How do you solve Divide a String Into Groups of Size k on LeetCode?

Divide a String Into Groups of Size k (LeetCode #2138) 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 how to handle the last group is crucial for solving this problem.

What is the brute force solution for Divide a String Into Groups of Size k?

The brute force approach for Divide a String Into Groups of Size k is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves manually iterating through the string and creating groups of size k. If the last group is smaller than k, we fill it with the specified fill character. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Divide a String Into Groups of Size k?

Divide a String Into Groups of Size k uses the following concepts: String, Simulation. The recommended patterns to study are: String Manipulation, Array Partitioning.

What are the interview tips for Divide a String Into Groups of Size k?

Always clarify the problem constraints and edge cases with the interviewer. Think aloud while coding to show your thought process. Test your solution with different inputs, especially edge cases.

What are common mistakes when solving Divide a String Into Groups of Size k?

Forgetting to pad the last group with the fill character when necessary. Not considering edge cases, such as when the string length is less than k.

#2138

Divide a String Into Groups of Size k

Easy

StringSimulationString ManipulationArray Partitioning

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)

The optimal approach calculates how many complete groups can be formed and handles the last group efficiently by checking the remaining characters in one pass.

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Algorithm

5 steps

1Step 1: Initialize an empty list to store the groups.
2Step 2: Calculate the number of complete groups using integer division of the string length by k.
3Step 3: Loop through the number of complete groups, appending each group of size k to the list.
4Step 4: For the last group, check if there are remaining characters. If not enough, fill with the fill character.
5Step 5: Return the list of groups.

solution.py9 lines

1def divideString(s, k, fill):
2    groups = []
3    n = len(s)
4    complete_groups = n // k
5    for i in range(complete_groups):
6        groups.append(s[i*k:(i+1)*k])
7    last_group = s[complete_groups*k:] + fill * (k - len(s[complete_groups*k:]))
8    groups.append(last_group)
9    return groups

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Complexity note: The time complexity is O(n) because we traverse the string a constant number of times, and the space complexity is O(n) due to storing the result in a list.

1Understanding how to handle the last group is crucial for solving this problem.
2Efficiently calculating the number of complete groups can save time and simplify the code.

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