LeetCode-1326. 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]

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

题解：

之前按点算的，一直不对，看讨论区是按区间算的，dp[1]代表0-1这个区间。

动态规划问题：先遍历每个喷头的喷射范围，然后计算范围内需要的喷头最小数。

class Solution {
public:
    int minTaps(int n, vector<int>& ranges) {
        vector<int> dp(n + 1, 1000);
        dp[0] = 0;
        for (int i = 0; i <= n; i++) {
            int left = max(0, i - ranges[i]);
            int right = min(n, i + ranges[i]);
            for (int j = left; j <= right; j++) {
                dp[j] = min(dp[j], dp[left] + 1);
            }
        }
        if (dp[n] >= 1000) {
            return -1;
        }
        return dp[n];
    }
};