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

  • 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;
        }
    };
    
  • print("Todo!")
    

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