1642. Furthest Building You Can Reach

Problem

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 <= 10^5

• 1 <= heights[i] <= 10^6

• 0 <= bricks <= 10^9

• 0 <= ladders <= heights.length

Solution (Java)

class Solution {
    public int furthestBuilding(int[] heights, int bricks, int ladders) {
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        int i = 0;
        // we'll assume to use ladders for the first l jumps and adjust it afterwards
        for (; i < heights.length - 1 && minHeap.size() < ladders; i++) {
            int diff = heights[i + 1] - heights[i];
            if (diff > 0) {
                minHeap.offer(diff);
            }
        }
        while (i < heights.length - 1) {
            int diff = heights[i + 1] - heights[i];
            if (diff > 0) {
                if (!minHeap.isEmpty() && minHeap.peek() < diff) {
                    bricks -= minHeap.poll();
                    minHeap.offer(diff);
                } else {
                    bricks -= diff;
                }
                if (bricks < 0) {
                    return i;
                }
            }
            i++;
        }
        return i;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).