What is the brute force solution for Maximum Difference Between Even and Odd Frequency I?

The brute force approach for Maximum Difference Between Even and Odd Frequency I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the frequency of each character in the string and then comparing every odd frequency with every even frequency to find the maximum difference. 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 Difference Between Even and Odd Frequency I?

Maximum Difference Between Even and Odd Frequency I uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Difference Between Even and Odd Frequency I?

Always clarify the problem constraints and edge cases before diving into coding. Think about the data structures that can simplify your solution, like frequency maps. Practice dry-running your code with sample inputs to catch any logical errors.

What are common mistakes when solving Maximum Difference Between Even and Odd Frequency I?

Not handling cases where there are no odd or even frequencies, even though the problem guarantees at least one of each. Confusing the logic for finding maximum and minimum values.

#3442

Maximum Difference Between Even and Odd Frequency I

Easy

Hash TableStringCountingHash MapArray

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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 a frequency map to efficiently find the maximum odd frequency and minimum even frequency in a single pass. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Create a frequency map to count occurrences of each character.
2Step 2: Initialize variables for maximum odd frequency and minimum even frequency.
3Step 3: Iterate through the frequency map to update the maximum odd and minimum even frequencies in one pass.
4Step 4: Return the difference between the maximum odd frequency and the minimum even frequency.

solution.py15 lines

1# Full working Python code
2from collections import Counter
3
4def max_diff(s):
5    freq = Counter(s)
6    max_odd = float('-inf')
7    min_even = float('inf')
8    
9    for count in freq.values():
10        if count % 2 == 1:
11            max_odd = max(max_odd, count)
12        else:
13            min_even = min(min_even, count)
14    
15    return max_odd - min_even

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the string once to build the frequency map and then through the frequency map, which is much smaller than the input size. The space complexity is O(n) due to the storage of character frequencies in the map.

1Identifying odd and even frequencies is crucial for solving the problem.
2Using a frequency map allows efficient counting and comparison.

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