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2185 - Counting Words With a Given Prefix (Easy)

https://leetcode.com/problems/counting-words-with-a-given-prefix/

Problem Statement​

You are given an array of strings words and a string pref.

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.

Approach 1: Iteration​

We can define the length of the prefix string to be n , then we need to grab the first n characters of each word and compare their substrings. In the case the word has length less than n, we skip the word.

Written by @heiheihang
def prefixCount(self, words: List[str], pref: str) -> int:
        
        #define the length of prefix
        n = len(pref)
        
        res = 0
        
        #iterate each word
        for s in words:
        
            #cannot get first n characters if it is shorter than n, so skip
            if( len(s) < n):
                continue
            #compare the first n characters
            if(s[:n] == pref):
                res += 1
                
        return res
Written by @wingkwong
class Solution {
public:
    int prefixCount(vector<string>& words, string pref) {
        int ans = 0;
        // string::find returns the first position 
        // of the first character of the first match
        for (auto s: words) ans += s.find(pref) == 0;
        return ans;
    }
};