2185 - Counting Words With a Given Prefix (Easy)
Problem Link
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 <= 1001 <= words[i].length, pref.length <= 100words[i]andprefconsist 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.
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 resclass 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;
}
};2183 - Count Array Pairs Divisible by K (Hard)
Author: @TBC | https://leetcode.com/problems/count-array-pairs-divisible-by-k/
2186 - Minimum Number of Steps to Make Two Strings Anagram II (Medium)
Author: @heiheihang, @wkw | https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram-ii/