[Leetcode] Problem 1642 - Furthest Building You Can Reach

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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

No.1

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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.

No.2

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

Output: 7

No.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

Code

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public int furthestBuilding(int[] heights, int bricks, int ladders) {
    PriorityQueue<Integer> minHeap = new PriorityQueue<>();

    for (int i = 1; i < heights.length; i++) {
        int diff = heights[i] - heights[i - 1];

        if (diff <= 0)
            continue;

        minHeap.offer(diff);

        if (minHeap.size() <= ladders)
            continue;

        bricks -= minHeap.poll();

        if (bricks < 0)
            return i - 1;
    }

    return heights.length - 1;
}