How do you solve Minimum Number of Taps to Open to Water a Garden on LeetCode?

Minimum Number of Taps to Open to Water a Garden (LeetCode #1326) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to represent the coverage of taps is crucial for optimizing the solution.

What is the brute force solution for Minimum Number of Taps to Open to Water a Garden?

The brute force approach for Minimum Number of Taps to Open to Water a Garden is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible combinations of taps to find the minimum number needed to cover the entire garden. This method is straightforward but inefficient, as it can take a long time for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Taps to Open to Water a Garden?

Minimum Number of Taps to Open to Water a Garden uses the following concepts: Array, Dynamic Programming, Greedy. The recommended patterns to study are: Greedy, Dynamic Programming, Interval Covering.

What are the interview tips for Minimum Number of Taps to Open to Water a Garden?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases and test your solution against them. Explain your thought process and reasoning while coding to demonstrate your understanding.

What are common mistakes when solving Minimum Number of Taps to Open to Water a Garden?

Failing to account for taps that don't cover any area, leading to incorrect assumptions about coverage. Not considering edge cases where the garden cannot be fully watered.

#1326

Minimum Number of Taps to Open to Water a Garden

Hard

ArrayDynamic ProgrammingGreedyGreedyDynamic ProgrammingInterval Covering

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a greedy approach to find the minimum number of taps needed. By prioritizing taps that cover the farthest distance, we can efficiently cover the entire garden without checking all combinations.

⚙️

Algorithm

3 steps

1Step 1: Create an array to represent the maximum right coverage for each position.
2Step 2: Populate this array based on the ranges of each tap.
3Step 3: Use a greedy approach to select taps that extend the coverage as far as possible, counting how many taps are needed.

solution.py17 lines

1def minTaps(n, ranges):
2    max_right = [0] * (n + 1)
3    for i in range(n + 1):
4        left = max(0, i - ranges[i])
5        right = min(n, i + ranges[i])
6        max_right[left] = max(max_right[left], right)
7    taps = 0
8    farthest = 0
9    current_end = 0
10    for i in range(n + 1):
11        farthest = max(farthest, max_right[i])
12        if i == current_end:
13            taps += 1
14            current_end = farthest
15            if current_end >= n:
16                return taps
17    return -1

ℹ

Complexity note: The time complexity is linear because we only traverse the garden once to calculate the maximum coverage and then once more to determine the number of taps needed. The space complexity is linear due to the additional array used to store maximum coverage.

1Understanding how to represent the coverage of taps is crucial for optimizing the solution.
2Greedy algorithms can often provide efficient solutions for interval covering problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.