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Posts tagged as “count_if”

花花酱 LeetCode 2185. Counting Words With a Given Prefix

By zxi on February 26, 2022

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.

Solution: Straight forward

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

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

C++

花花酱 LeetCode 1967. Number of Strings That Appear as Substrings in Word

By zxi on December 31, 2021

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: Brute Force

We can use count_if for 1-liner.

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

C++