What is the time complexity of Maximum Equal Frequency?

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

What data structures are used to solve Maximum Equal Frequency?

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

What are the key insights for solving Maximum Equal Frequency?

Insight 1: We need to keep track of the frequency of each number in the prefix and the frequency of those frequencies. Insight 2: We only need to check prefixes where, after removing one element, either all numbers appear once, one number appears once and the rest have the same frequency, one number appears one more time than the others, or all numbers except one have a frequency of 1.

Which companies ask LeetCode 1224. Maximum Equal Frequency in interviews?

This problem is frequently asked at American Express.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Maximum Equal Frequency - LeetCode 1224 Solution

Maximum Equal Frequency - Complete Solution Guide

Maximum Equal Frequency is LeetCode problem 1224, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an array nums of positive integers, return the longest possible length of an array prefix of nums , such that it is possible to remove exactly one element from this prefix so that every number that has appeared in it will have the same number of occurrences. If after removing one element there are no remaining elements, it's still considered that every appeared number has the same number of ocurrences (0). Example 1: Input: nums = [2,2,1,1,5,3,3,5] Output: 7 Explanation: For the subarray [

Detailed Explanation

The problem asks us to find the longest prefix of an array such that by removing exactly one element from that prefix, all the remaining elements have the same frequency (number of occurrences). If removing one element results in an empty array, that is also a valid case where all elements have the same frequency (zero). The input is an array of positive integers, and the output is the length of the longest valid prefix.

Solution Approach

The solution uses two hash maps: `counts` and `freqs`. `counts` stores the count of each number in the current prefix. `freqs` stores the number of frequencies that occur. For example, if the prefix is `[1, 1, 2, 2, 3]`, `counts` will be `{1: 2, 2: 2, 3: 1}` and `freqs` will be `{1: 1, 2: 2}` (one element appears once, and two elements appears twice). For each prefix, we update the `counts` and `freqs` maps. Then, we check if removing one element can result in all the remaining elements having the same frequency. We check for two main cases: 1. `freqs` contains only one entry (all elements have the same frequency). In this case, either every element appears once (f == 1) or only one element exists (n == 1). 2. `freqs` contains two entries. In this case, we check if one of the frequencies is one greater than the other and its count is 1, indicating that by removing one element, we can make all frequencies equal. We also check if a frequency of 1 exists with a count of 1, meaning removing the element that occurs once will equalize the other frequencies. The maximum length of a prefix satisfying the condition is tracked and returned.

Step-by-Step Algorithm

Step 1: Initialize two hash maps: `counts` to store the frequency of each number and `freqs` to store the frequency of the frequencies.
Step 2: Iterate through the input array `nums`.
Step 3: For each number `num`, update the `counts` and `freqs` hash maps.
Step 4: If the number was already in `counts` (i.e., it existed in previous part of array), reduce the frequency counter for its old frequency in the `freqs` map. Remove frequency from `freqs` if its count becomes 0.
Step 5: Increment the count for the current `num` in the `counts` map.
Step 6: Increment the count for the new frequency in the `freqs` map.
Step 7: Check if the `freqs` map satisfies either of the two conditions: having one entry or having two entries that satisfy the equal frequency condition after removing one element.
Step 8: If the current prefix satisfies the condition, update the `max_len`.
Step 9: Return the `max_len` after iterating through the entire array.

Key Insights

Insight 1: We need to keep track of the frequency of each number in the prefix and the frequency of those frequencies.
Insight 2: We only need to check prefixes where, after removing one element, either all numbers appear once, one number appears once and the rest have the same frequency, one number appears one more time than the others, or all numbers except one have a frequency of 1.
Insight 3: It's more efficient to iterate through the array and update the frequency maps dynamically rather than recomputing them for each prefix.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: American Express.

