3202. Find the Maximum Length of Valid Subsequence II
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Difficulty: medium Acceptance: 39% Topics: Array, Dynamic Programming
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Reference solution (spoiler · python)
# Time: O(n * k)
# Space: O(k)
# dp
class Solution(object):
def maximumLength(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
result = 0
for i in xrange(k):
dp = [0]*k
for x in nums:
dp[x%k] = dp[(i-x)%k]+1
result = max(result, max(dp))
return result
Solution from kamyu104/LeetCode-Solutions · MIT