README_EN.md

1876. Substrings of Size Three with Distinct Characters

中文文档

Description

A string is good if there are no repeated characters.

Given a string s​​​​​, return the number of good substrings of length three in s​​​​​​.

Note that if there are multiple occurrences of the same substring, every occurrence should be counted.

A substring is a contiguous sequence of characters in a string.

 

Example 1:

Input: s = "xyzzaz"
Output: 1
Explanation: There are 4 substrings of size 3: "xyz", "yzz", "zza", and "zaz". 
The only good substring of length 3 is "xyz".

Example 2:

Input: s = "aababcabc"
Output: 4
Explanation: There are 7 substrings of size 3: "aab", "aba", "bab", "abc", "bca", "cab", and "abc".
The good substrings are "abc", "bca", "cab", and "abc".

 

Constraints:

• 1 <= s.length <= 100
• s​​​​​​ consists of lowercase English letters.

Solutions

Python3

class Solution:
    def countGoodSubstrings(self, s: str) -> int:
        count, n = 0, len(s)
        for i in range(n - 2):
            count += s[i] != s[i + 1] and s[i] != s[i + 2] and s[i + 1] != s[i + 2]
        return count

Java

class Solution {
    public int countGoodSubstrings(String s) {
        int count = 0, n = s.length();
        for (int i = 0; i < n - 2; ++i) {
            char a = s.charAt(i), b = s.charAt(i + 1), c = s.charAt(i + 2);
            if (a != b && a != c && b != c) {
                ++count;
            }
        }
        return count;
    }
}

TypeScript

function countGoodSubstrings(s: string): number {
    const n: number = s.length;
    let count: number = 0;
    for (let i: number = 0; i < n - 2; ++i) {
        let a: string = s.charAt(i),
            b: string = s.charAt(i + 1),
            c: string = s.charAt(i + 2);
        if (a != b && a != c && b != c) {
            ++count;
        }
    }
    return count;
}

PHP

class Solution {
    /**
     * @param String $s
     * @return Integer
     */
    function countGoodSubstrings($s) {
        $cnt = 0;
        for ($i = 0; $i < strlen($s) - 2; $i++) {
            if (
                $s[$i] != $s[$i + 1] &&
                $s[$i] != $s[$i + 2] &&
                $s[$i + 1] != $s[$i + 2]
            ) {
                $cnt++;
            }
        }
        return $cnt++;
    }
}

...