1642. 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.

Constraints

• 1 <= heights.length <= 105

• 1 <= heights[i] <= 106

• 0 <= bricks <= 109

• 0 <= ladders <= heights.length

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int furthestBuilding(int[] heights, int bricks, int ladders) {
        PriorityQueue<Integer> pq = new PriorityQueue();
        int n = heights.length;
        for(int i = 1; i < n; i++) {
            int diff = heights[i] - heights[i-1];
            if(diff > 0) {
                pq.offer(diff);
            }
            if(pq.size() > ladders) {
                bricks -= pq.poll();
            }
            if(bricks < 0) {
                return i-1;
            }
        }
        
        return n-1;
    }
}

Follow up