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, 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

    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