# Solving Leetcode Interviews in Seconds with AI: Number of Substrings With Only 1s


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1513" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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'.  

	# Explanation
	Here's an efficient solution to count substrings with all 1's in a binary string:

*   **Count Consecutive 1s:** Iterate through the string, keeping track of the length of consecutive sequences of '1's.
*   **Calculate Substrings:** For each sequence of '1's of length *n*, the number of substrings with all 1's is *n*( *n* + 1) / 2.
*   **Modulo Arithmetic:** Apply the modulo operator to prevent integer overflow.

*   **Runtime Complexity:** O(n), where n is the length of the string. **Storage Complexity:** O(1).

	
	# Code
	```python
	def numSub(s: str) -> int:
    """
    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.
    """
    MOD = 10**9 + 7
    count = 0
    result = 0
    for char in s:
        if char == '1':
            count += 1
        else:
            result = (result + (count * (count + 1) // 2)) % MOD
            count = 0
    result = (result + (count * (count + 1) // 2)) % MOD
    return result
	```