1642. Furthest Building You Can Reach

Problem Statement

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

Intuition

Approach: 
Dp wont work here,
We try a Greedy approach

Till every index, find for ladder the maximum jumps and for minimum jumps 
We satisfy those with bricks

We use priority queue for this and when the ladder count exceed in pq
We satisfy with bricks

Links

https://leetcode.com/problems/furthest-building-you-can-reach/description/?envType=daily-question&envId=2024-02-17

Video Links

https://www.youtube.com/watch?v=35Z60cfjgKA&ab_channel=AryanMittal

Approach 1:

NlogN Greedy

C++

class Solution {
public:
    int furthestBuilding(vector<int>& arr, int bricks, int ladders) {
        int n = arr.size();
        priority_queue<int, vector<int>, greater<int>> pq;
        for(int i=1; i<arr.size(); i++){
            int diff = arr[i] - arr[i-1];
            if(diff < 0)
                continue;

            if(pq.size() < ladders)
                pq.push(diff);
            else{
                pq.push(diff);
                int top = pq.top();
                pq.pop();
                if(bricks >= top)
                    bricks -= top;
                else
                    return i-1;
            }
        } 

        return n-1;

    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++