<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> which represents the types of buildings along a street where:</p>

<ul>
	<li><code>s[i] = &#39;0&#39;</code> denotes that the <code>i<sup>th</sup></code> building is an office and</li>
	<li><code>s[i] = &#39;1&#39;</code> denotes that the <code>i<sup>th</sup></code> building is a restaurant.</li>
</ul>

<p>As a city official, you would like to <strong>select</strong> 3 buildings for random inspection. However, to ensure variety, <strong>no two consecutive</strong> buildings out of the <strong>selected</strong> buildings can be of the same type.</p>

<ul>
	<li>For example, given <code>s = &quot;0<u><strong>0</strong></u>1<u><strong>1</strong></u>0<u><strong>1</strong></u>&quot;</code>, we cannot select the <code>1<sup>st</sup></code>, <code>3<sup>rd</sup></code>, and <code>5<sup>th</sup></code> buildings as that would form <code>&quot;0<strong><u>11</u></strong>&quot;</code> which is <strong>not</strong> allowed due to having two consecutive buildings of the same type.</li>
</ul>

<p>Return <em>the <b>number of valid ways</b> to select 3 buildings.</em></p>

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

<pre>
<strong>Input:</strong> s = &quot;001101&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> 
The following sets of indices selected are valid:
- [0,2,4] from &quot;<u><strong>0</strong></u>0<strong><u>1</u></strong>1<strong><u>0</u></strong>1&quot; forms &quot;010&quot;
- [0,3,4] from &quot;<u><strong>0</strong></u>01<u><strong>10</strong></u>1&quot; forms &quot;010&quot;
- [1,2,4] from &quot;0<u><strong>01</strong></u>1<u><strong>0</strong></u>1&quot; forms &quot;010&quot;
- [1,3,4] from &quot;0<u><strong>0</strong></u>1<u><strong>10</strong></u>1&quot; forms &quot;010&quot;
- [2,4,5] from &quot;00<u><strong>1</strong></u>1<u><strong>01</strong></u>&quot; forms &quot;101&quot;
- [3,4,5] from &quot;001<u><strong>101</strong></u>&quot; forms &quot;101&quot;
No other selection is valid. Thus, there are 6 total ways.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;11100&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> It can be shown that there are no valid selections.
</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 numberOfWays(self, s: str) -> int:

        # 因为所有过程不超过3个，只有下面几种中间态
        # dp['01'] 就是目前拿到‘01’的情况有几种
        
        dp= {'0': 0, '1': 0, '01':0, '10':0, '010': 0, '101':0}

        for c in s:
            if c == '1':
                # 记住拿的时候不需要连着拿，所以不需要记录前面这个0 在哪里的，只需要记住目前拿到‘0’ 的次数，那么当前c=1, 那么当下拿到'01'的次数 就多了dp[0]次
                dp['01']+=dp['0']
                dp['101'] += dp['10']
                dp['1'] += 1
            if c== '0':
                dp['0'] += 1
                dp['10'] += dp['1']
                dp['010'] += dp['01']
        
        # 因为必须拿3个，且不能相邻重复，只能这两种情况
        return dp['010'] + dp['101']
        
17226