How do you solve Make Number of Distinct Characters Equal on LeetCode?

Make Number of Distinct Characters Equal (LeetCode #2531) 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 the distinct character counts is crucial for solving the problem efficiently.

What is the brute force solution for Make Number of Distinct Characters Equal?

The brute force approach for Make Number of Distinct Characters Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible pairs of characters from both strings and simulating a swap to see if the number of distinct characters becomes equal. This is straightforward but inefficient as it checks every possible combination. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Make Number of Distinct Characters Equal?

Make Number of Distinct Characters Equal uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Make Number of Distinct Characters Equal?

Always start with a brute-force solution to understand the problem better. Think about character frequency and distinct counts as key insights. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Make Number of Distinct Characters Equal?

Not considering the case where distinct counts are equal or differ by one. Failing to check if a valid swap exists when counts differ by two.

#2531

Make Number of Distinct Characters Equal

Medium

Hash TableStringCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Instead of checking all pairs, we can analyze the distinct character counts and determine if a swap can balance them. By counting the frequency of characters, we can derive the conditions under which a swap would equalize the distinct counts.

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Algorithm

5 steps

1Step 1: Count the distinct characters in both strings using frequency arrays.
2Step 2: Calculate the distinct counts for word1 and word2.
3Step 3: If the absolute difference between the distinct counts is greater than 2, return false (impossible to balance with one swap).
4Step 4: If the counts are equal or differ by 1, return true (one swap can balance them).
5Step 5: If they differ by 2, check if there exists a character in word1 that can be swapped with a character in word2.

solution.py9 lines

1def canMakeDistinctEqual(word1, word2):
2    from collections import Counter
3    count1 = Counter(word1)
4    count2 = Counter(word2)
5    distinct1 = len(count1)
6    distinct2 = len(count2)
7    if abs(distinct1 - distinct2) > 2:
8        return False
9    return distinct1 == distinct2 or abs(distinct1 - distinct2) == 1 or (distinct1 - distinct2 == 2 and any(c in count2 for c in count1)) or (distinct2 - distinct1 == 2 and any(c in count1 for c in count2))

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Complexity note: The time complexity is O(n) because we traverse each string once to count character frequencies. The space complexity is O(n) due to the space required for the frequency maps.

1Understanding the distinct character counts is crucial for solving the problem efficiently.
2A single swap can only balance the distinct counts if they differ by 0, 1, or 2.

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