How do you solve Count The Repetitions on LeetCode?

Count The Repetitions (LeetCode #466) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the relationship between the strings and how they can be formed is crucial.

What is the brute force solution for Count The Repetitions?

The brute force approach for Count The Repetitions is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate the string str1 by concatenating s1 n1 times and then check how many times we can find str2 (concatenated n2 times) within it. This approach is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count The Repetitions?

Count The Repetitions uses the following concepts: Two Pointers, String, Dynamic Programming. The recommended patterns to study are: Two Pointers, Dynamic Programming.

What are the interview tips for Count The Repetitions?

Clearly explain your thought process and how you arrived at your solution. Be prepared to discuss the time and space complexity of your approach. Practice explaining your code as if you were teaching someone else.

What are common mistakes when solving Count The Repetitions?

Students often try to create the full concatenated strings, leading to memory issues. Not considering the modulo operation when iterating through the strings can lead to incorrect results.

#466

Count The Repetitions

Hard

Two PointersStringDynamic ProgrammingTwo PointersDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of generating large strings, we can use a two-pointer technique to efficiently count how many times str2 can be formed from str1 without explicitly creating the full strings.

⚙️

Algorithm

3 steps

1Step 1: Use two pointers to traverse s1 and s2. Count how many characters from s2 can be matched in str1.
2Step 2: Keep track of how many full iterations of str2 can be formed from the characters matched in str1.
3Step 3: Repeat the process for the number of times str1 is repeated (n1).

solution.py13 lines

1def count_repetitions(s1, n1, s2, n2):
2    count = 0
3    s1_len, s2_len = len(s1), len(s2)
4    total_s2 = 0
5    i = 0
6    while total_s2 < n2:
7        for j in range(s1_len):
8            if s1[i % s1_len] == s2[total_s2 % s2_len]:
9                total_s2 += 1
10                if total_s2 % s2_len == 0:
11                    count += 1
12        i += 1
13    return count

ℹ

Complexity note: The time complexity is O(n) because we are effectively iterating through the characters of s1 and s2 without generating large strings, making it efficient.

1Understanding the relationship between the strings and how they can be formed is crucial.
2Using pointers to track positions in the strings can significantly reduce the time complexity.

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