Minimum Number of Taps to Open to Water a Garden

Problem

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.

Examples

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

Method 1 – Greedy Interval Covering

Intuition

This problem is similar to the Jump Game II or interval covering. For each tap, compute the interval it can water. Then, use a greedy approach to always open the tap that extends coverage the farthest among those starting before or at the current end.

Approach

1. For each tap, compute its watering interval [max(0, i - ranges[i]), min(n, i + ranges[i])].
2. For each position, keep track of the farthest right it can be watered by any tap starting at or before that position.
3. Use a greedy scan: at each step, open the tap that extends coverage the farthest.
4. If at any point coverage cannot be extended, return -1.
5. Continue until coverage reaches or exceeds n.

Code

C++

class Solution {
public:
    int minTaps(int n, vector<int>& ranges) {
        vector<int> maxReach(n+1, 0);
        for (int i = 0; i <= n; ++i) {
            int l = max(0, i - ranges[i]), r = min(n, i + ranges[i]);
            maxReach[l] = max(maxReach[l], r);
        }
        int ans = 0, end = 0, far = 0;
        for (int i = 0; i <= n; ++i) {
            if (i > far) return -1;
            if (i > end) {
                ans++;
                end = far;
            }
            far = max(far, maxReach[i]);
        }
        return ans;
    }
};

Go

func minTaps(n int, ranges []int) int {
    maxReach := make([]int, n+1)
    for i := 0; i <= n; i++ {
        l, r := max(0, i-ranges[i]), min(n, i+ranges[i])
        if maxReach[l] < r {
            maxReach[l] = r
        }
    }
    ans, end, far := 0, 0, 0
    for i := 0; i <= n; i++ {
        if i > far { return -1 }
        if i > end {
            ans++
            end = far
        }
        if maxReach[i] > far {
            far = maxReach[i]
        }
    }
    return ans
}
func max(a, b int) int { if a > b { return a }; return b }
func min(a, b int) int { if a < b { return a }; return b }

Java

class Solution {
    public int minTaps(int n, int[] ranges) {
        int[] maxReach = new int[n+1];
        for (int i = 0; i <= n; i++) {
            int l = Math.max(0, i - ranges[i]), r = Math.min(n, i + ranges[i]);
            maxReach[l] = Math.max(maxReach[l], r);
        }
        int ans = 0, end = 0, far = 0;
        for (int i = 0; i <= n; i++) {
            if (i > far) return -1;
            if (i > end) {
                ans++;
                end = far;
            }
            far = Math.max(far, maxReach[i]);
        }
        return ans;
    }
}

Kotlin

class Solution {
    fun minTaps(n: Int, ranges: IntArray): Int {
        val maxReach = IntArray(n+1)
        for (i in 0..n) {
            val l = maxOf(0, i - ranges[i])
            val r = minOf(n, i + ranges[i])
            maxReach[l] = maxOf(maxReach[l], r)
        }
        var ans = 0; var end = 0; var far = 0
        for (i in 0..n) {
            if (i > far) return -1
            if (i > end) {
                ans++
                end = far
            }
            far = maxOf(far, maxReach[i])
        }
        return ans
    }
}

Python

def minTaps(n: int, ranges: list[int]) -> int:
    maxReach = [0]*(n+1)
    for i in range(n+1):
        l, r = max(0, i-ranges[i]), min(n, i+ranges[i])
        maxReach[l] = max(maxReach[l], r)
    ans = end = far = 0
    for i in range(n+1):
        if i > far:
            return -1
        if i > end:
            ans += 1
            end = far
        far = max(far, maxReach[i])
    return ans

Rust

impl Solution {
    pub fn min_taps(n: i32, ranges: Vec<i32>) -> i32 {
        let n = n as usize;
        let mut max_reach = vec![0; n+1];
        for i in 0..=n {
            let l = i.saturating_sub(ranges[i] as usize);
            let r = (i + ranges[i] as usize).min(n);
            max_reach[l] = max_reach[l].max(r);
        }
        let (mut ans, mut end, mut far) = (0, 0, 0);
        for i in 0..=n {
            if i > far { return -1; }
            if i > end {
                ans += 1;
                end = far;
            }
            far = far.max(max_reach[i]);
        }
        ans as i32
    }
}

TypeScript

class Solution {
    minTaps(n: number, ranges: number[]): number {
        const maxReach = Array(n+1).fill(0);
        for (let i = 0; i <= n; i++) {
            const l = Math.max(0, i - ranges[i]), r = Math.min(n, i + ranges[i]);
            maxReach[l] = Math.max(maxReach[l], r);
        }
        let ans = 0, end = 0, far = 0;
        for (let i = 0; i <= n; i++) {
            if (i > far) return -1;
            if (i > end) {
                ans++;
                end = far;
            }
            far = Math.max(far, maxReach[i]);
        }
        return ans;
    }
}

Complexity

• ⏰ Time complexity: O(n), since we process each tap and scan the garden.
• 🧺 Space complexity: O(n), for the maxReach array.