Minimum Number of Taps to Open to Water a Garden

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]

思路：这题，转换成interval，就是1024，video stitching.

sort by start 解决问题：=> the minimum number of intervals to cover the whole range.

sort by end 解决问题: => the maximum number of non-overlapping intervals.

算法就是：sort by start, 因为前面的也要cover，首先找一个end最大的，然后在start <= end的所有interval中找end最大的。这样就是greedy的思想；如果nextfar >= T, 那么停止，已经找完了，如果nextfar == far说明走不动了，return-1;

class Solution {
    public int minTaps(int n, int[] ranges) {
        int[][] intervals = new int[n + 1][2];
        for(int i = 0; i <= n; i++) {
            intervals[i][0] = i - ranges[i];
            intervals[i][1] = i + ranges[i];
        }
        Arrays.sort(intervals, (a, b)->(a[0] != b[0] ? a[0] - b[0] : b[1] - a[1]));
        
        int i = 0;
        int far = 0;
        int count = 0;
        while(i < intervals.length) {
            int nextfar = far;
            while(i < intervals.length && intervals[i][0] <= far) {
                nextfar = Math.max(nextfar, intervals[i][1]);
                i++;
            }
            count++;
            if(nextfar >= n) {
                return count;
            } else if(nextfar == far) {
                return -1;
            }
            far = nextfar;
        }
        return -1;
    }
}