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

Description

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

Solutions

Solution 1: Greedy

We note that for all taps that can cover a certain left endpoint, choosing the tap that can cover the farthest right endpoint is optimal.

Therefore, we can preprocess the array \(ranges\). For the \(i\)-th tap, it can cover the left endpoint \(l = \max(0, i - ranges[i])\) and the right endpoint \(r = i + ranges[i]\). We calculate the position of the tap that can cover the left endpoint \(l\) with the farthest right endpoint and record it in the array \(last[i]\).

Then we define the following three variables:

Variable \(ans\) represents the final answer, i.e., the minimum number of taps;
Variable \(mx\) represents the farthest right endpoint that can currently be covered;
Variable \(pre\) represents the farthest right endpoint covered by the previous tap.

We traverse all positions in the range \([0, \ldots, n-1]\). For the current position \(i\), we use \(last[i]\) to update \(mx\), i.e., \(mx = \max(mx, last[i])\).

If \(mx \leq i\), it means the next position cannot be covered, so we return \(-1\).
If \(pre = i\), it means a new subinterval needs to be used, so we increment \(ans\) by \(1\) and update \(pre = mx\).

After the traversal, we return \(ans\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the length of the garden.

Similar problems:

Python3JavaC++GoTypeScriptRust

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

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;
    }
}

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

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
}

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;
}

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
    }
}

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

Description

Solutions

Solution 1: Greedy

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