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Formatted question description: https://leetcode.ca/all/1408.html

# 1408. String Matching in an Array (Easy)

Given an array of string words. Return all strings in words which is substring of another word in any order.

String words[i] is substring of words[j], if can be obtained removing some characters to left and/or right side of words[j].

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: []


Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 30
• words[i] contains only lowercase English letters.
• It's guaranteed that words[i] will be unique.

Related Topics:
String

## Solution 1. Brute force

For each words[i], check if it’s a substring of words[j] (0 <= j < N && i != j). Once find a match, we can push words[i] to result and continue to word[i + 1].

// OJ: https://leetcode.com/problems/string-matching-in-an-array/
// Time: O(N^2 * D)
// Space: O(1)
class Solution {
public:
vector<string> stringMatching(vector<string>& words) {
vector<string> ans;
int N = words.size();
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
if (i == j) continue;
if (words[j].find(words[i]) != string::npos) {
ans.push_back(words[i]);
break;
}
}
}
return ans;
}
};

• class Solution {
public List<String> stringMatching(String[] words) {
Arrays.sort(words, new Comparator<String>() {
public int compare(String word1, String word2) {
return word1.length() - word2.length();
}
});
List<String> list = new ArrayList<String>();
int length = words.length;
for (int i = 0; i < length; i++) {
String word1 = words[i];
boolean flag = false;
for (int j = i + 1; j < length; j++) {
String word2 = words[j];
if (word2.indexOf(word1) >= 0) {
flag = true;
break;
}
}
if (flag)
}
return list;
}
}

############

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])) {
break;
}
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/string-matching-in-an-array/
// Time: O(N^2 * D)
// Space: O(1)
class Solution {
public:
vector<string> stringMatching(vector<string>& words) {
vector<string> ans;
int N = words.size();
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
if (i == j) continue;
if (words[j].find(words[i]) != string::npos) {
ans.push_back(words[i]);
break;
}
}
}
return ans;
}
};

• class Solution:
def stringMatching(self, words: List[str]) -> List[str]:
ans = []
for i, w1 in enumerate(words):
for j, w2 in enumerate(words):
if i != j and w1 in w2:
ans.append(w1)
break
return ans

############

class Solution:
def stringMatching(self, words: List[str]) -> List[str]:
res = set()
for word in words:
for cur in words:
if cur in word and cur != word:
return list(res)

• 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
}

• function stringMatching(words: string[]): string[] {
const res: string[] = [];
for (const target of words) {
for (const word of words) {
if (word !== target && word.includes(target)) {
res.push(target);
break;
}
}
}
return res;
}


• impl Solution {
pub fn string_matching(words: Vec<String>) -> Vec<String> {
let mut res = Vec::new();
for target in words.iter() {
for word in words.iter() {
if word != target && word.contains(target) {
res.push(target.clone());
break;
}
}
}
res
}
}