Return the number of strings in words that contain pref as a prefix.

A prefix of a string s is any leading contiguous substring of s.

Example 1:

Input: words = ["pay","attention","practice","attend"], pref = "at"
Output: 2
Explanation: The 2 strings that contain "at" as a prefix are: "attention" and "attend".

Example 2:

Input: words = ["leetcode","win","loops","success"], pref = "code"
Output: 0
Explanation: There are no strings that contain "code" as a prefix.

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length, pref.length <= 100
• words[i] and pref consist of lowercase English letters.

Solution: Straight forward

We can use std::count_if and std::string::find.

Time complexity: O(n*l)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int prefixCount(vector<string>& words, string pref) {
    return count_if(begin(words), end(words), [&](const string& w) {
      return w.find(pref) == 0;
    });
  }
};

The post 花花酱 LeetCode 2185. Counting Words With a Given Prefix appeared first on Huahua's Tech Road.

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: Brute Force

We can use count_if for 1-liner.

Time complexity: O(m*n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int numOfStrings(vector<string>& patterns, string word) {
    return count_if(begin(patterns), end(patterns), [&](string& p) {
      return word.find(p) != string::npos;
    });
  }
};

The post 花花酱 LeetCode 1967. Number of Strings That Appear as Substrings in Word appeared first on Huahua's Tech Road.