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:

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

Output: 4

Explanation:

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

Example 2:

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] <= 10⁹
0 <= k <= min(nums.length, 25)

Solutions

Solution 1: Dynamic Programming

We define $f[i][h]$ as the length of the longest good subsequence ending with $nums[i]$ and having no more than $h$ indices satisfying the condition. Initially, $f[i][h] = 1$. The answer is $\max(f[i][k])$, where $0 \le i < n$.

We consider how to calculate $f[i][h]$. We can enumerate $0 \le j < i$, if $nums[i] = nums[j]$, then $f[i][h] = \max(f[i][h], f[j][h] + 1)$; otherwise, if $h > 0$, then $f[i][h] = \max(f[i][h], f[j][h - 1] + 1)$. That is:

$$ f[i][h]= \begin{cases} \max(f[i][h], f[j][h] + 1), & \textit{if } nums[i] = nums[j], \ \max(f[i][h], f[j][h - 1] + 1), & \textit{if } h > 0. \end{cases} $$

The final answer is $\max(f[i][k])$, where $0 \le i < n$.

The time complexity is $O(n^2 \times k)$, and the space complexity is $O(n \times k)$. Where $n$ is the length of the array nums.

Python3JavaC++GoTypeScript

class Solution:
    def maximumLength(self, nums: List[int], k: int) -> int:
        n = len(nums)
        f = [[1] * (k + 1) for _ in range(n)]
        ans = 0
        for i, x in enumerate(nums):
            for h in range(k + 1):
                for j, y in enumerate(nums[:i]):
                    if x == y:
                        f[i][h] = max(f[i][h], f[j][h] + 1)
                    elif h:
                        f[i][h] = max(f[i][h], f[j][h - 1] + 1)
            ans = max(ans, f[i][k])
        return ans

class Solution {
    public int maximumLength(int[] nums, int k) {
        int n = nums.length;
        int[][] f = new int[n][k + 1];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int h = 0; h <= k; ++h) {
                for (int j = 0; j < i; ++j) {
                    if (nums[i] == nums[j]) {
                        f[i][h] = Math.max(f[i][h], f[j][h]);
                    } else if (h > 0) {
                        f[i][h] = Math.max(f[i][h], f[j][h - 1]);
                    }
                }
                ++f[i][h];
            }
            ans = Math.max(ans, f[i][k]);
        }
        return ans;
    }
}

class Solution {
public:
    int maximumLength(vector<int>& nums, int k) {
        int n = nums.size();
        int f[n][k + 1];
        memset(f, 0, sizeof(f));
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int h = 0; h <= k; ++h) {
                for (int j = 0; j < i; ++j) {
                    if (nums[i] == nums[j]) {
                        f[i][h] = max(f[i][h], f[j][h]);
                    } else if (h) {
                        f[i][h] = max(f[i][h], f[j][h - 1]);
                    }
                }
                ++f[i][h];
            }
            ans = max(ans, f[i][k]);
        }
        return ans;
    }
};

func maximumLength(nums []int, k int) (ans int) {
    f := make([][]int, len(nums))
    for i := range f {
        f[i] = make([]int, k+1)
    }
    for i, x := range nums {
        for h := 0; h <= k; h++ {
            for j, y := range nums[:i] {
                if x == y {
                    f[i][h] = max(f[i][h], f[j][h])
                } else if h > 0 {
                    f[i][h] = max(f[i][h], f[j][h-1])
                }
            }
            f[i][h]++
        }
        ans = max(ans, f[i][k])
    }
    return
}

function maximumLength(nums: number[], k: number): number {
    const n = nums.length;
    const f: number[][] = Array.from({ length: n }, () => Array(k + 1).fill(0));
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        for (let h = 0; h <= k; ++h) {
            for (let j = 0; j < i; ++j) {
                if (nums[i] === nums[j]) {
                    f[i][h] = Math.max(f[i][h], f[j][h]);
                } else if (h) {
                    f[i][h] = Math.max(f[i][h], f[j][h - 1]);
                }
            }
            ++f[i][h];
        }
        ans = Math.max(ans, f[i][k]);
    }
    return ans;
}

Solution 2: Optimized Dynamic Programming

According to the state transition equation in Solution 1, if $nums[i] = nums[j]$, then we only need to get the maximum value of $f[j][h]$. We can maintain this with an array $mp$ of length $k + 1$. If $nums[i] \neq nums[j]$, we need to record the maximum value of $f[j][h - 1]$ corresponding to $nums[j]$, the maximum value and the second maximum value. We can maintain these with an array $g$ of length $k + 1$.

The time complexity is $O(n \times k)$, and the space complexity is $O(n \times k)$. Where $n$ is the length of the array nums.