LeetCode 1967. Number of Strings That Appear as Substrings in Word Solution in Java, C++, Python & More | Explanation + Code

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: Simulation

Traverse each string p in the array patterns and check if it is a substring of word. If it is, increment the answer by one.

After traversing, return the answer.

The time complexity is O(n × m), and the space complexity is O(1). Here, n and m are the lengths of patterns and word, respectively.

PythonJavaC++GoTypeScriptRust

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

class Solution { public int numOfStrings(String[] patterns, String word) { int ans = 0; for (String p : patterns) { if (word.contains(p)) { ++ans; } } return ans; } }(code-box)

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; } };(code-box)

func numOfStrings(patterns []string, word string) (ans int) { for _, p := range patterns { if strings.Contains(word, p) { ans++ } } return }(code-box)

function numOfStrings(patterns: string[], word: string): number { return patterns.filter(p => word.includes(p)).length; }(code-box)

impl Solution { pub fn num_of_strings(patterns: Vec<String>, word: String) -> i32 { patterns.iter().filter(|p| word.contains(&**p)).count() as i32 } }(code-box)