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),
(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.
\n\n\n
Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2020/10/27/q4.gif\")
\n\nInput: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1\nOutput: 4\nExplanation: Starting at building 0, you can follow these steps:\n- Go to building 1 without using ladders nor bricks since 4 >= 2.\n- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7.\n- Go to building 3 without using ladders nor bricks since 7 >= 6.\n- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9.\nIt is impossible to go beyond building 4 because you do not have any more bricks or ladders.\n\n\n
Example 2:
\n\n\nInput: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2\nOutput: 7\n\n\n
Example 3:
\n\n\nInput: heights = [14,3,19,3], bricks = 17, ladders = 0\nOutput: 3\n\n\n
\n
Constraints:
\n\n1 <= heights.length <= 1051 <= heights[i] <= 1060 <= bricks <= 1090 <= ladders <= heights.lengthCompiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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