# 1180. Count Substrings with Only One Distinct Letter

## Description

Given a string s, return the number of substrings that have only one distinct letter.

Example 1:

Input: s = "aaaba"
Output: 8
Explanation: The substrings with one distinct letter are "aaa", "aa", "a", "b".
"aaa" occurs 1 time.
"aa" occurs 2 times.
"a" occurs 4 times.
"b" occurs 1 time.
So the answer is 1 + 2 + 4 + 1 = 8.


Example 2:

Input: s = "aaaaaaaaaa"
Output: 55


Constraints:

• 1 <= s.length <= 1000
• s[i] consists of only lowercase English letters.

## Solutions

Solution 1: Two Pointers

We can use two pointers, where pointer $i$ points to the start of the current substring, and pointer $j$ moves to the right to the first position that is different from $s[i]$. Then, $[i,..j-1]$ is a substring with $s[i]$ as the only character, and its length is $j-i$. Therefore, the number of substrings with $s[i]$ as the only character is $\frac{(j-i+1)(j-i)}{2}$, which is added to the answer. Then, we set $i=j$ and continue to traverse until $i$ exceeds the range of string $s$.

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 countLetters(String s) {
int ans = 0;
for (int i = 0, n = s.length(); i < n;) {
int j = i;
while (j < n && s.charAt(j) == s.charAt(i)) {
++j;
}
ans += (1 + j - i) * (j - i) / 2;
i = j;
}
return ans;
}
}

• class Solution {
public:
int countLetters(string s) {
int ans = 0;
for (int i = 0, n = s.size(); i < n;) {
int j = i;
while (j < n && s[j] == s[i]) {
++j;
}
ans += (1 + j - i) * (j - i) / 2;
i = j;
}
return ans;
}
};

• class Solution:
def countLetters(self, s: str) -> int:
n = len(s)
i = ans = 0
while i < n:
j = i
while j < n and s[j] == s[i]:
j += 1
ans += (1 + j - i) * (j - i) // 2
i = j
return ans


• func countLetters(s string) int {
ans := 0
for i, n := 0, len(s); i < n; {
j := i
for j < n && s[j] == s[i] {
j++
}
ans += (1 + j - i) * (j - i) / 2
i = j
}
return ans
}

• function countLetters(s: string): number {
let ans = 0;
const n = s.length;
for (let i = 0; i < n; ) {
let j = i;
let cnt = 0;
while (j < n && s[j] === s[i]) {
++j;
ans += ++cnt;
}
i = j;
}
return ans;
}