# 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 10^4
• s only consists of a, b or 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 $O(1)$.

• 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
}