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Formatted question description: https://leetcode.ca/all/1759.html
1759. Count Number of Homogenous Substrings
Level
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.
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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; } }
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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
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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; } };
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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 }
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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; }
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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 } }
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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; } }