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):
1 2 3 4 5 6 7 | 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:
1 2 3 4 5 6 7 | 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:
1 2 3 | 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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