<p>You are given two <strong>0-indexed </strong>integer<strong> </strong>permutations <code>A</code> and <code>B</code> of length <code>n</code>.</p>

<p>A <strong>prefix common array</strong> of <code>A</code> and <code>B</code> is an array <code>C</code> such that <code>C[i]</code> is equal to the count of numbers that are present at or before the index <code>i</code> in both <code>A</code> and <code>B</code>.</p>

<p>Return <em>the <strong>prefix common array</strong> of </em><code>A</code><em> and </em><code>B</code>.</p>

<p>A sequence of <code>n</code> integers is called a&nbsp;<strong>permutation</strong> if it contains all integers from <code>1</code> to <code>n</code> exactly once.</p>

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

<pre>
<strong>Input:</strong> A = [1,3,2,4], B = [3,1,2,4]
<strong>Output:</strong> [0,2,3,4]
<strong>Explanation:</strong> At i = 0: no number is common, so C[0] = 0.
At i = 1: 1 and 3 are common in A and B, so C[1] = 2.
At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.
At i = 3: 1, 2, 3, and 4 are common in A and B, so C[3] = 4.
</pre>

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

<pre>
<strong>Input:</strong> A = [2,3,1], B = [3,1,2]
<strong>Output:</strong> [0,1,3]
<strong>Explanation:</strong> At i = 0: no number is common, so C[0] = 0.
At i = 1: only 3 is common in A and B, so C[1] = 1.
At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.
</pre>

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

<ul>
	<li><code>1 &lt;= A.length == B.length == n &lt;= 50</code></li>
	<li><code>1 &lt;= A[i], B[i] &lt;= n</code></li>
	<li><code>It is guaranteed that A and B are both a permutation of n integers.</code></li>
</ul>

class Solution:
    def findThePrefixCommonArray(self, A: List[int], B: List[int]) -> List[int]:
        l=[]
        for i in range(len(A)):
            l1=[]
            l2=[]
            for j in range(i+1):
                l1.append(A[j])
                l2.append(B[j])
            s1=set(l1)
            s2=set(l2)
            I=s1&s2
            l.append(len(I))
        return l
        

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