LeetCode #1408 — EASY

String Matching in an Array

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

The Problem

Problem Statement

Given an array of string words, return all strings in words that are a substring of another word. You can return the answer in any order.

Example 1:

Input: words = ["mass","as","hero","superhero"]
Output: ["as","hero"]
Explanation: "as" is substring of "mass" and "hero" is substring of "superhero".
["hero","as"] is also a valid answer.

Example 2:

Input: words = ["leetcode","et","code"]
Output: ["et","code"]
Explanation: "et", "code" are substring of "leetcode".

Example 3:

Input: words = ["blue","green","bu"]
Output: []
Explanation: No string of words is substring of another string.

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 30
words[i] contains only lowercase English letters.
All the strings of words are unique.

Step 01

Brute Force Baseline

Problem summary: Given an array of string words, return all strings in words that are a substring of another word. You can return the answer in any order.

Baseline thinking

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

Pattern signal: Array · String Matching

Example 1

["mass","as","hero","superhero"]

Example 2

["leetcode","et","code"]

Example 3

["blue","green","bu"]

Core Insight

What unlocks the optimal approach

Bruteforce to find if one string is substring of another or use KMP algorithm.

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 #1408: String Matching in an Array
class Solution {
    public List<String> stringMatching(String[] words) {
        List<String> ans = new ArrayList<>();
        int n = words.length;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i != j && words[j].contains(words[i])) {
                    ans.add(words[i]);
                    break;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1408: String Matching in an Array
func stringMatching(words []string) []string {
	ans := []string{}
	for i, w1 := range words {
		for j, w2 := range words {
			if i != j && strings.Contains(w2, w1) {
				ans = append(ans, w1)
				break
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #1408: String Matching in an Array
class Solution:
    def stringMatching(self, words: List[str]) -> List[str]:
        ans = []
        for i, s in enumerate(words):
            if any(i != j and s in t for j, t in enumerate(words)):
                ans.append(s)
        return ans

// Accepted solution for LeetCode #1408: String Matching in an Array
impl Solution {
    pub fn string_matching(words: Vec<String>) -> Vec<String> {
        let mut ans = Vec::new();
        let n = words.len();
        for i in 0..n {
            for j in 0..n {
                if i != j && words[j].contains(&words[i]) {
                    ans.push(words[i].clone());
                    break;
                }
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #1408: String Matching in an Array
function stringMatching(words: string[]): string[] {
    const ans: string[] = [];
    const n = words.length;
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n; ++j) {
            if (words[j].includes(words[i]) && i !== j) {
                ans.push(words[i]);
                break;
            }
        }
    }
    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^3)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × m) time

O(1) space

At each of the n starting positions in the text, compare up to m characters with the pattern. If a mismatch occurs, shift by one and restart. Worst case (e.g., searching "aab" in "aaaa...a") checks m characters at nearly every position: O(n × m).

KMP / Z-ALGO

O(n + m) time

O(m) space

KMP and Z-algorithm preprocess the pattern in O(m) to build a failure/Z-array, then scan the text in O(n) — never backtracking. Total: O(n + m). Rabin-Karp uses rolling hashes for O(n + m) expected time. All beat the O(n × m) brute force of checking every position.

Shortcut: Preprocessing avoids backtracking → O(n + m). The failure function is the key insight.

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.