1642. Furthest Building You Can Reach

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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.

Example 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.

Example 2:

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

Example 3:

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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

Hints/Notes

Solution

Language: C++

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