What is the time complexity of Minimum Number of Taps to Open to Water a Garden?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Number of Taps to Open to Water a Garden?

This problem uses Array, Dynamic Programming, Greedy. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Number of Taps to Open to Water a Garden?

Insight 1: Convert the problem into a covering problem. Each tap represents an interval, and we need to find the minimum number of intervals to cover the range [0, n]. Insight 2: Use a greedy approach. Sort or process the taps in a way that allows us to choose the tap that covers the most unwatered area at each step.

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Minimum Number of Taps to Open to Water a Garden - LeetCode 1326 Solution

Q: Which companies ask LeetCode 1326. Minimum Number of Taps to Open to Water a Garden in interviews?

This problem is frequently asked at Akuna Capital, BNY Mellon, DE Shaw, Flipkart, Intuit and 1 more companies.

Minimum Number of Taps to Open to Water a Garden - Complete Solution Guide

Minimum Number of Taps to Open to Water a Garden is LeetCode problem 1326, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 wa

Detailed Explanation

The problem asks us to find the minimum number of water taps to open in a one-dimensional garden to water the entire garden. The garden extends from point 0 to point n on the x-axis. There are n+1 taps located at points 0 to n. Each tap i has a range 'ranges[i]' that defines the area it can water: [i - ranges[i], i + ranges[i]]. The goal is to find the smallest number of taps that, when opened, can collectively water the entire interval [0, n]. If it's impossible to water the entire garden, return -1.

Solution Approach

The solution uses a greedy approach. First, it preprocesses the `ranges` array to create a `max_reach` array. `max_reach[i]` stores the farthest point that can be reached starting from index i. Then, it iterates through the garden from 0 to n, greedily selecting the tap that allows it to reach the farthest. The algorithm maintains `current_end` (the farthest point currently watered), `next_end` (the farthest point reachable from the current watered area), and `taps` (the number of taps used). If at any point it cannot extend the watered area (`next_end <= current_end`), it means the garden cannot be completely watered, and it returns -1.

Step-by-Step Algorithm

Step 1: Create a `max_reach` array of size n+1, initialized with zeros. This array will store the maximum reach from each starting point.
Step 2: Iterate through the `ranges` array. For each tap i with range `ranges[i]`, calculate the `start` and `end` points of its watering range. `start = max(0, i - ranges[i])` and `end = min(n, i + ranges[i])`. Update `max_reach[start]` to be the maximum of its current value and `end`.
Step 3: Initialize `taps = 0`, `current_end = 0`, `next_end = 0`, and `i = 0`. `current_end` represents the farthest point that has been watered so far. `next_end` represents the farthest point that can be reached from the current watered area.
Step 4: While `current_end < n`, iterate from `i` to `current_end`. In each inner iteration, update `next_end` to be the maximum of its current value and `max_reach[i]`. This finds the tap within the currently watered area that can reach the farthest.
Step 5: After the inner loop, if `next_end <= current_end`, it means no tap can extend the watered area, so return -1.
Step 6: Update `current_end = next_end` and increment `taps`.
Step 7: Continue the outer loop until `current_end >= n`, meaning the entire garden is watered. Return `taps`.

Key Insights

Insight 1: Convert the problem into a covering problem. Each tap represents an interval, and we need to find the minimum number of intervals to cover the range [0, n].
Insight 2: Use a greedy approach. Sort or process the taps in a way that allows us to choose the tap that covers the most unwatered area at each step.
Insight 3: The problem can be viewed as finding the minimum number of intervals that cover a range. This is a classical covering problem that can be efficiently solved using a greedy strategy after preprocessing the ranges.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Greedy.

Companies

Asked at: Akuna Capital, BNY Mellon, DE Shaw, Flipkart, Intuit, ServiceNow.

