Teaching Kids Programming – Substrings of Size Three with Distinct Characters


Teaching Kids Programming: Videos on Data Structures and Algorithms

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.

Hints:
Try using a set to find out the number of distinct characters in a substring.

Substrings of Size Three with Distinct Characters

Using a set and count if it has 3 elements (distinct letters):

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class Solution:
    def countGoodSubstrings(self, s: str) -> int:
        ans = 0
        for i in range(2, len(s)):
            if len(set(s[i-2:i+1])) == 3:
                ans += 1
        return ans
class Solution:
    def countGoodSubstrings(self, s: str) -> int:
        ans = 0
        for i in range(2, len(s)):
            if len(set(s[i-2:i+1])) == 3:
                ans += 1
        return ans

The time complexity is O(N) and space complexity is O(1).

We can also just compare directly to see if three letters are equal:

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

Same time and space complexity.

Also, one liner with Count:

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

–EOF (The Ultimate Computing & Technology Blog) —

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