available taps, hence return -1.
  • When the index exceeds currentReach, it indicates the need for a new tap. Increment usedTaps and update currentReach to furthestReach.
  • With each iteration, update furthestReach to the maximum of its current value and the coverage at the current index.
• Return the usedTaps as the final result once the loop completes, indicating the minimum taps required.

Overall, this solution effectively utilizes a greedy approach to ensure maximum coverage with minimal taps without exceeding the garden's length.

class Solution:
    def minimumTaps(self, garden_length: int, reach: List[int]) -> int:
        # Determine the maximum extent each position can be watered to
        waterable = [0] * (garden_length + 1)

        # Establish the watering reach from each tap position
        for position in range(len(reach)):
            left = max(0, position - reach[position])
            right = min(garden_length, position + reach[position])
            waterable[left] = max(waterable[left], right)
        
        # Initialize the tap operation metrics
        used_taps = 0
        current_max = 0
        next_max = 0

        # Loop through each position to ensure all can be watered
        for pos in range(garden_length + 1):
            if pos > next_max:
                # If the position cannot be reached, watering is not possible
                return -1
            if pos > current_max:
                # Use a new tap when exceeding the current reach
                used_taps += 1
                current_max = next_max
            # Update furthest position reachable with current taps
            next_max = max(next_max, waterable[pos])

        # Return total taps needed
        return used_taps

You are required to determine the minimum number of taps required to water an entire garden of a specified length. Each tap has a unique range, and your goal is to cover the entire garden using as few taps as possible.

In Python, utilize the minimumTaps function with parameters including garden_length and reach, defining the length of the garden and the list of maximum reaches from each tap, respectively.

Consider the following approach:

• Create an array waterable to store the furthest point that each position can be watered.
• Iterate over each tap, computing the maximum left and right extents to which the tap can water. Update the waterable array to reflect these extents.
• Initialize counters for taps used (used_taps), the current maximum distance covered (current_max), and the next possible furthest distance (next_max).
• Iterate from the start to the end of the garden, adjusting used_taps and extending the reach as necessary to ensure all positions are covered. If a position is found that exceeds next_max, it indicates the garden cannot be fully watered, and return -1.
• The output is the value of used_taps, representing the minimum number of taps required.

By following this approach strategically, ensure the entire garden is waterable using the minimum number of taps. The focus is on maximizing the reach from each tap and minimizing the number of taps activated. This solution provides an efficient way to address the problem of watering a garden with varying tap ranges.