LeetCode #2788 — EASY

Split Strings by Separator

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

Given an array of strings words and a character separator, split each string in words by separator.

Return an array of strings containing the new strings formed after the splits, excluding empty strings.

Notes

separator is used to determine where the split should occur, but it is not included as part of the resulting strings.
A split may result in more than two strings.
The resulting strings must maintain the same order as they were initially given.

Example 1:

Input: words = ["one.two.three","four.five","six"], separator = "."
Output: ["one","two","three","four","five","six"]
Explanation: In this example we split as follows:

"one.two.three" splits into "one", "two", "three"
"four.five" splits into "four", "five"
"six" splits into "six" 

Hence, the resulting array is ["one","two","three","four","five","six"].

Example 2:

Input: words = ["$easy$","$problem$"], separator = "$"
Output: ["easy","problem"]
Explanation: In this example we split as follows: 

"$easy$" splits into "easy" (excluding empty strings)
"$problem$" splits into "problem" (excluding empty strings)

Hence, the resulting array is ["easy","problem"].

Example 3:

Input: words = ["|||"], separator = "|"
Output: []
Explanation: In this example the resulting split of "|||" will contain only empty strings, so we return an empty array [].

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 20
characters in words[i] are either lowercase English letters or characters from the string ".,|$#@" (excluding the quotes)
separator is a character from the string ".,|$#@" (excluding the quotes)

Step 01

Brute Force Baseline

Problem summary: Given an array of strings words and a character separator, split each string in words by separator. Return an array of strings containing the new strings formed after the splits, excluding empty strings. Notes separator is used to determine where the split should occur, but it is not included as part of the resulting strings. A split may result in more than two strings. The resulting strings must maintain the same order as they were initially given.

Baseline thinking

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

Pattern signal: Array

Example 1

["one.two.three","four.five","six"]
"."

Example 2

["$easy$","$problem$"]
"$"

Example 3

["|||"]
"|"

Core Insight

What unlocks the optimal approach

Iterate over each string in the given array using a loop and perform string splitting based on the provided separator character.
Be sure not to return empty strings.

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 #2788: Split Strings by Separator
import java.util.regex.Pattern;

class Solution {
    public List<String> splitWordsBySeparator(List<String> words, char separator) {
        List<String> ans = new ArrayList<>();
        for (var w : words) {
            for (var s : w.split(Pattern.quote(String.valueOf(separator)))) {
                if (s.length() > 0) {
                    ans.add(s);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2788: Split Strings by Separator
func splitWordsBySeparator(words []string, separator byte) (ans []string) {
	for _, w := range words {
		for _, s := range strings.Split(w, string(separator)) {
			if s != "" {
				ans = append(ans, s)
			}
		}
	}
	return
}

# Accepted solution for LeetCode #2788: Split Strings by Separator
class Solution:
    def splitWordsBySeparator(self, words: List[str], separator: str) -> List[str]:
        return [s for w in words for s in w.split(separator) if s]

// Accepted solution for LeetCode #2788: Split Strings by Separator
/**
 * [2788] Split Strings by Separator
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn split_words_by_separator(words: Vec<String>, separator: char) -> Vec<String> {
        let mut result = Vec::new();

        for s in &words {
            for i in s.split(separator) {
                let word = String::from(i);

                if word.is_empty() {
                    continue;
                }

                result.push(word)
            }
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2788() {}
}

// Accepted solution for LeetCode #2788: Split Strings by Separator
function splitWordsBySeparator(words: string[], separator: string): string[] {
    return words.flatMap(w => w.split(separator).filter(s => s.length > 0));
}

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 × m)

Space

O(m)

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.