Minimum Number of Taps to Open to Water a Garden | LeetCode 1326

Hard

Array Dynamic Programming Greedy

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int minTaps(int n, int[] ranges) {
        int[] maxReach = new int[n + 1];
        for (int i = 0; i < ranges.length; i++) {
            int r = ranges[i];
            int start = Math.max(0, i - r);
            int end = Math.min(n, i + r);
            maxReach[start] = Math.max(maxReach[start], end);
        }

        int taps = 0;
        int currentEnd = 0;
        int nextEnd = 0;
        int i = 0;

        while (currentEnd < n) {
            while (i <= currentEnd) {
                nextEnd = Math.max(nextEnd, maxReach[i]);
                i++;
            }

            if (nextEnd <= currentEnd) {
                return -1;
            }

            currentEnd = nextEnd;
            taps++;
        }

        return taps;
    }
}

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Problem Info

Description

Problem 1326: Minimum Number of Taps to Open to Water a Garden - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy

Companies

Akuna CapitalBNY MellonDE ShawFlipkartIntuitServiceNow

Stats

Difficulty:Hard

Languages:4

Companies:6

Problem #:1326

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Create a `max_reach` array of size n+1, initialized with zeros. This array will store the maximum reach from each starting point.
2
Step 2: Iterate through the `ranges` array. For each tap i with range `ranges[i]`, calculate the `start` and `end` points of its watering range. `start = max(0, i - ranges[i])` and `end = min(n, i + ranges[i])`. Update `max_reach[start]` to be the maximum of its current value and `end`.
3
Step 3: Initialize `taps = 0`, `current_end = 0`, `next_end = 0`, and `i = 0`. `current_end` represents the farthest point that has been watered so far. `next_end` represents the farthest point that can be reached from the current watered area.
4
Step 4: While `current_end < n`, iterate from `i` to `current_end`. In each inner iteration, update `next_end` to be the maximum of its current value and `max_reach[i]`. This finds the tap within the currently watered area that can reach the farthest.
5
Step 5: After the inner loop, if `next_end <= current_end`, it means no tap can extend the watered area, so return -1.
6
Step 6: Update `current_end = next_end` and increment `taps`.
7
Step 7: Continue the outer loop until `current_end >= n`, meaning the entire garden is watered. Return `taps`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the two loops which iterate over the length of the garden. The space complexity is determined by the auxiliary `max_reach` array which is proportional to the length of the garden.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

We can solve this problem using dynamic programming. Let dp[i] be the minimum number of taps to water the garden up to point i. The transition would involve iterating through all taps j that cover point i and updating dp[i] = min(dp[i], dp[j-1] + 1).

Time: O(n^2)Space: O(n)

Trade-offs: Dynamic programming is less efficient than the greedy approach due to the nested loops. While it provides a solution, the greedy approach offers better performance in this specific problem.

✨

Key Insights

1Insight 1: Convert the problem into a covering problem. Each tap represents an interval, and we need to find the minimum number of intervals to cover the range [0, n].
2Insight 2: Use a greedy approach. Sort or process the taps in a way that allows us to choose the tap that covers the most unwatered area at each step.
3Insight 3: The problem can be viewed as finding the minimum number of intervals that cover a range. This is a classical covering problem that can be efficiently solved using a greedy strategy after preprocessing the ranges.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: If a tap has a range of 0, it doesn't cover any area. The algorithm correctly handles this because the `start` and `end` points will be the same, so the `max_reach` array will not be updated.
Edge case 2: If the garden cannot be watered completely, the algorithm returns -1. This is checked by ensuring `next_end` can always extend past `current_end`. If not, the algorithm recognizes that the watering cannot continue and returns -1.
Edge case 3: If the ranges array contains very large numbers, the algorithm handles this correctly because of the `min` and `max` functions when calculating start and end of tap coverage, ensuring the indexes are within the bounds of the garden length 'n'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the garden cannot be completely watered, which results in an infinite loop or incorrect output.

✗Common mistake 2: Incorrectly calculating the start and end points of the watering range for each tap, leading to inaccurate coverage calculations.

✗Common mistake 3: Not initializing the `max_reach` array correctly, which could lead to unexpected behavior and incorrect results.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine allocating cellular towers along a highway to provide coverage. Each tower has a certain coverage range, and the goal is to minimize the number of towers needed to cover the entire highway.

🏦Real-world scenario 2: Task scheduling. Suppose you have a set of tasks, each with a start and end time. You want to select the minimum number of workers to perform all the tasks, where each worker can only perform one task at a time. This is analogous to finding the minimum number of intervals that cover a given range.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Number of Taps to Open to Water a Garden"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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