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1759. Count Number of Homogenous Substrings

Description

Given a string s, return the number of homogenous substrings of s. Since the answer may be too large, return it modulo 109 + 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 <= 105
  • s consists of lowercase letters.

Solutions

Solution 1

Thinking

A homogenous substring lies inside a run of equal characters. A run of length \(cnt\) contributes \(cnt(cnt+1)/2\). \(n\le 10^5\), so scan by runs.

Two pointers bound each run, add the triangular number, and reduce modulo \(10^9+7\).

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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

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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 {
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;
    }
}

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int countHomogenous(char* s) {
    int MOD = 1e9 + 7;
    int ans = 0;
    for (int i = 0, j = 0; s[j]; j++) {
        if (s[i] != s[j]) {
            i = j;
        }
        ans = (ans + j - i + 1) % MOD;
    }
    return ans;
}

Solution 2

Thinking

Solution 1 sums each run at once. A linear scan can instead grow a run length while neighbours match, reset to \(1\) otherwise, and add the current length each step.

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class Solution:
    def countHomogenous(self, s: str) -> int:
        mod = 10**9 + 7
        ans = cnt = 1
        for a, b in pairwise(s):
            cnt = cnt + 1 if a == b else 1
            ans = (ans + cnt) % mod
        return ans

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

    public int countHomogenous(String s) {
        int n = s.length();
        int ans = 1, cnt = 1;
        for (int i = 1; i < n; ++i) {
            cnt = s.charAt(i) == s.charAt(i - 1) ? cnt + 1 : 1;
            ans = (ans + cnt) % MOD;
        }
        return ans;
    }
}

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

    int countHomogenous(string s) {
        int n = s.size();
        int ans = 1, cnt = 1;
        for (int i = 1; i < n; ++i) {
            cnt = s[i] == s[i - 1] ? cnt + 1 : 1;
            ans = (ans + cnt) % mod;
        }
        return ans;
    }
};

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func countHomogenous(s string) int {
    n := len(s)
    const mod int = 1e9 + 7
    ans, cnt := 1, 1
    for i := 1; i < n; i++ {
        if s[i] == s[i-1] {
            cnt++
        } else {
            cnt = 1
        }
        ans = (ans + cnt) % mod
    }
    return ans
}

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