1642. Furthest Building You Can Reach

You are given an integer array heights representing the heights of buildings, some bricks, and some ladders.

You start your journey from building 0 and move to the next building by possibly using bricks or ladders.

While moving from building i to building i+1 (0-indexed),

• If the current building’s height is greater than or equal to the next building’s height, you do not need a ladder or bricks.
• If the current building’s height is less than the next building’s height, you can either use one ladder or (h[i+1] - h[i]) bricks.

Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.

class Solution {
public:
    int furthestBuilding(vector<int>& h, int b, int l) {
        priority_queue<int, vector<int>, greater<int>> pq;
        int res = 0;
        for(int i = 0; i < h.size() - 1; i++) {
            if(h[i] >= h[i+1]) continue;
            pq.push(h[i+1] - h[i]);
            if(pq.size() > l) {
                if(pq.top() > b)
                    return i;
                b -= pq.top();
                pq.pop();
            }
        }
        return h.size() - 1;
    }
};