3490. Find The Maximum Length Of Valid Subsequence I¶

Difficulty: Medium

3490. Find the Maximum Length of Valid Subsequence I

Medium

You are given an integer array nums.

A subsequence sub of nums with length x is called valid if it satisfies:

(sub[0] + sub[1]) % 2 == (sub[1] + sub[2]) % 2 == ... == (sub[x - 2] + sub[x - 1]) % 2.

Return the length of the longest valid subsequence of nums.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [1,2,3,4]

Output: 4

Explanation:

The longest valid subsequence is [1, 2, 3, 4].

Example 2:

Input: nums = [1,2,1,1,2,1,2]

Output: 6

Explanation:

The longest valid subsequence is [1, 2, 1, 2, 1, 2].

Example 3:

Input: nums = [1,3]

Output: 2

Explanation:

The longest valid subsequence is [1, 3].

Constraints:

2 <= nums.length <= 2 * 10⁵
1 <= nums[i] <= 10⁷

Solution¶

class Solution {
    public int maximumLength(int[] nums) {
        int n = nums.length;
        ArrayList<Integer> first = new ArrayList<>();
        ArrayList<Integer> second = new ArrayList<>();
        first.add(nums[0]); second.add(nums[0]);
        for(int i = 1; i < n; i++) {
            int current = nums[i];
            if((current + first.get(first.size() - 1)) % 2 == 1)   
                first.add(current);
            if((current + second.get(second.size() - 1)) % 2 == 0) 
                second.add(current);
        }
        int res1 = Math.max(first.size(), second.size());
        if(n <= 1)
            return res1;
        first.clear(); second.clear();
        first.add(nums[1]); second.add(nums[1]);
        for(int i = 2; i < n; i++) {
            int current = nums[i];
            if((current + first.get(first.size() - 1)) % 2 == 1) 
                first.add(current);
            if((current + second.get(second.size() - 1)) % 2 == 0) 
                second.add(current);
        }
        int res2 = Math.max(first.size() , second.size());
        int ans = Math.max(res1, res2);
        int even = 0, odd = 0;
        for(int i = 0; i < n; i++) {
            if(nums[i] % 2 == 0) 
                even++;
            else if(nums[i] % 2 == 1) 
                odd++;
        }
        ans = Math.max(ans, even);
        ans = Math.max(ans, odd);
        return ans;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here