Introduction

In this blog post, we will explore how to solve the LeetCode problem "1326" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution to the garden watering problem:

• Greedy Approach: Convert the ranges into intervals. Sort the intervals based on their start points. Iterate through the intervals and greedily pick the interval that covers the maximum length of the garden from the current uncovered position.
• Interval Merging: If there's a gap in the garden that cannot be covered by any tap, return -1. The algorithm ensures complete coverage by selecting intervals that maximize reach.

• Edge Cases: Handle the case where the garden cannot be fully watered, which should return -1.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(n) to store the intervals.

Code

    def minTaps(n: int, ranges: list[int]) -> int:
    """
    Calculates the minimum number of taps to open to water the entire garden.

    Args:
        n: The length of the garden.
        ranges: A list of integers where ranges[i] represents the watering range of the i-th tap.

    Returns:
        The minimum number of taps needed, or -1 if the garden cannot be watered.
    """

    intervals = []
    for i in range(n + 1):
        intervals.append((i - ranges[i], i + ranges[i]))

    intervals.sort()

    taps = 0
    covered = 0
    i = 0
    while covered < n:
        max_reach = covered
        while i < len(intervals) and intervals[i][0] <= covered:
            max_reach = max(max_reach, intervals[i][1])
            i += 1

        if max_reach == covered:
            return -1  # Cannot water the garden

        taps += 1
        covered = max_reach

    return taps