How do you solve Count Caesar Cipher Pairs on LeetCode?

Count Caesar Cipher Pairs (LeetCode #3805) 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: Strings are similar based on cyclic shifts of characters.

What is the time complexity of Count Caesar Cipher Pairs?

The optimal solution for Count Caesar Cipher Pairs runs in O(n) time and uses O(n) space. We traverse the list once and use a hash map for counting, leading to linear time complexity.

What is the brute force solution for Count Caesar Cipher Pairs?

The brute force approach for Count Caesar Cipher Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every pair of words to see if they can be made similar by shifting letters. This involves comparing each word with every other word. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Caesar Cipher Pairs?

Count Caesar Cipher Pairs uses the following concepts: Array, Hash Table, Math, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Caesar Cipher Pairs?

Explain your thought process clearly. Use examples to illustrate your points. Practice coding without an IDE to simulate real interview conditions.

What are common mistakes when solving Count Caesar Cipher Pairs?

Not considering cyclic nature of alphabet. Incorrectly calculating pairs from frequency counts.

#3805

Count Caesar Cipher Pairs

Medium

ArrayHash TableMathStringCountingHash 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)

Normalize each word by converting it to a key based on relative character differences. Use a hash map to count occurrences of each key.

⚙️

Algorithm

3 steps

1Step 1: Normalize each word to a key representing its character differences.
2Step 2: Use a hash map to count how many times each key appears.
3Step 3: For each key, calculate the number of similar pairs using the formula k * (k - 1) / 2.

solution.py11 lines

1from collections import defaultdict
2
3def countPairs(words):
4    count = 0
5    freq = defaultdict(int)
6    for word in words:
7        key = tuple((ord(word[i]) - ord(word[0])) % 26 for i in range(len(word)))
8        freq[key] += 1
9    for k in freq.values():
10        count += k * (k - 1) // 2
11    return count

ℹ

Complexity note: We traverse the list once and use a hash map for counting, leading to linear time complexity.

1Strings are similar based on cyclic shifts of characters.
2Character differences can be normalized to create unique keys.

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