LeetCode #2958 — MEDIUM

Length of Longest Subarray With at Most K Frequency

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums and an integer k.

The frequency of an element x is the number of times it occurs in an array.

An array is called good if the frequency of each element in this array is less than or equal to k.

Return the length of the longest good subarray of nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,2,3,1,2,3,1,2], k = 2
Output: 6
Explanation: The longest possible good subarray is [1,2,3,1,2,3] since the values 1, 2, and 3 occur at most twice in this subarray. Note that the subarrays [2,3,1,2,3,1] and [3,1,2,3,1,2] are also good.
It can be shown that there are no good subarrays with length more than 6.

Example 2:

Input: nums = [1,2,1,2,1,2,1,2], k = 1
Output: 2
Explanation: The longest possible good subarray is [1,2] since the values 1 and 2 occur at most once in this subarray. Note that the subarray [2,1] is also good.
It can be shown that there are no good subarrays with length more than 2.

Example 3:

Input: nums = [5,5,5,5,5,5,5], k = 4
Output: 4
Explanation: The longest possible good subarray is [5,5,5,5] since the value 5 occurs 4 times in this subarray.
It can be shown that there are no good subarrays with length more than 4.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= k <= nums.length

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer k. The frequency of an element x is the number of times it occurs in an array. An array is called good if the frequency of each element in this array is less than or equal to k. Return the length of the longest good subarray of nums. A subarray is a contiguous non-empty sequence of elements within an array.

Baseline thinking

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

Pattern signal: Array · Hash Map · Sliding Window

Example 1

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

Example 2

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

Example 3

[5,5,5,5,5,5,5]
4

Core Insight

What unlocks the optimal approach

For each index <code>i</code>, find the rightmost index <code>j >= i</code> such that the frequency of each element in the subarray <code>[i, j]</code> is at most <code>k</code>.
We can use 2 pointers / sliding window to achieve it.

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 #2958: Length of Longest Subarray With at Most K Frequency
class Solution {
    public int maxSubarrayLength(int[] nums, int k) {
        Map<Integer, Integer> cnt = new HashMap<>();
        int ans = 0;
        for (int i = 0, j = 0; i < nums.length; ++i) {
            cnt.merge(nums[i], 1, Integer::sum);
            while (cnt.get(nums[i]) > k) {
                cnt.merge(nums[j++], -1, Integer::sum);
            }
            ans = Math.max(ans, i - j + 1);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2958: Length of Longest Subarray With at Most K Frequency
func maxSubarrayLength(nums []int, k int) (ans int) {
	cnt := map[int]int{}
	for i, j, n := 0, 0, len(nums); i < n; i++ {
		cnt[nums[i]]++
		for ; cnt[nums[i]] > k; j++ {
			cnt[nums[j]]--
		}
		ans = max(ans, i-j+1)
	}
	return
}

# Accepted solution for LeetCode #2958: Length of Longest Subarray With at Most K Frequency
class Solution:
    def maxSubarrayLength(self, nums: List[int], k: int) -> int:
        cnt = defaultdict(int)
        ans = j = 0
        for i, x in enumerate(nums):
            cnt[x] += 1
            while cnt[x] > k:
                cnt[nums[j]] -= 1
                j += 1
            ans = max(ans, i - j + 1)
        return ans

// Accepted solution for LeetCode #2958: Length of Longest Subarray With at Most K Frequency
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2958: Length of Longest Subarray With at Most K Frequency
// class Solution {
//     public int maxSubarrayLength(int[] nums, int k) {
//         Map<Integer, Integer> cnt = new HashMap<>();
//         int ans = 0;
//         for (int i = 0, j = 0; i < nums.length; ++i) {
//             cnt.merge(nums[i], 1, Integer::sum);
//             while (cnt.get(nums[i]) > k) {
//                 cnt.merge(nums[j++], -1, Integer::sum);
//             }
//             ans = Math.max(ans, i - j + 1);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2958: Length of Longest Subarray With at Most K Frequency
function maxSubarrayLength(nums: number[], k: number): number {
    const cnt: Map<number, number> = new Map();
    let ans = 0;
    for (let i = 0, j = 0; i < nums.length; ++i) {
        cnt.set(nums[i], (cnt.get(nums[i]) ?? 0) + 1);
        for (; cnt.get(nums[i])! > k; ++j) {
            cnt.set(nums[j], cnt.get(nums[j])! - 1);
        }
        ans = Math.max(ans, i - j + 1);
    }
    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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

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.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.