How do you solve Thousand Separator on LeetCode?

Thousand Separator (LeetCode #1556) 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 string manipulation is crucial for this problem.

What is the time complexity of Thousand Separator?

The optimal solution for Thousand Separator runs in O(n) time and uses O(n) space. The time complexity is O(n) since we only traverse the string once, and the space complexity is O(n) for the result string.

What is the brute force solution for Thousand Separator?

The brute force approach for Thousand Separator is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves converting the integer to a string and then iterating through it to insert dots at the appropriate positions. This is straightforward but inefficient for larger numbers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Thousand Separator?

Thousand Separator uses the following concepts: String. The recommended patterns to study are: String Manipulation, Two Pointers.

What are the interview tips for Thousand Separator?

Think about how to minimize operations on strings. Practice building strings in reverse to avoid multiple passes. Always consider edge cases in your solution.

What are common mistakes when solving Thousand Separator?

Not considering edge cases like very small or very large numbers. Overusing string reversals which can lead to inefficiencies.

#1556

Thousand Separator

Easy

StringString ManipulationTwo Pointers

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)

The optimal solution builds the result string in a single pass without reversing the string multiple times. It counts characters and appends dots directly.

⚙️

Algorithm

5 steps

1Step 1: Convert the integer n to a string.
2Step 2: Initialize an empty result string and a counter for digits.
3Step 3: Iterate through the string from the end to the beginning.
4Step 4: Append each character to the result and increment the counter.
5Step 5: If the counter reaches 3, append a dot and reset the counter.

solution.py10 lines

1def thousand_separator(n):
2    s = str(n)
3    result = ''
4    count = 0
5    for i in range(len(s) - 1, -1, -1):
6        if count > 0 and count % 3 == 0:
7            result += '.'
8        result += s[i]
9        count += 1
10    return result[::-1]

ℹ

Complexity note: The time complexity is O(n) since we only traverse the string once, and the space complexity is O(n) for the result string.

1Understanding string manipulation is crucial for this problem.
2Counting characters and managing positions can optimize the solution.

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