<p>You are given an integer array <code>heights</code> representing the heights of buildings, some <code>bricks</code>, and some <code>ladders</code>.</p>

<p>You start your journey from building <code>0</code> and move to the next building by possibly using bricks or ladders.</p>

<p>While moving from building <code>i</code> to building <code>i+1</code> (<strong>0-indexed</strong>),</p>

<ul>
	<li>If the current building&#39;s height is <strong>greater than or equal</strong> to the next building&#39;s height, you do <strong>not</strong> need a ladder or bricks.</li>
	<li>If the current building&#39;s height is <b>less than</b> the next building&#39;s height, you can either use <strong>one ladder</strong> or <code>(h[i+1] - h[i])</code> <strong>bricks</strong>.</li>
</ul>

<p><em>Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/27/q4.gif" style="width: 562px; height: 561px;" />
<pre>
<strong>Input:</strong> heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1
<strong>Output:</strong> 4
<strong>Explanation:</strong> Starting at building 0, you can follow these steps:
- Go to building 1 without using ladders nor bricks since 4 &gt;= 2.
- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 &lt; 7.
- Go to building 3 without using ladders nor bricks since 7 &gt;= 6.
- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 &lt; 9.
It is impossible to go beyond building 4 because you do not have any more bricks or ladders.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
<strong>Output:</strong> 7
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> heights = [14,3,19,3], bricks = 17, ladders = 0
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= heights.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= heights[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>0 &lt;= bricks &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= ladders &lt;= heights.length</code></li>
</ul>

import heapq
class Solution:
    def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:

        # https://leetcode.com/problems/furthest-building-you-can-reach/solutions/4739684/dp-binary-search-greedy-all-approaches-explained/

#         //https://leetcode.com/problems/furthest-building-you-can-reach/solutions/4739509/why-dp-fails-here-greedy-approach-min-heap/

# // we will greedily first use ladders and store the diffs in the pq..
# // so in the future we will know if we can replace earlier used ladder to bricks.

    ## first fill the ladder heap.  

    ## there is an approach of binary search also, with time comlexity 
    ## of  n* logn * logn   
    ## explained in the above link.. search space.. 0 to n - 1.
    ## for using isValid function... sort that subarray... using bricks in starting and ladder at the end.
        
        h = []

        for i in range(len(heights)-1):

            if heights[i+1] <= heights[i]:
                continue
            else:
                diff = heights[i+1] - heights[i]
                if len(h) < ladders:
                    heapq.heappush(h, diff)
                else:
                    if len(h) == 0 or h[0] > diff:
                        bricks -= diff
                    else:
                        t = heapq.heappop(h)
                        bricks -= t
                        heapq.heappush(h, diff)
                    if bricks < 0:
                        return i
        
        return len(heights) - 1


    # the below code is slightly wrong and can give wrong answer in some cases.

    ## because after filling the ladder heap.. we are only replacing when diff > bricks...

    ## which is wrong.. ideally the ladder heap should have all diffs in decreasing order (larger to smaller).. till the size of ladder.

    ## but here we will miss out on some cases... when the heap is filled in not in that order.

        # h = []

        # for i in range(1, len(heights)):

        #     if heights[i] <= heights[i-1]:
        #         continue
        #     else:
        #         diff = heights[i] - heights[i-1]
        #         if ladders > 0:
        #             ladders -= 1
        #             heapq.heappush(h, diff)
                
        #         elif diff <= bricks:
        #             bricks -= diff

        #         elif len(h) > 0:
        #             t = heapq.heappop(h)
        #             if t <= bricks:
        #                 bricks -= t
        #             else:
        #                 return i-1
        #         else:
        #             return i-1
                
        # return len(heights) - 1

                

                
       



















        # for i in range(len(heights)-1):
        #     diff = heights[i+1] - heights[i]
        #     if diff <= 0:
        #         continue
        #     if (len(h) < ladders):
        #         heapq.heappush(h, diff)
        #     else:
            
        #         if (len(h) == 0 or h[0] >= diff):
                
        #             bricks -= diff
        #         else:
        #             t = heapq.heappop(h)
        #             bricks -= t
        #             heapq.heappush(h, diff)
        #     if bricks < 0:
        #         return i
        #     # print('h, bricks ', h, bricks, diff, i)

        # return len(heights) -1;
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