Maximum Equal Frequency | LeetCode 1224

Hard

Array Hash Table

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

Solution Code

Available in:

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

class Solution {
    public int maxEqualFreq(int[] nums) {
        Map<Integer, Integer> counts = new HashMap<>();
        Map<Integer, Integer> freqs = new HashMap<>();
        int max_len = 0;

        for (int i = 0; i < nums.length; i++) {
            int num = nums[i];
            int c = counts.getOrDefault(num, 0);
            if (c > 0) {
                freqs.put(c, freqs.get(c) - 1);
                if (freqs.get(c) == 0) {
                    freqs.remove(c);
                }
            }

            c += 1;
            counts.put(num, c);
            freqs.put(c, freqs.getOrDefault(c, 0) + 1);

            if (freqs.size() == 1) {
                Map.Entry<Integer, Integer> entry = freqs.entrySet().iterator().next();
                int f = entry.getKey();
                int n = entry.getValue();
                if (f == 1 || n == 1) {
                    max_len = i + 1;
                }
            } else if (freqs.size() == 2) {
                Integer[] f = new Integer[2];
                Integer[] n = new Integer[2];
                int index = 0;
                for (Map.Entry<Integer, Integer> entry : freqs.entrySet()) {
                    f[index] = entry.getKey();
                    n[index] = entry.getValue();
                    index++;
                }

                if ((f[0] + 1 == f[1] && n[1] == 1) ||
                    (f[1] + 1 == f[0] && n[0] == 1) ||
                    (f[0] == 1 && n[0] == 1) ||
                    (f[1] == 1 && n[1] == 1)) {
                    max_len = i + 1;
                }
            }
        }

        return max_len;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1224: Maximum Equal Frequency - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

American Express

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1224

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

9 steps

1
Step 1: Initialize two hash maps: `counts` to store the frequency of each number and `freqs` to store the frequency of the frequencies.
2
Step 2: Iterate through the input array `nums`.
3
Step 3: For each number `num`, update the `counts` and `freqs` hash maps.
4
Step 4: If the number was already in `counts` (i.e., it existed in previous part of array), reduce the frequency counter for its old frequency in the `freqs` map. Remove frequency from `freqs` if its count becomes 0.
5
Step 5: Increment the count for the current `num` in the `counts` map.
6
Step 6: Increment the count for the new frequency in the `freqs` map.
7
Step 7: Check if the `freqs` map satisfies either of the two conditions: having one entry or having two entries that satisfy the equal frequency condition after removing one element.
8
Step 8: If the current prefix satisfies the condition, update the `max_len`.
9
Step 9: Return the `max_len` after iterating through the entire array.

💡

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 dominated by the single loop that iterates through the input array. The space complexity is determined by the hash maps, which can grow up to the size of the input array in the worst-case scenario where all elements are unique.

📖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 prefixes and, for each prefix, iterate through all possible elements to remove. Then, check if the remaining elements have the same frequency.

Time: O(n^2 * m), where n is the length of the array and m is the length of prefix (maximum of n)Space: O(n)

Trade-offs: The brute force approach is much simpler to understand but significantly less efficient. It's suitable for very small input sizes, but quickly becomes impractical for larger inputs.

✨

Key Insights

1Insight 1: We need to keep track of the frequency of each number in the prefix and the frequency of those frequencies.
2Insight 2: We only need to check prefixes where, after removing one element, either all numbers appear once, one number appears once and the rest have the same frequency, one number appears one more time than the others, or all numbers except one have a frequency of 1.
3Insight 3: It's more efficient to iterate through the array and update the frequency maps dynamically rather than recomputing them for each prefix.

📚

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!

Edge Cases

Edge case 1: Array with all the same elements. The solution correctly identifies this case and returns the correct length.
Edge case 2: Array where the first element appears only once, and all others appear the same number of times. The solution accounts for this case and ensures correct calculation of maximum length.
Edge case 3: Empty array (while the problem statement mentions the constraint nums.length >= 2, it's good practice to consider an empty array). The provided solutions inherently handles this case by iterating 0 times, so the initialization value of max_len = 0 will be correctly returned.
Edge case 4: Array where all elements are distinct. The solution also correctly handles this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the case where removing an element results in an empty array (frequency 0).

✗Common mistake 2: Not correctly updating the `freqs` hash map after modifying the `counts` hash map.

✗Common mistake 3: Overlooking one of the conditions for two entries in `freqs` to be valid.

✅

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 website traffic data to identify the longest period where, after removing a single type of request, all remaining request types have the same number of occurrences. This could help in load balancing.

🏦Real-world scenario 2: Processing sensor data to find the longest sequence where removing a single outlier ensures that all remaining sensor readings have the same frequency within a specified range, indicating a stable and reliable system.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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 23, 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

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

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 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.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Maximum Equal Frequency"?

For this hard-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?