Description
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 <= 1001 <= words[i].length <= 30words[i] contains only lowercase English letters.- All the strings of
words are unique.
Solutions
Solution 1: Brute Force Enumeration
We directly enumerate all strings words[i], and check whether it is a substring of other strings. If it is, we add it to the answer.
The time complexity is O(n3), and the space complexity is O(n). Where n is the length of the string array.
PythonJavaC++GoTypeScriptRust
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(code-box)
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;
}
}(code-box)
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 && words[j].find(words[i]) != string::npos) {
ans.push_back(words[i]);
break;
}
}
}
return ans;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
