# 1967. Number of Strings That Appear as Substrings in Word

## Description

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] and word consist of lowercase English letters.

## Solutions

• class Solution {
public int numOfStrings(String[] patterns, String word) {
int ans = 0;
for (String p : patterns) {
if (word.contains(p)) {
++ans;
}
}
return ans;
}
}

• class Solution {
public:
int numOfStrings(vector<string>& patterns, string word) {
int ans = 0;
for (auto& p : patterns) {
ans += word.find(p) != string::npos;
}
return ans;
}
};

• class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum(p in word for p in patterns)


• func numOfStrings(patterns []string, word string) (ans int) {
for _, p := range patterns {
if strings.Contains(word, p) {
ans++
}
}
return
}

• function numOfStrings(patterns: string[], word: string): number {
let ans = 0;
for (const p of patterns) {
if (word.includes(p)) {
++ans;
}
}
return ans;
}