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
(0indexed),
 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 (0indexed) 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:
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 <= 10^{5}
1 <= heights[i] <= 10^{6}
0 <= bricks <= 10^{9}
0 <= ladders <= heights.length

