Given an array of strings patterns
and a string word
, return the number of strings in patterns
that exist as a substring in word
.
A substring is a contiguous sequence of characters within a string.
Example 1:
Input: patterns = ["a","abc","bc","d"], word = "abc" Output: 3 Explanation: - "a" appears as a substring in "abc". - "abc" appears as a substring in "abc". - "bc" appears as a substring in "abc". - "d" does not appear as a substring in "abc". 3 of the strings in patterns appear as a substring in word.
Example 2:
Input: patterns = ["a","b","c"], word = "aaaaabbbbb" Output: 2 Explanation: - "a" appears as a substring in "aaaaabbbbb". - "b" appears as a substring in "aaaaabbbbb". - "c" does not appear as a substring in "aaaaabbbbb". 2 of the strings in patterns appear as a substring in word.
Example 3:
Input: patterns = ["a","a","a"], word = "ab" Output: 3 Explanation: Each of the patterns appears as a substring in word "ab".
Constraints:
1 <= patterns.length <= 100
1 <= patterns[i].length <= 100
1 <= word.length <= 100
patterns[i]
andword
consist of lowercase English letters.
class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum(1 for p in patterns if p in word)
class Solution {
public int numOfStrings(String[] patterns, String word) {
int res = 0;
for (String p : patterns) {
if (word.contains(p)) {
++res;
}
}
return res;
}
}
function numOfStrings(patterns: string[], word: string): number {
let ans = 0;
for (let pattern of patterns) {
if (word.includes(pattern)) {
ans++;
}
}
return ans;
};
class Solution {
public:
int numOfStrings(vector<string> &patterns, string word) {
int res = 0;
for (auto p : patterns)
if (word.find(p) != string::npos)
++res;
return res;
}
};
func numOfStrings(patterns []string, word string) int {
res := 0
for _, p := range patterns {
if strings.Contains(word, p) {
res++
}
}
return res
}