Furthest Building You Can Reach

Description

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.

Example 1:

Input: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1 Output: 4 Explanation: Starting at building 0, you can follow these steps:

• Go to building 1 without using ladders nor bricks since 4 >= 2.
• Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7.
• Go to building 3 without using ladders nor bricks since 7 >= 6.
• Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9. It is impossible to go beyond building 4 because you do not have any more bricks or ladders.

Example 2:

Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2 Output: 7

Example 3:

Input: heights = [14,3,19,3], bricks = 17, ladders = 0 Output: 3

Constraints:

• 1 <= heights.length <= 105
• 1 <= heights[i] <= 106
• 0 <= bricks <= 109
• 0 <= ladders <= heights.length

Code

Time Complexity: $O(n\log l)$, Space Complexity: $O(l)$

int ladders = l

class Solution {
public:
    int furthestBuilding(vector<int>& heights, int bricks, int ladders) {
        int i;
        int n = heights.size();
        priority_queue<int, vector<int>, greater<int>> min_heap;
        for(i = 0; i < n - 1; i++) {
            if(heights[i + 1] > heights[i]) {
                int climb = heights[i + 1] - heights[i];
                min_heap.push(climb);
                if(min_heap.size() > ladders) {
                    auto n = min_heap.top();
                    min_heap.pop();
                    bricks -= n;
                }
 
                if(bricks < 0) 
                    break;
            }
        }
 
        return i;
    }
};

Source

• Furthest Building You Can Reach - LeetCode