How do you solve Maximum Equal Frequency on LeetCode?

Maximum Equal Frequency (LeetCode #1224) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding frequency distributions is crucial for this problem.

What is the brute force solution for Maximum Equal Frequency?

The brute force approach for Maximum Equal Frequency is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible prefix of the array and determining if removing one element from that prefix can lead to equal frequencies for the remaining elements. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Equal Frequency?

Maximum Equal Frequency uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Equal Frequency?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process and how you plan to optimize your solution as you go. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Maximum Equal Frequency?

Failing to correctly handle the edge cases where all frequencies are the same or only one number exists. Not maintaining the frequency counts properly while iterating through the array.

#1224

Maximum Equal Frequency

Hard

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages frequency counting and the properties of frequency distributions to determine the longest valid prefix in linear time. By tracking the frequency of numbers and their counts, we can efficiently check the conditions for equal frequency after removing one element.

⚙️

Algorithm

4 steps

1Step 1: Initialize frequency maps to count occurrences of numbers and their frequencies.
2Step 2: Iterate through the array and update the frequency maps as you go.
3Step 3: For each prefix, check the conditions for valid frequency distributions using the frequency maps.
4Step 4: Update the maximum length of valid prefixes found.

solution.py26 lines

1# Full working Python code
2
3def maxEqualFreq(nums):
4    freq = {}
5    count = {}
6    max_length = 0
7    for i, num in enumerate(nums):
8        if num in freq:
9            count[freq[num]] -= 1
10            if count[freq[num]] == 0:
11                del count[freq[num]]
12            freq[num] += 1
13        else:
14            freq[num] = 1
15        count[freq[num]] = count.get(freq[num], 0) + 1
16        if len(count) == 1:
17            key = next(iter(count))
18            if key == 1 or count[key] == 1:
19                max_length = max(max_length, i + 1)
20        elif len(count) == 2:
21            keys = list(count.keys())
22            if (count[keys[0]] == 1 and (keys[0] - 1 == keys[1] or keys[0] == 1)) or 
23                (count[keys[1]] == 1 and (keys[1] - 1 == keys[0] or keys[1] == 1))) :
24                max_length = max(max_length, i + 1)
25    return max_length
26

ℹ

Complexity note: The time complexity is O(n) since we only traverse the array once while maintaining frequency counts. The space complexity is O(n) due to the storage of frequency maps.

1Understanding frequency distributions is crucial for this problem.
2The ability to efficiently update and check conditions on frequency maps can significantly reduce time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.