What is the time complexity of Minimum Operations to Make Character Frequencies Equal?

The optimal solution for Minimum Operations to Make Character Frequencies Equal runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the string a few times, and the space complexity is O(n) due to the frequency map storing character counts.

What is the brute force solution for Minimum Operations to Make Character Frequencies Equal?

The brute force approach for Minimum Operations to Make Character Frequencies Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying all possible combinations of operations to make the character frequencies equal. This means we will check every possible frequency and calculate how many operations are needed to achieve that frequency for all characters. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Make Character Frequencies Equal?

Minimum Operations to Make Character Frequencies Equal uses the following concepts: Hash Table, String, Dynamic Programming, Counting, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Make Character Frequencies Equal?

Always start with a clear understanding of the problem and constraints. Think about how you can reduce the number of operations by focusing on character frequencies. Practice explaining your thought process as you code, as communication is key in interviews.

What are common mistakes when solving Minimum Operations to Make Character Frequencies Equal?

Not considering all possible target frequencies when calculating operations. Failing to optimize the frequency counting step, leading to unnecessary complexity.

#3389

Minimum Operations to Make Character Frequencies Equal

Hard

Hash TableStringDynamic ProgrammingCountingEnumerationHash 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)

Instead of checking every possible frequency, we can directly calculate the operations needed based on the frequency counts. We can use a frequency count array and determine the operations needed to balance the frequencies more efficiently.

⚙️

Algorithm

4 steps

1Step 1: Count the frequency of each character in the string using an array of size 26.
2Step 2: Calculate the total number of characters and the number of unique frequencies.
3Step 3: For each unique frequency, calculate the number of operations needed to convert all character counts to that frequency.
4Step 4: Return the minimum operations found.

solution.py20 lines

1from collections import Counter
2
3def min_operations_optimal(s):
4    freq = Counter(s)
5    count_freq = [0] * 26
6    for count in freq.values():
7        count_freq[count] += 1
8    total_chars = len(s)
9    min_operations = float('inf')
10    
11    for target_freq in range(1, total_chars + 1):
12        operations = 0
13        for i in range(1, 26):
14            if count_freq[i] > 0:
15                if i < target_freq:
16                    operations += (target_freq - i) * count_freq[i]
17                else:
18                    operations += (i - target_freq) * count_freq[i]
19        min_operations = min(min_operations, operations)
20    return min_operations

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string a few times, and the space complexity is O(n) due to the frequency map storing character counts.

1Understanding character frequency is crucial to solving this problem efficiently.
2Operations can be thought of as balancing the character counts to a common frequency.

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