# 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 $\text{nums}$, and $k$ is the given positive integer.

• class Solution {
public int maximumLength(int[] nums, int k) {
int[][] f = new int[k][k];
int ans = 0;
for (int x : nums) {
x %= k;
for (int j = 0; j < k; ++j) {
int y = (j - x + k) % k;
f[x][y] = f[y][x] + 1;
ans = Math.max(ans, f[x][y]);
}
}
return ans;
}
}

• class Solution {
public:
int maximumLength(vector<int>& nums, int k) {
int f[k][k];
memset(f, 0, sizeof(f));
int ans = 0;
for (int x : nums) {
x %= k;
for (int j = 0; j < k; ++j) {
int y = (j - x + k) % k;
f[x][y] = f[y][x] + 1;
ans = max(ans, f[x][y]);
}
}
return ans;
}
};

• class Solution:
def maximumLength(self, nums: List[int], k: int) -> int:
f = [[0] * k for _ in range(k)]
ans = 0
for x in nums:
x %= k
for j in range(k):
y = (j - x + k) % k
f[x][y] = f[y][x] + 1
ans = max(ans, f[x][y])
return ans


• func maximumLength(nums []int, k int) (ans int) {
f := make([][]int, k)
for i := range f {
f[i] = make([]int, k)
}
for _, x := range nums {
x %= k
for j := 0; j < k; j++ {
y := (j - x + k) % k
f[x][y] = f[y][x] + 1
ans = max(ans, f[x][y])
}
}
return
}

• function maximumLength(nums: number[], k: number): number {
const f: number[][] = Array.from({ length: k }, () => Array(k).fill(0));
let ans: number = 0;
for (let x of nums) {
x %= k;
for (let j = 0; j < k; ++j) {
const y = (j - x + k) % k;
f[x][y] = f[y][x] + 1;
ans = Math.max(ans, f[x][y]);
}
}
return ans;
}