What is the time complexity of Length of Longest Subarray With at Most K Frequency?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Length of Longest Subarray With at Most K Frequency?

This problem uses Array, Hash Table, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Length of Longest Subarray With at Most K Frequency?

Insight 1: The core idea is to use a sliding window to maintain a 'good' subarray. We expand the window to the right, adding elements, and shrink it from the left whenever an element's frequency exceeds `k`. Insight 2: A hash map (or dictionary) is essential to efficiently track the frequency of each element within the current window.

Which companies ask LeetCode 2958. Length of Longest Subarray With at Most K Frequency in interviews?

This problem is frequently asked at Citadel, MakeMyTrip.

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Length of Longest Subarray With at Most K Frequency - LeetCode 2958 Solution

Length of Longest Subarray With at Most K Frequency - Complete Solution Guide

Length of Longest Subarray With at Most K Frequency is LeetCode problem 2958, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 va

Detailed Explanation

The problem asks us to find the length of the longest subarray within a given array `nums` where the frequency of each element in the subarray is less than or equal to a given integer `k`. A 'good' subarray is defined as one where no element appears more than `k` times. We need to return the maximum length of such a 'good' subarray.

Solution Approach

The provided solution uses the sliding window technique. A window is maintained by two pointers, `left` and `right`. The `right` pointer expands the window one element at a time. The frequency of each element in the window is tracked using a hash map. If adding an element to the window causes the frequency of any element to exceed `k`, the `left` pointer is advanced until the frequency constraint is satisfied again. The maximum window length encountered during this process is the answer.

Step-by-Step Algorithm

Step 1: Initialize a hash map `freq` to store the frequencies of elements within the current window. Initialize `ans` to 0 (to store the maximum length), and `left` to 0 (the starting index of the window).
Step 2: Iterate through the input array `nums` with the `right` pointer.
Step 3: For each element at index `right`, increment its frequency in the `freq` hash map.
Step 4: While the frequency of the current element `nums[right]` is greater than `k`, decrement the frequency of the element at the `left` pointer in the `freq` hash map and increment the `left` pointer.
Step 5: Update `ans` with the maximum of its current value and the current window length (`right - left + 1`).
Step 6: After iterating through all elements, return `ans`.

Key Insights

Insight 1: The core idea is to use a sliding window to maintain a 'good' subarray. We expand the window to the right, adding elements, and shrink it from the left whenever an element's frequency exceeds `k`.
Insight 2: A hash map (or dictionary) is essential to efficiently track the frequency of each element within the current window.
Insight 3: The sliding window technique allows us to explore all possible subarrays in O(n) time, where n is the length of the input array.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Sliding Window.

Companies

Asked at: Citadel, MakeMyTrip.

Length of Longest Subarray With at Most K Frequency | LeetCode 2958

Medium

Array Hash Table Sliding Window

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int maxSubarrayLength(int[] nums, int k) {
        Map<Integer, Integer> freq = new HashMap<>();
        int ans = 0;
        int left = 0;
        for (int right = 0; right < nums.length; right++) {
            freq.put(nums[right], freq.getOrDefault(nums[right], 0) + 1);
            while (freq.get(nums[right]) > k) {
                freq.put(nums[left], freq.get(nums[left]) - 1);
                left++;
            }
            ans = Math.max(ans, right - left + 1);
        }
        return ans;
    }
}

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Problem Info

Description

Problem 2958: Length of Longest Subarray With at Most K Frequency - Detailed explanation and solution approaches.

Topics

Array Hash Table Sliding Window

Companies

CitadelMakeMyTrip

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2958

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize a hash map `freq` to store the frequencies of elements within the current window. Initialize `ans` to 0 (to store the maximum length), and `left` to 0 (the starting index of the window).
2
Step 2: Iterate through the input array `nums` with the `right` pointer.
3
Step 3: For each element at index `right`, increment its frequency in the `freq` hash map.
4
Step 4: While the frequency of the current element `nums[right]` is greater than `k`, decrement the frequency of the element at the `left` pointer in the `freq` hash map and increment the `left` pointer.
5
Step 5: Update `ans` with the maximum of its current value and the current window length (`right - left + 1`).
6
Step 6: After iterating through all elements, return `ans`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because the sliding window traverses the array at most twice (once with the right pointer and at most once with the left pointer). The space complexity is O(n) because, in the worst-case scenario, we might need to store the frequency of every element of the input array in the hash map.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible subarrays and check if each subarray is 'good'.

Time: O(n^3) or O(n^2) depending on how frequency calculation is implemented in inner loops. O(n) to compute each sub-array * O(n^2) to generate all sub-arrays.Space: O(n) in worst case, for frequency counting in sub-array

Trade-offs: The brute force approach is simple to understand but extremely inefficient. It should be avoided due to its high time complexity. It could be useful if the array sizes are extremely small.

✨

Key Insights

1Insight 1: The core idea is to use a sliding window to maintain a 'good' subarray. We expand the window to the right, adding elements, and shrink it from the left whenever an element's frequency exceeds `k`.
2Insight 2: A hash map (or dictionary) is essential to efficiently track the frequency of each element within the current window.
3Insight 3: The sliding window technique allows us to explore all possible subarrays in O(n) time, where n is the length of the input array.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: Empty input array. The provided code handles this implicitly because the loop won't execute, and the initial value of `ans` (0) will be returned, which is correct.
Edge case 2: `k` is greater than or equal to the length of the input array. In this case, the entire array is a valid subarray, and the code correctly returns the length of the input array.
Edge case 3: All elements in `nums` are the same. The sliding window will grow until the frequency exceeds `k`, then shrink until the frequency is `k`, and continue doing so. The correct result will be obtained.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to shrink the window when the frequency of an element exceeds `k`. This will lead to incorrect results.

✗Common mistake 2: Not initializing the frequency hash map correctly, or not handling the case where an element's frequency becomes zero after shrinking the window.

✗Common mistake 3: Not understanding how the sliding window works and thus incorrectly managing the `left` and `right` pointers.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing network traffic to detect denial-of-service (DoS) attacks. We can use the sliding window to monitor the frequency of requests from a particular IP address and flag IP addresses exceeding a certain threshold (`k`).

🏦Real-world scenario 2: Monitoring website activity to identify potential bot activity. The sliding window can be used to track the frequency of actions (e.g., page views, clicks) from a specific user within a given time frame. Users with abnormally high activity frequencies could be flagged as bots.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 18, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Length of Longest Subarray With at Most K Frequency"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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