Question

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

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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

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
  }
}