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1358. Number of Substrings Containing All Three Characters
Description
Given a string s consisting only of characters a, b and c.
Return the number of substrings containing at least one occurrence of all these characters a, b and c.
Example 1:
Input: s = "abcabc" Output: 10 Explanation: The substrings containing at least one occurrence of the characters a, b and c are "abc", "abca", "abcab", "abcabc", "bca", "bcab", "bcabc", "cab", "cabc" and "abc" (again).
Example 2:
Input: s = "aaacb" Output: 3 Explanation: The substrings containing at least one occurrence of the characters a, b and c are "aaacb", "aacb" and "acb".
Example 3:
Input: s = "abc" Output: 1
Constraints:
3 <= s.length <= 5 x 10^4sonly consists of a, b or c characters.
Solutions
Solution 1: Single Pass
We use an array $d$ of length $3$ to record the most recent occurrence of the three characters, initially all set to $-1$.
We traverse the string $s$. For the current position $i$, we first update $d[s[i]]=i$, then the number of valid strings is $\min(d[0], d[1], d[2]) + 1$, which is accumulated to the answer.
The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
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class Solution { public int numberOfSubstrings(String s) { int[] d = new int[] {-1, -1, -1}; int ans = 0; for (int i = 0; i < s.length(); ++i) { char c = s.charAt(i); d[c - 'a'] = i; ans += Math.min(d[0], Math.min(d[1], d[2])) + 1; } return ans; } } -
class Solution { public: int numberOfSubstrings(string s) { int d[3] = {-1, -1, -1}; int ans = 0; for (int i = 0; i < s.size(); ++i) { d[s[i] - 'a'] = i; ans += min(d[0], min(d[1], d[2])) + 1; } return ans; } }; -
class Solution: def numberOfSubstrings(self, s: str) -> int: d = {"a": -1, "b": -1, "c": -1} ans = 0 for i, c in enumerate(s): d[c] = i ans += min(d["a"], d["b"], d["c"]) + 1 return ans -
func numberOfSubstrings(s string) (ans int) { d := [3]int{-1, -1, -1} for i, c := range s { d[c-'a'] = i ans += min(d[0], min(d[1], d[2])) + 1 } return }