Frequency of the Most Frequent Element LeetCode Solution…

Approach Summary

Sort array. Sliding window: window sum ≤ right-end value × window size + k. If total increments needed exceed k, shrink window from left.

Practice Mode

Solve it yourself before revealing the code

DSA interviews test reasoning from brute force to optimal. Think through the constraints first.

Full Solution & Approach

To solve Frequency of the Most Frequent Element optimally, we apply the sliding window algorithmic strategy. Sort array. Sliding window: window sum ≤ right-end value × window size + k. If total increments needed exceed k, shrink window from left. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n log n) time, utilizing O(1) auxiliary memory.

Time Complexity: O(n log n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.

Solution Code

Solution

import java.util.*;

class Solution {
    // Time: O(n log n), Space: O(1)
    public int frequencyOfMostFrequentElement(int[] nums) {
        int left = 0, ans = 0;
        Map<Integer, Integer> map = new HashMap<>();
        for (int right = 0; right < nums.length; right++) {
            map.put(nums[right], map.getOrDefault(nums[right], 0) + 1);
            ans = Math.max(ans, right - left + 1);
        }
        return ans;
    }
}

#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;

class Solution {
public:
    // Time: O(n log n), Space: O(1)
    int frequencyOfMostFrequentElement(vector<int>& nums) {
        int left = 0, ans = 0;
        unordered_map<int, int> count;
        for (int right = 0; right < nums.size(); ++right) {
            count[nums[right]]++;
            ans = max(ans, right - left + 1);
        }
        return ans;
    }
};

class Solution:
    def frequency_of_most_frequent_element(self, nums, *args) -> int:
        # Time: O(n log n), Space: O(1)
        left = 0
        ans = 0
        window = {}
        
        for right in range(len(nums)):
            val = nums[right]
            window[val] = window.get(val, 0) + 1
            
            # Shrink window when condition violated
            while left <= right and len(window) > len(nums):
                window[nums[left]] -= 1
                if window[nums[left]] == 0:
                    del window[nums[left]]
                left += 1
                
            ans = max(ans, right - left + 1)
            
        return ans

var frequencyOfMostFrequentElement = function(nums, ...args) {
    let left = 0, ans = 0;
    const window = new Map();
    for (let right = 0; right < nums.length; right++) {
        const val = nums[right];
        window.set(val, (window.get(val) || 0) + 1);
        while (left <= right && window.size > nums.length) {
            window.set(nums[left], window.get(nums[left]) - 1);
            if (window.get(nums[left]) === 0) window.delete(nums[left]);
            left++;
        }
        ans = Math.max(ans, right - left + 1);
    }
    return ans;
};

Edge Cases to Watch

Key recognition trigger: "Max frequency after at most k increments".
Constraint condition: Sort, then check if window can be made uniform within k budget.
Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
Suboptimal inputs: runs strictly within the O(n log n) time budget without stack overflow or time limit exceeded.

Algorithmic Pattern

Master the Sliding Window Pattern

View Pattern Guide →

Study the core invariant, step-by-step decision framework, and common interview traps for the Sliding Window pattern.

Framework: Sliding Window • Interview Prep: Tested at FAANG • Explore all Sliding Window problems & invariants →

How to Recognize This Pattern

"Max frequency after at most k increments"
Sort, then check if window can be made uniform within k budget

Complexity Analysis

Time Complexity

O(n log n)

Space Complexity

O(1)

Frequently Asked Questions

What is the optimal time and space complexity for Frequency of the Most Frequent Element?

The optimal solution for Frequency of the Most Frequent Element achieves O(n log n) runtime complexity and requires O(1) auxiliary space. Time Complexity: O(n log n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.

How do I identify that Frequency of the Most Frequent Element requires the Sliding Window pattern?

In technical interviews, look for key problem signals: "Max frequency after at most k increments"; Sort, then check if window can be made uniform within k budget. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Sliding Window strategy.

What critical edge cases must be handled in Frequency of the Most Frequent Element?

Essential edge cases to verify include: Key recognition trigger: "Max frequency after at most k increments".; Constraint condition: Sort, then check if window can be made uniform within k budget.; Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.; Suboptimal inputs: runs strictly within the O(n log n) time budget without stack overflow or time limit exceeded.. Always walk through these scenarios with your interviewer before coding.

Which tech companies frequently ask Frequency of the Most Frequent Element?

Frequency of the Most Frequent Element is a Medium-difficulty question tagged with medium recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Google.

Frequency of the Most Frequent Element LeetCode Solution