How do you solve Determine if Two Strings Are Close on LeetCode?

Determine if Two Strings Are Close (LeetCode #1657) 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: Both strings must have the same unique characters to be close.

What is the brute force solution for Determine if Two Strings Are Close?

The brute force approach for Determine if Two Strings Are Close is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute-force approach checks all possible character arrangements and transformations to see if one string can be converted into another. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Determine if Two Strings Are Close?

Determine if Two Strings Are Close uses the following concepts: Hash Table, String, Sorting, Counting. The recommended patterns to study are: Hash Map, Counting.

What are the interview tips for Determine if Two Strings Are Close?

Always start by checking for edge cases, such as different lengths. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice counting character frequencies and using data structures like HashMaps or Counters.

What are common mistakes when solving Determine if Two Strings Are Close?

Failing to check if the lengths of the strings are equal before proceeding. Not considering the character frequency distribution when checking for closeness.

#1657

Determine if Two Strings Are Close

Medium

Hash TableStringSortingCountingHash MapCounting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n)
Space	O(n)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution leverages frequency counts of characters and their occurrences. By ensuring that both strings have the same unique characters and the same frequency distribution, we can determine if they are close without generating permutations.

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Algorithm

3 steps

1Step 1: Count the frequency of each character in both strings.
2Step 2: Check if both strings have the same set of unique characters.
3Step 3: Check if the frequency counts can be rearranged to match each other.

solution.py6 lines

1from collections import Counter
2
3def areClose(word1, word2):
4    if len(word1) != len(word2): return False
5    count1, count2 = Counter(word1), Counter(word2)
6    return sorted(count1.values()) == sorted(count2.values()) and set(count1.keys()) == set(count2.keys())

ℹ

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 storage of frequency counts.

1Both strings must have the same unique characters to be close.
2The frequency of characters must be rearrangeable to match between the two strings.

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