LeetCode #1326 — HARD

Minimum Number of Taps to Open to Water a Garden

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Greedy

Example 1

5
[3,4,1,1,0,0]

Example 2

3
[0,0,0,0]

Step 02

Core Insight

What unlocks the optimal approach

Create intervals of the area covered by each tap, sort intervals by the left end.
We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right.
What if there is a gap between intervals that is not covered ? we should stop and return -1 as there is some interval that cannot be covered.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1326: Minimum Number of Taps to Open to Water a Garden
class Solution {
    public int minTaps(int n, int[] ranges) {
        int[] last = new int[n + 1];
        for (int i = 0; i < n + 1; ++i) {
            int l = Math.max(0, i - ranges[i]), r = i + ranges[i];
            last[l] = Math.max(last[l], r);
        }
        int ans = 0, mx = 0, pre = 0;
        for (int i = 0; i < n; ++i) {
            mx = Math.max(mx, last[i]);
            if (mx <= i) {
                return -1;
            }
            if (pre == i) {
                ++ans;
                pre = mx;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1326: Minimum Number of Taps to Open to Water a Garden
func minTaps(n int, ranges []int) (ans int) {
	last := make([]int, n+1)
	for i, x := range ranges {
		l, r := max(0, i-x), i+x
		last[l] = max(last[l], r)
	}
	var pre, mx int
	for i, j := range last[:n] {
		mx = max(mx, j)
		if mx <= i {
			return -1
		}
		if pre == i {
			ans++
			pre = mx
		}
	}
	return
}

# Accepted solution for LeetCode #1326: Minimum Number of Taps to Open to Water a Garden
class Solution:
    def minTaps(self, n: int, ranges: List[int]) -> int:
        last = [0] * (n + 1)
        for i, x in enumerate(ranges):
            l, r = max(0, i - x), i + x
            last[l] = max(last[l], r)

        ans = mx = pre = 0
        for i in range(n):
            mx = max(mx, last[i])
            if mx <= i:
                return -1
            if pre == i:
                ans += 1
                pre = mx
        return ans

// Accepted solution for LeetCode #1326: Minimum Number of Taps to Open to Water a Garden
impl Solution {
    #[allow(dead_code)]
    pub fn min_taps(n: i32, ranges: Vec<i32>) -> i32 {
        let mut last = vec![0; (n + 1) as usize];
        let mut ans = 0;
        let mut mx = 0;
        let mut pre = 0;

        // Initialize the last vector
        for (i, &r) in ranges.iter().enumerate() {
            if (i as i32) - r >= 0 {
                last[((i as i32) - r) as usize] =
                    std::cmp::max(last[((i as i32) - r) as usize], (i as i32) + r);
            } else {
                last[0] = std::cmp::max(last[0], (i as i32) + r);
            }
        }

        for i in 0..n as usize {
            mx = std::cmp::max(mx, last[i]);
            if mx <= (i as i32) {
                return -1;
            }
            if pre == (i as i32) {
                ans += 1;
                pre = mx;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #1326: Minimum Number of Taps to Open to Water a Garden
function minTaps(n: number, ranges: number[]): number {
    const last = new Array(n + 1).fill(0);
    for (let i = 0; i < n + 1; ++i) {
        const l = Math.max(0, i - ranges[i]);
        const r = i + ranges[i];
        last[l] = Math.max(last[l], r);
    }
    let ans = 0;
    let mx = 0;
    let pre = 0;
    for (let i = 0; i < n; ++i) {
        mx = Math.max(mx, last[i]);
        if (mx <= i) {
            return -1;
        }
        if (pre == i) {
            ++ans;
            pre = mx;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.