How do you solve Majority Frequency Characters on LeetCode?

Majority Frequency Characters (LeetCode #3692) 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: Character frequency can be efficiently counted using hash maps.

What is the time complexity of Majority Frequency Characters?

The optimal solution for Majority Frequency Characters runs in O(n) time and uses O(n) space. The complexity is linear as we process the string and groups in a single pass.

What is the brute force solution for Majority Frequency Characters?

The brute force approach for Majority Frequency Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. Count frequencies of each character, then check each frequency group for distinct characters. This method is inefficient as it compares groups repeatedly. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Majority Frequency Characters?

Majority Frequency Characters uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Majority Frequency Characters?

Explain your thought process clearly. Consider edge cases, like empty strings or single characters. Practice coding with time constraints.

What are common mistakes when solving Majority Frequency Characters?

Not handling ties correctly between frequency groups. Forgetting to consider all characters in the string.

#3692

Majority Frequency Characters

Easy

Hash TableStringCountingHash 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)

Count character frequencies and group them efficiently. Use a single pass to determine the majority frequency group based on size and frequency.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each character using a hash map.
2Step 2: Create a second hash map to group characters by their frequencies.
3Step 3: Find the group with the maximum size; if tied, choose the higher frequency.

solution.py9 lines

1from collections import Counter
2
3def majority_frequency(s):
4    freq = Counter(s)
5    groups = {}
6    for char, count in freq.items():
7        groups.setdefault(count, set()).add(char)
8    max_group = max(groups.items(), key=lambda x: (len(x[1]), x[0]))
9    return ''.join(max_group[1])

ℹ

Complexity note: The complexity is linear as we process the string and groups in a single pass.

1Character frequency can be efficiently counted using hash maps.
2Grouping characters by frequency allows for quick comparisons.

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