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Formatted question description: https://leetcode.ca/all/1759.html

# 1759. Count Number of Homogenous Substrings

Medium

## Description

Given a string s, return the number of homogenous substrings of s. Since the answer may be too large, return it modulo 10^9 + 7.

A string is homogenous if all the characters of the string are the same.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = “abbcccaa”

Output: 13

Explanation: The homogenous substrings are listed as below:

“a” appears 3 times.

“aa” appears 1 time.

“b” appears 2 times.

“bb” appears 1 time.

“c” appears 3 times.

“cc” appears 2 times.

“ccc” appears 1 time.

3 + 1 + 2 + 1 + 3 + 2 + 1 = 13.

Example 2:

Input: s = “xy”

Output: 2

Explanation: The homogenous substrings are “x” and “y”.

Example 3:

Input: s = “zzzzz”

Output: 15

Constraints:

• 1 <= s.length <= 10^5
• s consists of lowercase letters.

## Solution

First, find all longest substrings such that all characters in the substring are the same. Next, calculate the count of each longest substring. If a longest substring have length n, then the number of homogeneous substrings in such a substring is n * (n + 1) / 2. Finally, return the total count.

• class Solution {
public int countHomogenous(String s) {
final int MODULO = 1000000007;
long totalCount = 0;
int length = s.length();
char prev = '0';
int consecutive = 0;
for (int i = 0; i < length; i++) {
char c = s.charAt(i);
if (c == prev)
consecutive++;
else {
long curCount = (long) consecutive * (consecutive + 1) / 2 % MODULO;
totalCount = (totalCount + curCount) % MODULO;
consecutive = 1;
prev = c;
}
}
long curCount = (long) consecutive * (consecutive + 1) / 2 % MODULO;
totalCount = (totalCount + curCount) % MODULO;
return (int) totalCount;
}
}

############

class Solution {
private static final int MOD = (int) 1e9 + 7;

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

• class Solution:
def countHomogenous(self, s: str) -> int:
mod = 10**9 + 7
i, n = 0, len(s)
ans = 0
while i < n:
j = i
while j < n and s[j] == s[i]:
j += 1
cnt = j - i
ans += (1 + cnt) * cnt // 2
ans %= mod
i = j
return ans

############

# 1759. Count Number of Homogenous Substrings
# https://leetcode.com/problems/count-number-of-homogenous-substrings

class Solution:
def countHomogenous(self, s: str) -> int:
n = len(s)
M = 10 ** 9 + 7
res = 0

mp = Counter(s)

i = 0
while i < n:
j = i + 1
while j < n and s[i] == s[j]:
res += j - i
j += 1

i = j

return (res + sum(mp.values())) % M


• class Solution {
public:
const int mod = 1e9 + 7;

int countHomogenous(string s) {
int n = s.size();
long ans = 0;
for (int i = 0, j = 0; i < n; i = j) {
j = i;
while (j < n && s[j] == s[i]) ++j;
int cnt = j - i;
ans += 1ll * (1 + cnt) * cnt / 2;
ans %= mod;
}
return ans;
}
};

• func countHomogenous(s string) (ans int) {
n := len(s)
const mod int = 1e9 + 7
for i, j := 0, 0; i < n; i = j {
j = i
for j < n && s[j] == s[i] {
j++
}
cnt := j - i
ans += (1 + cnt) * cnt / 2
ans %= mod
}
return
}

• function countHomogenous(s: string): number {
const mod = 1e9 + 7;
const n = s.length;
let ans = 0;
for (let i = 0, j = 0; j < n; j++) {
if (s[i] !== s[j]) {
i = j;
}
ans = (ans + j - i + 1) % mod;
}
return ans;
}


• impl Solution {
pub fn count_homogenous(s: String) -> i32 {
const MOD: usize = 1e9 as usize + 7;
let s = s.as_bytes();
let n = s.len();
let mut ans = 0;
let mut i = 0;
for j in 0..n {
if s[i] != s[j] {
i = j;
}
ans = (ans + j - i + 1) % MOD;
}
ans as i32
}
}


• public class Solution {
public int CountHomogenous(string s) {
long MOD = 1000000007;
long ans = 0;
for (int i = 0, j = 0; i < s.Length; i = j) {
j = i;
while (j < s.Length && s[j] == s[i]) {
++j;
}
int cnt = j - i;
ans += (long) (1 + cnt) * cnt / 2;
ans %= MOD;
}
return (int) ans;
}
}