Furthest Building You Can Reach
Time complexity: $O(n\log k)$
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
|
#include <iostream>
#include <vector>
#include <queue>
class Solution {
public:
int furthestBuilding(std::vector<int>& heights, int bricks, int ladders) {
int n = heights.size();
std::priority_queue<int, std::vector<int>, std::greater<int>> pq;
for (int i = 0; i < n - 1; ++i) {
int diff = heights[i + 1] - heights[i];
if (diff > 0) {
pq.push(diff);
if (pq.size() > ladders) {
bricks -= pq.top();
pq.pop();
}
if (bricks < 0) {
return i;
}
}
}
return n - 1;
}
};
|
