3202. Find the Maximum Length of Valid Subsequence II
Description
You are given an integer array nums
and a positive integer k
.
A subsequence sub
of nums
with length x
is called valid if it satisfies:
(sub[0] + sub[1]) % k == (sub[1] + sub[2]) % k == ... == (sub[x - 2] + sub[x - 1]) % k.
Return the length of the longest valid subsequence of nums
.
Example 1:
Input: nums = [1,2,3,4,5], k = 2
Output: 5
Explanation:
The longest valid subsequence is [1, 2, 3, 4, 5]
.
Example 2:
Input: nums = [1,4,2,3,1,4], k = 3
Output: 4
Explanation:
The longest valid subsequence is [1, 4, 1, 4]
.
Constraints:
2 <= nums.length <= 103
1 <= nums[i] <= 107
1 <= k <= 103
Solutions
Solution 1: Dynamic Programming
Based on the problem description, we know that for a subsequence $a_1, a_2, a_3, \cdots, a_x$, if it satisfies $(a_1 + a_2) \bmod k = (a_2 + a_3) \bmod k$, then $a_1 \bmod k = a_3 \bmod k$. This means that the result of taking modulo $k$ for all odd-indexed elements is the same, and the result for all even-indexed elements is the same as well.
We can solve this problem using dynamic programming. Define the state $f[x][y]$ as the length of the longest valid subsequence where the last element modulo $k$ equals $x$, and the second to last element modulo $k$ equals $y$. Initially, $f[x][y] = 0$.
Iterate through the array $nums$, and for each number $x$, we get $x = x \bmod k$. Then, we can enumerate the sequences where two consecutive numbers modulo $j$ yield the same result, where $j \in [0, k)$. Thus, the previous number modulo $k$ would be $y = (j - x + k) \bmod k$. At this point, $f[x][y] = f[y][x] + 1$.
The answer is the maximum value among all $f[x][y]$.
The time complexity is $O(n \times k)$, and the space complexity is $O(k^2)$. Here, $n$ is the length of the array $\textit{nums}$, and $k$ is the given positive integer.
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