3176. Find the Maximum Length of a Good Subsequence I

# Time:  O(n * k)
# Space: O(n * k)

import collections


# dp
class Solution(object):
    def maximumLength(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        lookup = {x:i for i, x in enumerate(set(nums))}
        dp = [[0]*len(lookup) for _ in xrange(k+1)]
        result = [0]*(k+1)
        for x in nums:
            x = lookup[x]
            for i in reversed(xrange(k+1)):
                dp[i][x] = max(dp[i][x], result[i-1] if i-1 >= 0 else 0)+1
                result[i] = max(result[i], dp[i][x])
        return result[k]


# Time:  O(n * k)
# Space: O(n * k)
import collections


# dp
class Solution2(object):
    def maximumLength(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        dp = [collections.defaultdict(int) for _ in xrange(k+1)]
        result = [0]*(k+1)
        for x in nums:
            for i in reversed(xrange(k+1)):
                dp[i][x] = max(dp[i][x], result[i-1] if i-1 >= 0 else 0)+1
                result[i] = max(result[i], dp[i][x])
        return result[k]


# Time:  O(n^2 * k)
# Space: O(n * k)
# dp
class Solution(object):
    def maximumLength(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        dp = [[0]*(k+1) for _ in xrange(len(nums))]
        result = 0
        for i in xrange(len(nums)):
            dp[i][0] = 1
            for l in xrange(k+1):
                for j in xrange(i):
                    dp[i][l] = max(dp[i][l], dp[j][l]+1 if nums[j] == nums[i] else 1, dp[j][l-1]+1 if l-1 >= 0 else 1)
                result = max(result, dp[i][l])
        return result

Solution from kamyu104/LeetCode-Solutions · MIT