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

Problem Description

In this problem, you have a garden represented as a one-dimensional line that starts at point 0 and ends at point n. Along this line, there are n + 1 taps at positions 0 through n. Each tap has a range associated with it, given by the array ranges. The range ranges[i]—which is associated with the tap at position i—indicates that this tap can water the garden from position i - ranges[i] to i + ranges[i].

Your objective is to figure out the minimum number of taps that need to be opened to water the entire length of the garden (from point 0 to point n). If it's not possible to water the whole garden with the given taps, the result should be -1.

Intuition

The problem is essentially about finding the minimum number of intervals (tap ranges) needed to cover a continuous segment (the garden). This is similar to the classic interval covering problem.

To solve this problem, we can use a greedy approach:

1. We start by transforming the problem into an interval covering problem. We'll create an array last where each index represents the starting point of an interval in the garden, and the value at each index represents the farthest point that can be watered from this interval. This is done by iterating over the ranges array and determining for each tap the leftmost and rightmost points it can water (i - ranges[i] and i + ranges[i] respectively).
2. We then apply a greedy method to find the minimum number of intervals needed to cover the entire garden.
   • Initialize counters for the current max right end of intervals mx, the previous furthest right end pre, and the number of taps ans.
   • Iterate over the garden from 0 to n. At each point, we update mx to be the maximum of mx and the furthest we can reach from the current position (found in last[i]). If at any point mx is less than or equal to i, it means there is a gap in coverage, and we cannot water the entire garden—return -1.
   • Whenever we reach the point pre (which is the end of the coverage from the selected taps), it means we need to select a new tap. Increment ans and set pre to mx.
3. By the end of the iteration, ans will hold the minimum number of taps needed to water the entire garden.

This approach ensures that at each step, we are extending the watering range as much as possible with the current number of taps, and we only add a new tap when necessary to cover new ground.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The solution provided uses a greedy algorithm to minimize the number of taps opened to cover the garden. Here's a detailed walkthrough of the implementation:

1. Create the last array: A list called last is initialized with 0 values and has a length of n + 1. This array will hold the maximum distance that can be watered starting from each position. The goal is to use this array to find the best tap to cover the longest distance from each point.

2. Populate the last array: We loop through the ranges array with the index i and range x. The variables l and r represent the leftmost (i - x) and rightmost (i + x) points that can be watered by tap i. We need to make sure that l does not go below 0, hence we use max(0, i - x). For each position l, we save the farthest point r that can be reached, by setting last[l] to max(last[l], r). This ensures that for each starting position l, we have the tap that waters the farthest.

3. Iterating through the garden: We initialize three variables ans, mx, and pre to track the number of taps used (ans), the current maximal right boundary (mx), and the previous maximal right boundary (pre). Now, we iterate through the garden from 0 to n.

4. Select taps and track coverage: During each iteration, we update mx to be the furthest point we can reach from the current point i, based on last[i]. If at any point mx is less than or equal to i, it means there is a gap, and the entire garden can't be watered, so we return -1. If we reach the pre point (where the current taps' coverage ends), it means we need to open another tap to extend the coverage, therefore we increment ans and update pre to mx.

5. Greedy selection: Each time we select a tap, we make the greedy choice by picking the tap that extends our coverage to the rightmost point possible. This is done by always updating mx with the maximum distance we can cover from i. Only when we have to move beyond pre do we lock in our choice of a tap and increment ans since we know up to that point we have covered with the minimum number of taps.

6. Returning the answer: Once the loop is complete, we have either returned -1 if the garden can't be watered completely, or we have the minimum number of taps required in ans, which we return as the final result.

In summary, by transforming the problem into finding the optimal coverage for each interval and then iteratively expanding these intervals using a greedy approach, we efficiently find the minimum number of taps needed to cover the whole garden represented by the x-axis from 0 to n.

Example Walkthrough

Let's illustrate the greedy solution approach with a small example. Suppose our garden is represented by a line of length n = 5 and we have 6 taps positioned at 0 through 5. The ranges array provided is [1, 0, 2, 1, 2, 0], representing the range each tap can water on both sides.

Step-by-Step Solution:

1. Create the last array: We create an array last of length n + 1, which gives us last = [0, 0, 0, 0, 0, 0].

2. Populate the last array: We process the provided ranges:

   • Tap 0 has range 1: it can water the garden from max(0, 0-1) to 0+1, hence we update last[0] = 1.
   • Tap 1 has range 0: it can only water its own position, so no update is necessary (last[1] remains 0).
   • Tap 2 has range 2: it can water from max(0, 2-2) to 2+2, so we update last[0] = 4 (since this tap offers a better range than tap 0).
   • Tap 3 has range 1: waters from max(0, 3-1) to 3+1, we update last[2] = 4.
   • Tap 4 has range 2: waters from max(0, 4-2) to 4+2, we update last[2] = 6.
   • Tap 5 has range 0: waters only its position, no update. After updating, last = [4, 0, 6, 4, 6, 0].

3. Iterating through the garden: We initialize the counters ans = 0, mx = 0, and pre = 0.

4. Select taps and track coverage: We iterate from i = 0 to i = 5 (position n in the garden):

   • At i = 0: mx = max(0, 4) = 4. No gap, continue.
   • At i = 1: mx = 4. Still no gap, continue.
   • At i = 2: mx = max(4, 6) = 6. As i = pre, we select this tap and update ans = 1, pre = mx = 6.
   • At i = 3, 4 and 5: we don't need to update mx or ans since we've already covered the garden with pre = 6.

5. Returning the answer: Since we've iterated through the whole garden without encountering a gap, the final answer is ans = 1. Only one tap (at position 2) is needed to water the entire garden from 0 to 5.

Solution Implementation

Time and Space Complexity

Time Complexity

The given Python code involves a single pass to transform the ranges list into the last list, followed by another single pass to find the minimum number of taps needed. Therefore, this algorithm runs in two linear passes over an array of length n + 1.

• The first pass is constructing the last array, which involves iterating over the ranges list once, resulting in O(n) complexity, where n is the length of the ranges list.
• The second pass is the loop that determines the minimum number of taps by iterating once through the last array, which is also O(n) complexity.

Combining both passes, the overall time complexity remains O(n) because they are sequential, not nested.

Space Complexity

The space complexity is determined by the additional space used by the algorithm besides the input. Here, the last array of size n + 1 is an auxiliary space used to store the farthest point that can be reached from each index. Therefore, the space complexity for the last array is O(n).

No other data structures that are dependent on the size of the input are used, so the total space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.