What is the time complexity of Count Elements With Maximum Frequency?

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

What data structures are used to solve Count Elements With Maximum Frequency?

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

What are the key insights for solving Count Elements With Maximum Frequency?

Insight 1: Frequency counting is crucial. We need a way to efficiently determine how many times each element appears in the array. Insight 2: A hash map (or dictionary in Python) is a suitable data structure to store the frequency of each element. Alternatively, since the input numbers are constrained to be between 1 and 100, a simple array can be used as a frequency counter.

Which companies ask LeetCode 3005. Count Elements With Maximum Frequency in interviews?

This problem is frequently asked at CRED, Capgemini.

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Count Elements With Maximum Frequency - LeetCode 3005 Solution

Count Elements With Maximum Frequency - Complete Solution Guide

Count Elements With Maximum Frequency is LeetCode problem 3005, a Easy 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 array nums consisting of positive integers. Return the total frequencies of elements in nums such that those elements all have the maximum frequency . The frequency of an element is the number of occurrences of that element in the array. Example 1: Input: nums = [1,2,2,3,1,4] Output: 4 Explanation: The elements 1 and 2 have a frequency of 2 which is the maximum frequency in the array. So the number of elements in the array with maximum frequency is 4. Example 2: Input: nums = [1

Detailed Explanation

The problem asks you to find the total number of elements in an array that have the maximum frequency. The input is an array of positive integers (`nums`). The output is a single integer representing the sum of the counts of all elements that appear with the highest frequency in the array. For example, if the array is `[1, 2, 2, 3, 1, 4]`, the elements 1 and 2 both appear twice (maximum frequency). Therefore, the output should be 4 (2 occurrences of 1 + 2 occurrences of 2).

Solution Approach

The provided solutions use a frequency counting approach. First, they count the occurrences of each element in the input array. Then, they find the maximum frequency among all elements. Finally, they iterate through the frequency counts and sum the counts of elements that have the maximum frequency. The solutions use either a hash map (Python, Java) or a fixed-size array (C++, C) as a frequency counter. The fixed-size array approach is optimized for this problem due to the constraint that numbers are between 1 and 100.

Step-by-Step Algorithm

Step 1: Initialize a frequency counter (hash map or array).
Step 2: Iterate through the input array `nums`. For each element, increment its count in the frequency counter.
Step 3: Find the maximum frequency among all elements in the frequency counter.
Step 4: Iterate through the frequency counter again. Sum the counts of all elements that have a frequency equal to the maximum frequency found in Step 3.
Step 5: Return the sum calculated in Step 4.

Key Insights

Insight 1: Frequency counting is crucial. We need a way to efficiently determine how many times each element appears in the array.
Insight 2: A hash map (or dictionary in Python) is a suitable data structure to store the frequency of each element. Alternatively, since the input numbers are constrained to be between 1 and 100, a simple array can be used as a frequency counter.
Insight 3: After finding the maximum frequency, we need to iterate through the frequency counts again to sum the counts of elements having that maximum frequency.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Counting.

Companies

Asked at: CRED, Capgemini.

Count Elements With Maximum Frequency | LeetCode 3005

Easy

Array Hash Table Counting

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

Solution Code

Available in:

class Solution {
    public int maxFrequencyElements(int[] nums) {
        int[] counts = new int[101];
        int maxFreq = 0;
        for (int num : nums) {
            counts[num]++;
            maxFreq = Math.max(maxFreq, counts[num]);
        }
        int result = 0;
        for (int i = 1; i <= 100; i++) {
            if (counts[i] == maxFreq) {
                result += counts[i];
            }
        }
        return result;
    }
}

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Problem Info

Description

Problem 3005: Count Elements With Maximum Frequency - Detailed explanation and solution approaches.

Topics

Array Hash Table Counting

Companies

CREDCapgemini

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:3005

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

5 steps

1
Step 1: Initialize a frequency counter (hash map or array).
2
Step 2: Iterate through the input array `nums`. For each element, increment its count in the frequency counter.
3
Step 3: Find the maximum frequency among all elements in the frequency counter.
4
Step 4: Iterate through the frequency counter again. Sum the counts of all elements that have a frequency equal to the maximum frequency found in Step 3.
5
Step 5: Return the sum calculated in Step 4.

💡

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 dominant operations are the two iterations. The first iteration has a time complexity of O(n). The second iteration's complexity depends on the size of the frequency counter; O(n) for the hashmap and O(1) for the fixed-size array. Space is used to store the frequency counts. The hash map can use space proportional to the number of unique elements, up to O(n) in the worst case. The fixed-size array uses a constant amount of space.

📖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

Using a `std::map` in C++ (or similar sorted map structures)

Instead of a `std::vector`, we could use a `std::map` to store frequencies. This would automatically keep the counts sorted by element value, making it slightly less efficient than the vector approach but potentially easier to read.

Time: O(n log n) due to map insertions and lookupsSpace: O(n)

Trade-offs: This approach is less efficient than the vector approach if the range of input numbers is known and small, as in this problem. It would be more suitable for problems where the range of numbers is large or unknown.

✨

Key Insights

1Insight 1: Frequency counting is crucial. We need a way to efficiently determine how many times each element appears in the array.
2Insight 2: A hash map (or dictionary in Python) is a suitable data structure to store the frequency of each element. Alternatively, since the input numbers are constrained to be between 1 and 100, a simple array can be used as a frequency counter.
3Insight 3: After finding the maximum frequency, we need to iterate through the frequency counts again to sum the counts of elements having that maximum frequency.

📚

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: Empty input array. The solutions handle this gracefully; the frequency counter will be empty, resulting in a return value of 0.
Edge case 2: All elements are the same. The maximum frequency will be equal to the array's length, and the function will correctly return the array's length.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the input array is empty.

✗Common mistake 2: Incorrectly calculating or handling the maximum frequency. A common error would be to not iterate through all elements of the frequency array to find the maximum frequency.

✅

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 most frequently accessed pages or features. This would help determine which features to prioritize or optimize.

🏦Real-world scenario 2: Processing sensor data to find the most frequent readings, potentially indicating a particular state or condition of the 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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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 4, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

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

@InterviewProJun 30

❓

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Count Elements With Maximum Frequency"?

For this easy-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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