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Formatted question description: https://leetcode.ca/all/1326.html

1326. Minimum Number of Taps to Open to Water a Garden (Hard)

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.

Example 3:

Input: n = 7, ranges = [1,2,1,0,2,1,0,1]
Output: 3

Example 4:

Input: n = 8, ranges = [4,0,0,0,0,0,0,0,4]
Output: 2

Example 5:

Input: n = 8, ranges = [4,0,0,0,4,0,0,0,4]
Output: 1

 

Constraints:

• 1 <= n <= 10^4
• ranges.length == n + 1
• 0 <= ranges[i] <= 100

Related Topics:
Dynamic Programming, Greedy

Solution 1.

// OJ: https://leetcode.com/problems/minimum-number-of-taps-to-open-to-water-a-garden/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int minTaps(int n, vector<int>& ranges) {
        vector<pair<int, int>> v;
        for (int i = 0; i < ranges.size(); ++i)
            v.emplace_back(max(0, i - ranges[i]),  min(n, i + ranges[i]));
        sort(v.begin(), v.end());
        int end = 0, newEnd = 0, ans = 0;
        for (int i = 0; i < v.size() && end != n; ++i) {
            if (v[i].first > end) return -1;
            newEnd = max(newEnd, v[i].second);
            if (i + 1 == v.size() || v[i + 1].first > end) {
                end = newEnd;
                ++ans;
            }
        }
        return ans;
    }
};

• Java
• C++
• Python

• class Solution {
      public int minTaps(int n, int[] ranges) {
          int[][] minMaxArray = new int[n + 1][2];
          for (int i = 0; i <= n; i++) {
              int range = ranges[i];
              int min = Math.max(i - range, 0);
              int max = Math.min(i + range - 1, n - 1);
              minMaxArray[i][0] = min;
              minMaxArray[i][1] = max;
          }
          boolean[] covered = new boolean[n];
          int coverCount = 0;
          int openCount = 0;
          boolean[] used = new boolean[n + 1];
          boolean flag = true;
          while (flag) {
              flag = false;
              for (int i = 0; i < n; i++) {
                  if (covered[i])
                      continue;
                  int maxLengthIndex = -1;
                  int maxLength = 0;
                  for (int j = 0; j <= n; j++) {
                      if (used[j])
                          continue;
                      int[] array = minMaxArray[j];
                      int min = array[0], max = array[1];
                      if (min <= i && max >= i) {
                          int length = 0;
                          for (int k = min; k <= max; k++) {
                              if (!covered[k])
                                  length++;
                          }
                          if (length > maxLength) {
                              maxLengthIndex = j;
                              maxLength = length;
                          }
                      }
                  }
                  if (maxLengthIndex >= 0) {
                      used[maxLengthIndex] = true;
                      int min = minMaxArray[maxLengthIndex][0], max = minMaxArray[maxLengthIndex][1];
                      for (int j = min; j <= max; j++) {
                      	if (!covered[j]) {
                      	    covered[j] = true;
                              coverCount++;
                      	}
                      }
                      flag = true;
                      openCount++;
                  }
              }
          }
          return coverCount == n ? openCount : -1;
      }
  }
  

• // OJ: https://leetcode.com/problems/minimum-number-of-taps-to-open-to-water-a-garden/
  // Time: O(NlogN)
  // Space: O(N)
  class Solution {
  public:
      int minTaps(int n, vector<int>& ranges) {
          vector<pair<int, int>> v;
          for (int i = 0; i < ranges.size(); ++i)
              v.emplace_back(max(0, i - ranges[i]),  min(n, i + ranges[i]));
          sort(v.begin(), v.end());
          int end = 0, newEnd = 0, ans = 0;
          for (int i = 0; i < v.size() && end != n; ++i) {
              if (v[i].first > end) return -1;
              newEnd = max(newEnd, v[i].second);
              if (i + 1 == v.size() || v[i + 1].first > end) {
                  end = newEnd;
                  ++ans;
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def minTaps(self, n: int, ranges: List[int]) -> int:
          last = [0] * (n + 1)
          for i, v in enumerate(ranges):
              l, r = max(0, i - v), min(n, i + v)
              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
  
  
  

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