<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:
        l = []
        for c in s:
            l.append(int(c))

        n = len(l)
        if(sum(l)==0):
            return ((n-1)*(n-2)//2)%(10**9 + 7)

        if(not(sum(l)%3==0)):
            return 0

        pre = [0 for i in range(n)]
        if(l[0]==1):
            pre[0] = 1

        for i in range(1, n):
            if(l[i]==1):
                pre[i] = pre[i-1] + 1

            else:
                pre[i] = pre[i-1]

        ends = {}
        for i in range(n):
            if(pre[i] not in ends):
                ends[pre[i]] = [i, i]

            else:
                ends[pre[i]][1] = i

        one_third = sum(l)*1//3
        two_third = sum(l)*2//3
        # print(pre)
        print(ends)
        return ((ends[one_third][1]-ends[one_third][0]+1)*(ends[two_third][1]-ends[two_third][0]+1))%(10**9 + 7)
20100