What is the time complexity of Find the Longest Equal Subarray?

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

What data structures are used to solve Find the Longest Equal Subarray?

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

What are the key insights for solving Find the Longest Equal Subarray?

Insight 1: We can group the indices of each number in the array. This allows us to efficiently find potential equal subarrays for each number. Insight 2: A sliding window approach is suitable for finding the longest equal subarray for each number, considering the constraint of deleting at most 'k' elements.

Which companies ask LeetCode 2831. Find the Longest Equal Subarray in interviews?

This problem is frequently asked at Palo Alto Networks.

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Find the Longest Equal Subarray - LeetCode 2831 Solution

Find the Longest Equal Subarray - Complete Solution Guide

Find the Longest Equal Subarray is LeetCode problem 2831, 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 a 0-indexed integer array nums and an integer k . A subarray is called equal if all of its elements are equal. Note that the empty subarray is an equal subarray. Return the length of the longest possible equal subarray after deleting at most k elements from nums . A subarray is a contiguous, possibly empty sequence of elements within an array. Example 1: Input: nums = [1,3,2,3,1,3], k = 3 Output: 3 Explanation: It's optimal to delete the elements at index 2 and index 4. After delet

Detailed Explanation

The problem asks us to find the length of the longest 'equal subarray' within a given array `nums`, after deleting at most `k` elements. An 'equal subarray' is a subarray where all elements are the same. We need to find the optimal way to delete elements to maximize the length of such a subarray.

Solution Approach

The solution uses a sliding window approach. First, it creates a map (dictionary or hash table) to store the indices of each number in the input array. Then, for each number, it iterates through its indices using a sliding window. The window expands to the right and shrinks from the left as needed, ensuring the number of deletions needed to make the window an 'equal subarray' is always less than or equal to `k`. During each step, the maximum length of a valid 'equal subarray' is updated.

Step-by-Step Algorithm

Step 1: Create a dictionary (or hash map) `positions` to store the indices of each number in the input array `nums`. The keys of the dictionary are the numbers in `nums`, and the values are lists of their indices in ascending order.
Step 2: Iterate through the values (lists of indices) in the `positions` dictionary.
Step 3: For each list of indices, initialize a left pointer `left` to 0.
Step 4: Iterate through the indices with a right pointer `right` from 0 to the end of the indices list.
Step 5: Calculate the 'cost' of making the subarray from `indices[left]` to `indices[right]` equal. The cost is calculated as `(indices[right] - indices[left] + 1) - (right - left + 1)`. This represents the number of elements we would need to delete to make it an 'equal subarray'.
Step 6: While the cost is greater than `k`, move the left pointer `left` one step to the right to reduce the cost.
Step 7: Update the `max_length` with the current window size (`right - left + 1`) if it's larger than the current `max_length`.
Step 8: After iterating through all indices, return the final `max_length`.

Key Insights

Insight 1: We can group the indices of each number in the array. This allows us to efficiently find potential equal subarrays for each number.
Insight 2: A sliding window approach is suitable for finding the longest equal subarray for each number, considering the constraint of deleting at most 'k' elements.
Insight 3: For a given window of indices for a specific number, the 'cost' of making it an equal subarray is the number of elements we need to delete. This cost is equal to (end index - start index + 1) - (number of elements in the window), which simplifies to (indices[right] - indices[left] + 1) - (right - left + 1).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Palo Alto Networks.

Find the Longest Equal Subarray | LeetCode 2831

Medium

Array Hash Table Binary Search Sliding Window

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public int longestEqualSubarray(List<Integer> nums, int k) {
        Map<Integer, List<Integer>> positions = new HashMap<>();
        for (int i = 0; i < nums.size(); i++) {
            int num = nums.get(i);
            if (!positions.containsKey(num)) {
                positions.put(num, new ArrayList<>());
            }
            positions.get(num).add(i);
        }

        int max_length = 0;

        for (List<Integer> indices : positions.values()) {
            int left = 0;
            for (int right = 0; right < indices.size(); right++) {
                int cost = (indices.get(right) - indices.get(left) + 1) - (right - left + 1);

                while (cost > k) {
                    left += 1;
                    cost = (indices.get(right) - indices.get(left) + 1) - (right - left + 1);
                }

                max_length = Math.max(max_length, right - left + 1);
            }
        }

        return max_length;
    }
}

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Problem Info

Description

Problem 2831: Find the Longest Equal Subarray - Detailed explanation and solution approaches.

