Question

1967. Number of Strings That Appear as Substrings in Word

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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

Solution 1

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

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;
  }
};

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;
}