Introduction

In this blog post, we will explore how to solve the LeetCode problem "1573" 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, you can split s into 3 non-empty strings s1, s2, and s3 where s1 + s2 + s3 = s. Return the number of ways s can be split such that the number of ones is the same in s1, s2, and s3. Since the answer may be too large, return it modulo 109 + 7. Example 1: Input: s = "10101" Output: 4 Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters '1'. "1|010|1" "1|01|01" "10|10|1" "10|1|01" Example 2: Input: s = "1001" Output: 0 Example 3: Input: s = "0000" Output: 3 Explanation: There are three ways to split s in 3 parts. "0|0|00" "0|00|0" "00|0|0" Constraints: 3 <= s.length <= 105 s[i] is either '0' or '1'.

Explanation

Here's the breakdown of the solution:

• Count Ones: Calculate the total number of ones in the string. If the count isn't divisible by 3, no valid splits exist.
• Find Split Points: If the total ones are divisible by 3, iterate through the string to find the indices where the cumulative count of ones equals total_ones / 3 and 2 * total_ones / 3. These are potential split points for s1 and s2.

• Calculate Combinations: Count the possible split positions for s1 and s2 independently. The product of these counts gives the number of valid ways to split the string. Handle the special case where the total number of ones is zero carefully.

• Runtime Complexity: O(n), where n is the length of the string.

• Storage Complexity: O(1).

Code

    def numWays(s: str) -> int:
    n = len(s)
    ones = s.count('1')
    MOD = 10**9 + 7

    if ones % 3 != 0:
        return 0

    if ones == 0:
        return ((n - 1) * (n - 2) // 2) % MOD

    ones_per_part = ones // 3
    count1 = 0
    first_split_ways = 0
    second_split_ways = 0

    ones_count = 0
    for i in range(n):
        if s[i] == '1':
            ones_count += 1
        if ones_count == ones_per_part:
            first_split_ways += 1
        elif ones_count == 2 * ones_per_part:
            second_split_ways += 1

    return (first_split_ways * second_split_ways) % MOD