DataStructures-Algorithms/Leetcode Daily Challenge/April-2021/26. Furthest Building You Can Reach.cpp at master · sachuverma/DataStructures-Algorithms · GitHub

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/*
Furthest Building You Can Reach
===============================

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

Hint #1
Assume the problem is to check whether you can reach the last building or not.

Hint #2
You'll have to do a set of jumps, and choose for each one whether to do it using a rope or bricks. It's always optimal to use ropes in the largest jumps.

Hint #3
Iterate on the buildings, maintaining the largest r jumps and the sum of the remaining ones so far, and stop whenever this sum exceeds b.
*/

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