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1513. Number of Substrings With Only 1s

Description

Given a binary string s, return the number of substrings with all characters 1's. Since the answer may be too large, return it modulo 109 + 7.

 

Example 1:

Input: s = "0110111"
Output: 9
Explanation: There are 9 substring in total with only 1's characters.
"1" -> 5 times.
"11" -> 3 times.
"111" -> 1 time.

Example 2:

Input: s = "101"
Output: 2
Explanation: Substring "1" is shown 2 times in s.

Example 3:

Input: s = "111111"
Output: 21
Explanation: Each substring contains only 1's characters.

 

Constraints:

  • 1 <= s.length <= 105
  • s[i] is either '0' or '1'.

Solutions

Solution 1

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class Solution:
    def numSub(self, s: str) -> int:
        ans = cnt = 0
        for c in s:
            if c == "1":
                cnt += 1
            else:
                cnt = 0
            ans += cnt
        return ans % (10**9 + 7)

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class Solution {
    public int numSub(String s) {
        final int mod = (int) 1e9 + 7;
        int ans = 0, cnt = 0;
        for (int i = 0; i < s.length(); ++i) {
            cnt = s.charAt(i) == '1' ? cnt + 1 : 0;
            ans = (ans + cnt) % mod;
        }
        return ans;
    }
}

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class Solution {
public:
    int numSub(string s) {
        int ans = 0, cnt = 0;
        const int mod = 1e9 + 7;
        for (char& c : s) {
            cnt = c == '1' ? cnt + 1 : 0;
            ans = (ans + cnt) % mod;
        }
        return ans;
    }
};

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func numSub(s string) (ans int) {
    const mod = 1e9 + 7
    cnt := 0
    for _, c := range s {
        if c == '1' {
            cnt++
        } else {
            cnt = 0
        }
        ans = (ans + cnt) % mod
    }
    return
}

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function numSub(s: string): number {
    const mod = 10 ** 9 + 7;
    let ans = 0;
    let cnt = 0;
    for (const c of s) {
        cnt = c == '1' ? cnt + 1 : 0;
        ans = (ans + cnt) % mod;
    }
    return ans;
}

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