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Diffstat (limited to 'minimum_sideways_jumps.py')
-rw-r--r-- | minimum_sideways_jumps.py | 37 |
1 files changed, 37 insertions, 0 deletions
diff --git a/minimum_sideways_jumps.py b/minimum_sideways_jumps.py new file mode 100644 index 0000000..01e1cd6 --- /dev/null +++ b/minimum_sideways_jumps.py @@ -0,0 +1,37 @@ +# 1824. Minimum Sideway Jumps + +""" +there is a 3 lane road of length 'n' that consists of 'n + 1' points labeled +from 0 to 'n'. a frog starts at point 0 in the second lane and wants to jump +to point 'n'. however there could be obstacles along the way. you are given +an array 'obstacles' of length 'n + 1' where each 'obstacles[i]' (ranging +from 0 to 3) describes an obstacle on the lane 'obstacles[i]' at point 'i'. +if 'obstacles[i] == 0' there are no obstacles at point 'i'. there will be at +most one obstacles in the 3 lanes at each point. the frog can only travel +from point 'i' to point 'i + 1' on the same lane if there is not an obstacle +on the lane at point 'i + 1'. return the minimum number of side jumps the +frog needs to reach any lane at point 'n' starting from lane 2 at point 0. +""" + + +class Solution(object): + def minSideJumps(self, obstacles): + """ + :type obstacles: List[int] + :rtype: int + """ + dp = [1, 0, 1] + for i in obstacles: + if i: + dp[i - 1] = float("inf") + for j in range(3): + if i != j + 1: + dp[j] = min(dp[j], dp[(j + 1) % 3] + 1, dp[(j + 2) % 3] + 1) + return min(dp) + + +if __name__ == "__main__": + obj = Solution() + print(obj.minSideJumps(obstacles=[0, 1, 2, 3, 0])) + print(obj.minSideJumps(obstacles=[0, 1, 1, 3, 3, 0])) + print(obj.minSideJumps(obstacles=[0, 2, 1, 0, 3, 0])) |