##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1967.html

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

Easy

## 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.

## Solution

Loop over patterns and for each element in pattern, check whether it is a substring of word. Return the number of elements that are substrings of word.

• class Solution {
public int numOfStrings(String[] patterns, String word) {
int count = 0;
for (String pattern : patterns) {
if (word.indexOf(pattern) >= 0)
count++;
}
return count;
}
}

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

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

• // OJ: https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/
// Time: O(MN)
// Space: O(1)
class Solution {
public:
int numOfStrings(vector<string>& A, string word) {
int ans = 0;
for (auto &s : A) {
if (word.find(s) != string::npos) ++ans;
}
return ans;
}
};

• class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum(1 for p in patterns if p in word)

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

# 1967. Number of Strings That Appear as Substrings in Word
# https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/

class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
res = 0

for p in patterns:
if p in word:
res += 1

return res


• 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;
}