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Furthest Building You Can Reach
Bookmark
Brute Force
Min-Heap
Input
Example 1
Example 2
Only Bricks
Custom
heights
=
[4, 2, 7, 6, 9, 14, 12]
,
bricks
=
5
,
ladders
=
1
bricks
0
/5
ladders left =
1
4
0
2
1
7
2
6
3
9
4
14
5
12
6
i
dfs(
0
) with
5
bricks and
1
ladders
bricks
0
/5
ladders left =
1
4
0
2
1
7
2
6
3
9
4
14
5
12
6
i
dfs(
0
) with
5
bricks and
1
ladders
bricks
0
/5
ladders left =
1
4
0
2
1
7
2
6
3
9
4
14
5
12
6
i
bricks
5
/5
ladders left =
1
4
0
2
1
7
2
6
3
9
4
14
5
12
6
+5
i
climb ≤ 0 →
walk across for free
bricks
5
/5
ladders left =
1
4
0
2
1
7
2
6
3
9
4
14
5
12
6
+5
i
try bricks:
0
available for a climb of
3
bricks
5
/5
ladders left =
0
4
0
2
1
7
2
6
3
9
4
14
5
12
6
+5
i
bricks
0
/5
ladders left =
1
4
0
2
1
7
2
6
3
9
4
14
5
12
6
i
try a ladder:
1
left
bricks
0
/5
ladders left =
0
4
0
2
1
7
2
6
3
9
4
14
5
12
6
i
dfs(
3
) with
5
bricks and
0
ladders
bricks
3
/5
ladders left =
0
4
0
2
1
7
2
6
3
9
4
14
5
12
6
+3
i
pay
3
bricks, used
3
of 5
bricks
3
/5
ladders left =
0
4
0
2
1
7
2
6
3
9
4
14
5
12
6
+3
i
try a ladder:
0
left
bricks
0
/5
ladders left =
1
4
0
2
1
7
2
6
3
9
4
14
5
12
6
i
Furthest building = 4 (out of resources)
algo
master
.
io
Step:
dfs(0, bricks 5, ladders 1)
0 / 43
Input
Example 1
Example 2
Only Bricks
Custom
heights
=
[4, 2, 7, 6, 9, 14, 12]
,
bricks
=
5
,
ladders
=
1
0 / 43
algo
master
.
io
Step:
dfs(0, bricks 5, ladders 1)
