Minimum Number of Taps to Open to Water a Garden

Hard

Intuit

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Topics:

Greedy AlgorithmsArrays

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

Minimum Number of Taps to Open to Water a Garden

Hard

Intuit

1 view

Topics:

Greedy AlgorithmsArrays

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

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the length of the input array `ranges` and the maximum value of any element within it?
If it is impossible to water the entire garden, what value should be returned?
Can a range's value be zero, meaning it only waters the position of the tap itself?
Is it possible for any of the range values to be negative, and if so, how should those be handled?
If multiple combinations of taps can water the entire garden with the minimum number of taps, is any one of them acceptable?

Brute Force Solution

Approach

To find the fewest taps needed to water the entire garden using the brute force method, we'll explore every possible combination of taps. This involves trying out all the different subsets of taps and figuring out if any of those combinations cover the whole garden.

Here's how the algorithm would work step-by-step:

First, consider using no taps at all and check if the whole garden gets watered. It won't, but we need to check it!
Then, try using only one tap at a time. Check if any single tap can water the entire garden. If one can, we're done!
If not, try all possible pairs of taps. See if any combination of two taps can water the entire garden.
Continue this process by trying all combinations of three taps, four taps, and so on, until you've tried using all the taps at once.
For each combination of taps, determine the area of the garden that combination covers.
Keep track of the smallest number of taps needed to fully water the garden.
Once you've explored all possible combinations, the smallest number you kept track of is the answer.

Code Implementation

def min_taps_brute_force(garden_length, ranges):    min_taps_needed = float('inf')
    for i in range(1 << len(ranges)):        taps_used_count = 0
        covered_area = [False] * garden_length
        # Iterate through each tap to check if it's used in the current combination
        for tap_index in range(len(ranges)):            if (i >> tap_index) & 1:
                taps_used_count += 1
                reach = ranges[tap_index]
                tap_start = max(0, tap_index - reach)
                tap_end = min(garden_length, tap_index + reach + 1)

                # Mark the area covered by the current tap
                for j in range(tap_start, tap_end):
                    covered_area[j] = True

        # Check if the current combination covers the entire garden
        if all(covered_area):
            min_taps_needed = min(min_taps_needed, taps_used_count)
    # If no combination covers the garden, return -1
    if min_taps_needed == float('inf'):        return -1
    else:
        return min_taps_needed

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(2^n * n) – The brute force solution explores every possible subset of taps. For 'n' taps, there are 2^n possible subsets. For each subset, we need to iterate through all the garden positions, which takes O(n) time in the worst case to determine if the garden is fully watered by the selected taps. Therefore, the total time complexity is O(2^n * n), dominated by the exponential growth of subsets.

Space Complexity

O(1) – The brute force approach described explores combinations of taps without storing all combinations simultaneously. It calculates the area covered for each combination and tracks the minimum number of taps needed so far. Therefore, the algorithm only needs to store a few integer variables to keep track of the current combination being tested, the covered area, and the minimum taps found so far, regardless of the number of taps N. The space used remains constant and does not scale with the input size.

Optimal Solution

Approach

The goal is to find the fewest sprinklers needed to water the entire garden. We'll use a greedy approach: at each step, we pick the sprinkler that waters furthest to the right, as long as it helps us extend our watered area.

Here's how the algorithm would work step-by-step:

Imagine you're walking along the garden from left to right, and you want to figure out which sprinkler to activate at each point.
For each point in the garden, find all the sprinklers that can reach that point.
Out of those sprinklers, choose the one that reaches furthest to the right. This is the 'greediest' choice because it covers the most unwatered area.
Once you've chosen a sprinkler, you know that everything it waters is now covered. So, advance your position to the furthest point watered by that sprinkler.
Repeat this process until you've reached the end of the garden. If at any point, you can't find a sprinkler to reach your current position, it means you can't water the entire garden.
The number of sprinklers you activated is the minimum number needed to water the whole garden.

Code Implementation

def min_taps(garden_length, ranges):    max_reach = [0] * (garden_length + 1)

    for i, current_range in enumerate(ranges):
        left = max(0, i - current_range)
        right = min(garden_length, i + current_range)
        max_reach[left] = max(max_reach[left], right)

    taps_needed = 0
    current_reach = 0
    next_reach = 0

    for i in range(garden_length):
        # If we can't advance, the garden can't be watered.
        if next_reach <= i and max_reach[i] == i:
            return -1

        next_reach = max(next_reach, max_reach[i])

        # When current reach equals i, we must open a tap
        if current_reach == i:
            taps_needed += 1
            current_reach = next_reach

    return taps_needed

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates conceptually across the garden's length 'n'. At each position, it considers sprinklers that can reach that position. In the worst-case scenario, for each position, it may have to examine all 'n' sprinklers to find the one that reaches furthest to the right. This results in a nested loop-like behavior where, for each of the 'n' positions in the garden, a maximum of 'n' sprinklers are potentially checked. Hence, the time complexity approximates n*n operations, which simplifies to O(n^2).

Space Complexity

O(N) – The algorithm implicitly requires creating an array or list of size N+1 (call it 'reachable') to store the maximum reachable position for each garden position i from 0 to N. This array is used to determine the furthest a sprinkler can reach for any given position in the garden. Therefore, the auxiliary space used is directly proportional to the size of the garden, N, leading to O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty input array n = 0	Return 0 since no garden exists to water.
All ranges are 0, meaning no tap waters any part of the garden	Return -1 since it is impossible to water the entire garden.
One tap covers the entire garden range [0, n]	Return 1, as only one tap is needed.
Maximum-sized input array n near the integer limit and large ranges	Ensure the interval array construction and subsequent greedy selection avoids potential integer overflow.
No tap at the start of the garden (index 0)	Return -1, as there's no way to water the initial part of the garden without a tap at or near index 0.
Taps are sparsely distributed such that there are unwatered sections	Return -1, indicating that the garden cannot be fully watered with the given taps.
Taps with overlapping coverage; greedy selection must prioritize the tap that extends furthest	The greedy approach should correctly choose the tap that covers the largest unwatered segment.
Input contains negative or fractional watering ranges	Assume the input is non-negative; explicitly handle cases of negative range values, either by rejecting input or interpreting them as 0.