LeetCode #3176 — MEDIUM

Find the Maximum Length of a Good Subsequence I

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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)

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Dynamic Programming

Example 1

[1,2,1,1,3]
2

Example 2

[1,2,3,4,5,1]
0

Core Insight

What unlocks the optimal approach

The absolute values in <code>nums</code> don’t really matter. So we can remap the set of values to the range <code>[0, n - 1]</code>.
Let <code>dp[i][j]</code> be the length of the longest subsequence till index <code>j</code> with at most <code>i</code> positions such that <code>seq[i] != seq[i + 1]</code>.
For each value <code>x</code> from left to right, update <code>dp[i][x] = max(dp[i][x] + 1, dp[i - 1][y] + 1)</code>, where <code>y != x</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3176: Find the Maximum Length of a Good Subsequence I
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;
    }
}

// Accepted solution for LeetCode #3176: Find the Maximum Length of a Good Subsequence I
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
}

# Accepted solution for LeetCode #3176: Find the Maximum Length of a Good Subsequence I
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

// Accepted solution for LeetCode #3176: Find the Maximum Length of a Good Subsequence I
/**
 * [3176] Find the Maximum Length of a Good Subsequence I
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn maximum_length(nums: Vec<i32>, k: i32) -> i32 {
        use std::collections::HashMap;

        let k = k as usize;
        let mut map = HashMap::with_capacity(nums.len());
        let mut mx = vec![0; k + 2];

        for x in nums {
            let entry = map.entry(x).or_insert(vec![0; k + 1]);

            for j in (0..=k).rev() {
                entry[j] = entry[j].max(mx[j]) + 1;
                mx[j + 1] = mx[j + 1].max(entry[j]);
            }
        }

        mx[k + 1]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3176() {
        assert_eq!(4, Solution::maximum_length(vec![1, 2, 1, 1, 3], 2));
        assert_eq!(2, Solution::maximum_length(vec![1, 2, 3, 4, 5, 1], 0));
    }
}

// Accepted solution for LeetCode #3176: Find the Maximum Length of a Good Subsequence I
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2 × k)

Space

O(n × k)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.