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
Solution 1: Greedy
Reduce to 1024. Video Stitching
Time complexity: O(nlogn)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: int minTaps(int n, vector<int>& ranges) { vector<pair<int, int>> t; // O(n) reduction for (int i = 0; i <= n; ++i) t.emplace_back(max(0, i - ranges[i]), min(i + ranges[i], n)); // 1024. Video Stiching sort(begin(t), end(t)); int ans = 0; int i = 0; int l = 0; int e = 0; while (e < n) { // Extend to the right most w.r.t t[i].first <= l while (i <= n && t[i].first <= l) e = max(e, t[i++].second); if (l == e) return -1; // Can not extend l = e; ++ans; } return ans; } }; |
Solution 2: Greedy
Reduce to 45. Jump Game II
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: int minTaps(int n, vector<int>& ranges) { vector<int> nums(ranges.size()); for (int i = 0; i <= n; ++i) { int s = max(0, i - ranges[i]); nums[s] = max(nums[s], i + ranges[i]); } // 45. Jump Game II int steps = 0; int l = 0; int e = 0; for (int i = 0; i <= n; ++i) { if (i > e) return -1; if (i > l) { ++steps; l = e; } e = max(e, nums[i]); } return steps; } }; |
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