LeetCode #3005 — EASY

Count Elements With Maximum Frequency

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

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 = [1,2,3,4,5]
Output: 5
Explanation: All elements of the array have a frequency of 1 which is the maximum.
So the number of elements in the array with maximum frequency is 5.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[1,2,2,3,1,4]

Example 2

[1,2,3,4,5]

Core Insight

What unlocks the optimal approach

Find frequencies of all elements of the array.
Find the elements that have the maximum frequencies and count their total occurrences.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3005: Count Elements With Maximum Frequency
class Solution {
    public int maxFrequencyElements(int[] nums) {
        int[] cnt = new int[101];
        for (int x : nums) {
            ++cnt[x];
        }
        int ans = 0, mx = -1;
        for (int x : cnt) {
            if (mx < x) {
                mx = x;
                ans = x;
            } else if (mx == x) {
                ans += x;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3005: Count Elements With Maximum Frequency
func maxFrequencyElements(nums []int) (ans int) {
	cnt := [101]int{}
	for _, x := range nums {
		cnt[x]++
	}
	mx := -1
	for _, x := range cnt {
		if mx < x {
			mx, ans = x, x
		} else if mx == x {
			ans += x
		}
	}
	return
}

# Accepted solution for LeetCode #3005: Count Elements With Maximum Frequency
class Solution:
    def maxFrequencyElements(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        mx = max(cnt.values())
        return sum(x for x in cnt.values() if x == mx)

// Accepted solution for LeetCode #3005: Count Elements With Maximum Frequency
impl Solution {
    pub fn max_frequency_elements(nums: Vec<i32>) -> i32 {
        let mut cnt = [0; 101];
        for &x in &nums {
            cnt[x as usize] += 1;
        }
        let mut ans = 0;
        let mut mx = -1;
        for &x in &cnt {
            if mx < x {
                mx = x;
                ans = x;
            } else if mx == x {
                ans += x;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #3005: Count Elements With Maximum Frequency
function maxFrequencyElements(nums: number[]): number {
    const cnt: number[] = Array(101).fill(0);
    for (const x of nums) {
        ++cnt[x];
    }
    let [ans, mx] = [0, -1];
    for (const x of cnt) {
        if (mx < x) {
            mx = x;
            ans = x;
        } else if (mx === x) {
            ans += x;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.