<p>Given a binary string <code>s</code>, you can split <code>s</code> into 3 <strong>non-empty</strong> strings <code>s1</code>, <code>s2</code>, and <code>s3</code> where <code>s1 + s2 + s3 = s</code>.</p>

<p>Return the number of ways <code>s</code> can be split such that the number of ones is the same in <code>s1</code>, <code>s2</code>, and <code>s3</code>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;10101&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are four ways to split s in 3 parts where each part contain the same number of letters &#39;1&#39;.
&quot;1|010|1&quot;
&quot;1|01|01&quot;
&quot;10|10|1&quot;
&quot;10|1|01&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;1001&quot;
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;0000&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are three ways to split s in 3 parts.
&quot;0|0|00&quot;
&quot;0|00|0&quot;
&quot;00|0|0&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution:
    def numWays(self, s: str) -> int:
        
        N_ones = 0
        for c in s:
            N_ones += 1 if c == "1" else 0
        if N_ones % 3 != 0:
            return 0
        N_ones = N_ones // 3
        if N_ones == 0:
            N = len(s)-2
            return (1 + N) * N // 2 % (10**9 + 7)

        def dfs(i=0, cur_group=0, cur_count=0):
            if cur_group == 2:
                return 1
            ni = i
            start_i = -1
            while ni < len(s) and cur_count <= N_ones:
                if s[ni] == "1":
                    cur_count += 1
                if start_i == -1 and cur_count == N_ones:
                    start_i = ni
                ni += 1
            ni -= 1 # move back one step as we add one to ni before the check

            return (ni-start_i) * dfs(ni, cur_group+1)
                
        return dfs() % (10**9 + 7)
18700