Topics

Array Hash Table Binary Search Sliding Window

Companies

Palo Alto Networks

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2831

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

8 steps

1
Step 1: Create a dictionary (or hash map) `positions` to store the indices of each number in the input array `nums`. The keys of the dictionary are the numbers in `nums`, and the values are lists of their indices in ascending order.
2
Step 2: Iterate through the values (lists of indices) in the `positions` dictionary.
3
Step 3: For each list of indices, initialize a left pointer `left` to 0.
4
Step 4: Iterate through the indices with a right pointer `right` from 0 to the end of the indices list.
5
Step 5: Calculate the 'cost' of making the subarray from `indices[left]` to `indices[right]` equal. The cost is calculated as `(indices[right] - indices[left] + 1) - (right - left + 1)`. This represents the number of elements we would need to delete to make it an 'equal subarray'.
6
Step 6: While the cost is greater than `k`, move the left pointer `left` one step to the right to reduce the cost.
7
Step 7: Update the `max_length` with the current window size (`right - left + 1`) if it's larger than the current `max_length`.
8
Step 8: After iterating through all indices, return the final `max_length`.

💡

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 outer loop iterates through the unique elements of nums which is bounded by n. The inner loop iterates at most the number of indices which could be n. The while loop's amortized time complexity is O(n) since the left pointer can only move to the right at most n times.

📖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

Binary Search

For each possible value in nums, we can use binary search to find the start and end positions of subarrays that meet the given criteria of at most k deletions. This approach would require precomputing prefix sums or using a similar data structure to efficiently calculate the number of elements to delete within a given range.

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

Trade-offs: This approach could be useful if the number of unique elements in `nums` is large, and binary search provides more efficient access to indices. However, it adds complexity and may not be as intuitive as the sliding window method.

✨

Key Insights

1Insight 1: We can group the indices of each number in the array. This allows us to efficiently find potential equal subarrays for each number.
2Insight 2: A sliding window approach is suitable for finding the longest equal subarray for each number, considering the constraint of deleting at most 'k' elements.
3Insight 3: For a given window of indices for a specific number, the 'cost' of making it an equal subarray is the number of elements we need to delete. This cost is equal to (end index - start index + 1) - (number of elements in the window), which simplifies to (indices[right] - indices[left] + 1) - (right - left + 1).

📚

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!

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sliding Window

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

Edge Cases

Edge case 1: Empty array or `k` equals the array size. The loop won't execute, and the initial `max_length` (0) will be returned, which is the correct answer for an empty array. If k is greater than or equal to the array length and the array is not empty, the algorithm will correctly find the length of the largest equal subarray after deleting all the other elements.
Edge case 2: `nums` contains only one distinct element. The outer loop will run only once, and the sliding window will span the entire index list. Since no deletions are needed, the entire array's length will be returned.
Edge case 3: All elements in `nums` are distinct and k is 0. The sliding window will only have length 1, and the algorithm will find the longest equal subarray has length 1, as expected.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the cost calculation correctly. Ensure the formula (indices[right] - indices[left] + 1) - (right - left + 1) is properly implemented to determine the number of deletions required.

✗Common mistake 2: Failing to handle the edge cases where the array is empty or k is large enough to delete all elements. The initial value of maxLength ensures this will be handled correctly.

✗Common mistake 3: Incorrectly updating the left pointer in the sliding window, leading to incorrect cost calculations or missed optimal subarray lengths.

✅

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: Data compression. Finding the longest 'equal subarray' can be related to finding patterns or repeated sequences in data that can be compressed by representing the sequence with a single value and a repetition count.

🏦Real-world scenario 2: Signal processing. Identifying the longest period of a stable signal (i.e., a consistent value) in a noisy environment, where k represents the tolerance for noise or errors that can be ignored.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: October 10, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

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

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find the Longest Equal Subarray"?

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.

🤖 Need help with this problem?