LeetCode #2138 — EASY

Divide a String Into Groups of Size k

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

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 exists) and concatenating all the groups in order, the resultant string should be s.

Given the string s, the size of each group k and the character fill, return a string array denoting the composition of every group s has been divided into, using the above procedure.

Example 1:

Input: s = "abcdefghi", k = 3, fill = "x"
Output: ["abc","def","ghi"]
Explanation:
The first 3 characters "abc" form the first group.
The next 3 characters "def" form the second group.
The last 3 characters "ghi" form the third group.
Since all groups can be completely filled by characters from the string, we do not need to use fill.
Thus, the groups formed are "abc", "def", and "ghi".

Example 2:

Input: s = "abcdefghij", k = 3, fill = "x"
Output: ["abc","def","ghi","jxx"]
Explanation:
Similar to the previous example, we are forming the first three groups "abc", "def", and "ghi".
For the last group, we can only use the character 'j' from the string. To complete this group, we add 'x' twice.
Thus, the 4 groups formed are "abc", "def", "ghi", and "jxx".

Constraints:

1 <= s.length <= 100
s consists of lowercase English letters only.
1 <= k <= 100
fill is a lowercase English letter.

Step 01

Brute Force Baseline

Problem summary: 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 exists) and concatenating all the groups in order, the resultant string should be s. Given the string s, the size of each group k and the character fill, return a string array denoting the composition of every group s has been divided into, using the above procedure.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"abcdefghi"
3
"x"

Example 2

"abcdefghij"
3
"x"

Core Insight

What unlocks the optimal approach

Using the length of the string and k, can you count the number of groups the string can be divided into?
Try completing each group using characters from the string. If there aren’t enough characters for the last group, use the fill character to complete the group.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2138: Divide a String Into Groups of Size k
class Solution {
    public String[] divideString(String s, int k, char fill) {
        int n = s.length();
        String[] ans = new String[(n + k - 1) / k];
        if (n % k != 0) {
            s += String.valueOf(fill).repeat(k - n % k);
        }
        for (int i = 0; i < ans.length; ++i) {
            ans[i] = s.substring(i * k, (i + 1) * k);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2138: Divide a String Into Groups of Size k
func divideString(s string, k int, fill byte) (ans []string) {
	n := len(s)
	if n%k != 0 {
		s += strings.Repeat(string(fill), k-n%k)
	}
	for i := 0; i < len(s)/k; i++ {
		ans = append(ans, s[i*k:(i+1)*k])
	}
	return
}

# Accepted solution for LeetCode #2138: Divide a String Into Groups of Size k
class Solution:
    def divideString(self, s: str, k: int, fill: str) -> List[str]:
        return [s[i : i + k].ljust(k, fill) for i in range(0, len(s), k)]

// Accepted solution for LeetCode #2138: Divide a String Into Groups of Size k
struct Solution;

impl Solution {
    pub fn divide_string(s: String, k: i32, fill: char) -> Vec<String> {
        let k = k as usize;
        let mut answer = s.chars().collect::<Vec<char>>();
        answer.resize(s.len() + (k - s.len() % k) % k, fill);
        answer
            .chunks(k as usize)
            .map(|c| c.iter().collect::<String>())
            .collect::<Vec<String>>()
    }
}

#[test]
fn test() {
    let s = "abcdefghi".to_string();
    let k = 3;
    let fill = 'x';
    let res = vec_string!["abc", "def", "ghi"];
    assert_eq!(Solution::divide_string(s, k, fill), res);

    let s = "abcdefghij".to_string();
    let k = 3;
    let fill = 'x';
    let res = vec_string!["abc", "def", "ghi", "jxx"];
    assert_eq!(Solution::divide_string(s, k, fill), res);
}

// Accepted solution for LeetCode #2138: Divide a String Into Groups of Size k
function divideString(s: string, k: number, fill: string): string[] {
    const ans: string[] = [];
    for (let i = 0; i < s.length; i += k) {
        ans.push(s.slice(i, i + k).padEnd(k, fill));
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.