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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;
}

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