<p>Given two positive integers <code>num1</code> and <code>num2</code>, find the positive integer <code>x</code> such that:</p>

<ul>
	<li><code>x</code> has the same number of set bits as <code>num2</code>, and</li>
	<li>The value <code>x XOR num1</code> is <strong>minimal</strong>.</li>
</ul>

<p>Note that <code>XOR</code> is the bitwise XOR operation.</p>

<p>Return <em>the integer </em><code>x</code>. The test cases are generated such that <code>x</code> is <strong>uniquely determined</strong>.</p>

<p>The number of <strong>set bits</strong> of an integer is the number of <code>1</code>&#39;s in its binary representation.</p>

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

<pre>
<strong>Input:</strong> num1 = 3, num2 = 5
<strong>Output:</strong> 3
<strong>Explanation:</strong>
The binary representations of num1 and num2 are 0011 and 0101, respectively.
The integer <strong>3</strong> has the same number of set bits as num2, and the value <code>3 XOR 3 = 0</code> is minimal.
</pre>

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

<pre>
<strong>Input:</strong> num1 = 1, num2 = 12
<strong>Output:</strong> 3
<strong>Explanation:</strong>
The binary representations of num1 and num2 are 0001 and 1100, respectively.
The integer <strong>3</strong> has the same number of set bits as num2, and the value <code>3 XOR 1 = 2</code> is minimal.
</pre>

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

<ul>
	<li><code>1 &lt;= num1, num2 &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minimizeXor(self, num1: int, num2: int) -> int:
        def binary(num):
            arr = []
            while(num!=0):
                arr.append(num%2)
                num = num//2
            return arr
        
        b_num1 = binary(num1)
        b_num2 = binary(num2)

        k = sum(b_num2)
        
        x = [0 for i in range(max(len(b_num2), len(b_num1)))]
        for i in range(len(b_num1)-1 ,-1, -1):
            if(k>0 and b_num1[i] == 1):
                x[i] = 1
                k -= 1
            if(k==0):
                break
        for i in range(len(x)):
            if(k>0 and x[i]!= 1):
                x[i] = 1
                k-=1
            if(k==0):
                break
        def binToNum(arr):
            ans = 0
            temp = 1
            for i in range(len(arr)):
                ans += arr[i]*temp
                temp = temp*2
            return ans
        # print(b_num1, b_num2, x)
        return binToNum(x)

        
43