3176. Find the Maximum Length of a Good Subsequence I

Posted on Edited on In

3176. Find the Maximum Length of a Good Subsequence I

Description

You are given an integer array nums and a non-negative integer k. A sequence of integers seq is called good if there are at most k indices i in the range [0, seq.length - 2] such that seq[i] != seq[i + 1].

Return the maximum possible length of a good subsequence of nums.

Example 1:

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

Output: 4

Explanation:

The maximum length subsequence is [1,2,1,1,3].

Example 2:

1
2
3
Input: nums = [1,2,3,4,5,1], k = 0

Output: 2

Explanation:

The maximum length subsequence is [1,2,3,4,5,1].

Constraints:

  • 1 <= nums.length <= 500
  • 1 <= nums[i] <= 10^9
  • 0 <= k <= min(nums.length, 25)

Hints/Notes

  • Biweenly Contest 132
  • dp

Solution

Language: C++

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
class Solution {
public:
    vector<vector<int>> dp;

    int maximumLength(vector<int>& nums, int k) {
        int size = nums.size();
        dp.resize(size, vector<int>(k + 1, INT_MIN));
        int res = 0;
        for (int i = 0; i < nums.size(); i++) {
            res = max(res, traverse(i, k, nums));
        }
        return res;
    }

    int traverse(int curIndex, int remaining, vector<int>& nums) {
        if (remaining < 0) {
            return 0;
        }
        if (curIndex == nums.size()) {
            return 0;
        }
        if (curIndex == nums.size() - 1) {
            return 1;
        }
        if (dp[curIndex][remaining] != INT_MIN) {
            return dp[curIndex][remaining];
        }
        int res = 0;
        for (int i = curIndex + 1; i < nums.size(); i++) {
            if (nums[curIndex] == nums[i]) {
                res = max(res, traverse(i, remaining, nums) + 1);
            } else {
                res = max(res, traverse(i, remaining - 1, nums) + 1);
            }
        }
        dp[curIndex][remaining] = res;
        return res;
    }
};