What is the time complexity of Frequency of the Most Frequent Element?

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

What data structures are used to solve Frequency of the Most Frequent Element?

This problem uses Array, Binary Search, Greedy, Sliding Window, Sorting, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Frequency of the Most Frequent Element?

Insight 1: Sorting the array is crucial. By sorting, we can focus on making smaller elements equal to larger elements more efficiently. Insight 2: A sliding window technique is effective because it allows us to explore different subarrays (potential frequencies) while ensuring the total cost of incrementing elements within the window stays within the allowed `k` operations.

Which companies ask LeetCode 1838. Frequency of the Most Frequent Element in interviews?

This problem is frequently asked at CRED, Deutsche Bank, PhonePe, Pony.ai.

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Frequency of the Most Frequent Element - LeetCode 1838 Solution

Frequency of the Most Frequent Element - Complete Solution Guide

Frequency of the Most Frequent Element is LeetCode problem 1838, 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

The frequency of an element is the number of times it occurs in an array. You are given an integer array nums and an integer k . In one operation, you can choose an index of nums and increment the element at that index by 1 . Return the maximum possible frequency of an element after performing at most k operations . Example 1: Input: nums = [1,2,4], k = 5 Output: 3 Explanation: Increment the first element three times and the second element two times to make nums = [4,4,4]. 4 has a frequency of 3

Detailed Explanation

The problem asks us to find the maximum frequency of any element in the input array `nums` after performing at most `k` increment operations. We can choose any index in `nums` and increment the element at that index by 1. The goal is to maximize the number of times a single element appears in the array after these operations.

Solution Approach

The solution sorts the input array `nums` in ascending order. It then uses a sliding window to find the largest possible window (subarray) where the elements can be made equal to the rightmost element using at most `k` operations. The window expands to the right and contracts from the left when the operations needed to make all elements in the window equal to the rightmost element exceed `k`. The length of the largest valid window represents the maximum possible frequency.

Step-by-Step Algorithm

Step 1: Sort the `nums` array in ascending order.
Step 2: Initialize `left` pointer to 0, `current_sum` to 0, and `max_freq` to 0.
Step 3: Iterate through the `nums` array with the `right` pointer.
Step 4: Add the current element `nums[right]` to `current_sum`.
Step 5: While the cost to make all elements in the window `[left, right]` equal to `nums[right]` (which is `(right - left + 1) * nums[right] - current_sum`) is greater than `k`:
Step 6: Subtract `nums[left]` from `current_sum` and increment the `left` pointer to shrink the window from the left.
Step 7: Update `max_freq` to the maximum of its current value and the current window size `(right - left + 1)`.
Step 8: After the loop finishes, return `max_freq`.

Key Insights

Insight 1: Sorting the array is crucial. By sorting, we can focus on making smaller elements equal to larger elements more efficiently.
Insight 2: A sliding window technique is effective because it allows us to explore different subarrays (potential frequencies) while ensuring the total cost of incrementing elements within the window stays within the allowed `k` operations.
Insight 3: To maximize the frequency, we try to make all numbers within the sliding window equal to the rightmost (largest) element of the window, as this requires the least number of operations.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Greedy, Sliding Window, Sorting, Prefix Sum.

Companies

Asked at: CRED, Deutsche Bank, PhonePe, Pony.ai.

Frequency of the Most Frequent Element | LeetCode 1838

Medium

Array Binary Search Greedy Sliding Window Sorting Prefix Sum

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maxFrequency(int[] nums, int k) {
        Arrays.sort(nums);
        int left = 0;
        long current_sum = 0;
        int max_freq = 0;
        
        for (int right = 0; right < nums.length; right++) {
            current_sum += nums[right];
            
            while ((long)(right - left + 1) * nums[right] - current_sum > k) {
                current_sum -= nums[left];
                left += 1;
            }
            
            max_freq = Math.max(max_freq, right - left + 1);
            
        }
        
        return max_freq;
    }
}

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Problem Info

Description

Problem 1838: Frequency of the Most Frequent Element - Detailed explanation and solution approaches.

Topics

Array Binary Search Greedy Sliding Window Sorting Prefix Sum

Companies

CREDDeutsche BankPhonePePony.ai

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:1838

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: Sort the `nums` array in ascending order.
2
Step 2: Initialize `left` pointer to 0, `current_sum` to 0, and `max_freq` to 0.
3
Step 3: Iterate through the `nums` array with the `right` pointer.
4
Step 4: Add the current element `nums[right]` to `current_sum`.
5
Step 5: While the cost to make all elements in the window `[left, right]` equal to `nums[right]` (which is `(right - left + 1) * nums[right] - current_sum`) is greater than `k`:
6
Step 6: Subtract `nums[left]` from `current_sum` and increment the `left` pointer to shrink the window from the left.
7
Step 7: Update `max_freq` to the maximum of its current value and the current window size `(right - left + 1)`.
8
Step 8: After the loop finishes, return `max_freq`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the sorting step, which takes O(n log n) time. The sliding window part takes O(n) time in the worst case. Therefore, the overall time complexity is O(n log n). The space complexity is O(1) because we're using a constant amount of extra space, regardless of the input size. In C, qsort may use O(log n) stack space in worst case, but it is still typically classified as O(1) auxilliary 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

Binary Search

For each possible length of the subarray, we can use binary search to find if it is possible to form such a subarray with k operations. Time complexity is O(n log n) * O(n), space complexity is O(1)

Time: O(n^2 log n)Space: O(1)

Trade-offs: Slightly more complex implementation with the same or worse performance, not recommended

✨

Key Insights

1Insight 1: Sorting the array is crucial. By sorting, we can focus on making smaller elements equal to larger elements more efficiently.
2Insight 2: A sliding window technique is effective because it allows us to explore different subarrays (potential frequencies) while ensuring the total cost of incrementing elements within the window stays within the allowed `k` operations.
3Insight 3: To maximize the frequency, we try to make all numbers within the sliding window equal to the rightmost (largest) element of the window, as this requires the least number of operations.

📚

Key Terms Explained

For Students

Array

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

Binary Search

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sliding Window

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input array. The solution implicitly handles this correctly as the loop won't execute and 0 will be returned.
Edge case 2: `k` is very large. The sliding window will expand as much as possible within the bounds of the array, correctly calculating the maximum frequency.
Edge case 3: `k` is 0. The solution will find the longest sequence of identical elements in the array after sorting.
Edge case 4: All elements are the same. The solution correctly identifies the length of the array as the maximum frequency.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the array. Without sorting, the sliding window approach won't work correctly to find the most efficient way to increment elements.

✗Common mistake 2: Integer overflow. The cost calculation `(right - left + 1) * nums[right]` and `current_sum` can potentially overflow if `nums[right]` or `k` are large. Using `long` data type for the sum and intermediate calculation is crucial, especially in Java and C++.

✗Common mistake 3: Not understanding the sliding window shrinking condition. The window must shrink when the cost exceeds `k`, and understanding how to update `current_sum` is important for correctness.

✅

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: Resource allocation. Imagine allocating computational resources (CPU cycles or memory) to different processes. You might have a limited budget (k) of resources to distribute, and you want to maximize the number of processes that can achieve a certain minimum performance level (frequency).

🏦Real-world scenario 2: Inventory management. Suppose you have different product lines with varying demand. You have a limited budget (k) to increase the production of some product lines to meet a target demand level. This algorithm helps determine the maximum number of product lines you can bring up to the target demand, subject to your budget constraint.

💡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: July 9, 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.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

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 "Frequency of the Most Frequent Element"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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