2023-08-31 Daily Challenge
Today I have done leetcode's August LeetCoding Challenge with cpp
.
August LeetCoding Challenge 31
Description
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.
Constraints:
1 <= n <= 104
ranges.length == n + 1
0 <= ranges[i] <= 100
Solution
class Solution {
public:
int minTaps(int n, vector<int>& ranges) {
vector<int> right(n + 1);
for(int i = 0; i <= n; ++i) {
if(!ranges[i]) continue;
int left = max(i - ranges[i], 0);
if(right[left] < i + ranges[i]) {
right[left] = i + ranges[i];
}
}
int count = 0;
int rightMost = 0;
int end = 0;
for(int i = 0; i <= n; ++i) {
if(end < i) {
if(rightMost <= end) return -1;
end = rightMost;
count += 1;
}
rightMost = max(rightMost, right[i]);
}
return count + (end < n);
}
};
// Accepted
// 38/38 cases passed (14 ms)
// Your runtime beats 85.18 % of cpp submissions
// Your memory usage beats 55.15 % of cpp submissions (14.6 MB)