Question

1513. Number of Substrings With Only 1s

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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

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)

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;
  }
}

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;
  }
};

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
}

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;
}