Divide a String Into Groups of Size k - Complete Solution Guide

Divide a String Into Groups of Size k is LeetCode problem 2138, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A string s can be partitioned into groups of size k using the following procedure: The first group consists of the first k characters of the string, the second group consists of the next k characters of the string, and so on. Each element can be a part of exactly one group. For the last group, if the string does not have k characters remaining, a character fill is used to complete the group. Note that the partition is done so that after removing the fill character from the last group (if it exis

Detailed Explanation

The problem asks you to divide a given string `s` into groups of size `k`. If the length of `s` is not a multiple of `k`, the last group is padded with a specified character `fill` to reach the size `k`. The function should return an array of strings, where each string represents a group.

Solution Approach

The provided solutions use a similar iterative approach. They iterate through the input string `s` in chunks of size `k`. For each chunk, they check if its length is less than `k`. If it is, they append the `fill` character until the chunk's length becomes `k`. Finally, each chunk (group) is added to the result array which is then returned.

Step-by-Step Algorithm

1. Initialize an empty array `result` to store the groups.
2. Iterate through the string `s` with a step of `k` (using a `for` loop with `i += k`).
3. Extract a substring of length `k` starting from index `i` (or less if the end of the string is reached).
4. If the length of the substring is less than `k`, pad it with the `fill` character until its length is `k`.
5. Append the resulting group (substring) to the `result` array.
6. Return the `result` array.

Key Insights

• The problem involves string manipulation and handling of edge cases where the string length is not divisible by `k`.
• Iterating through the string with a step size of `k` is a straightforward approach.
• Using string builders or similar methods can improve efficiency by avoiding repeated string concatenations.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Simulation.

Companies

Asked at: Canonical.