<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:
    # To arrive at cnt1 small xor, try to turn off some bits from num1
    # If there are still left bits to set, try to set them from the least significant bit
        cnt1, cnt2 = bin(num1).count('1'), bin(num2).count('1')
        if cnt1 == cnt2:
            return num1
        res = num1
        for i in range(32):
            if cnt1 > cnt2 and (1 << i) & num1 > 0:
                res ^= 1 << i # pick x equals the bigger cnt2 bits of num1, so that we can XOR the bigger cnt2 bits of num1, and keep the smaller cnt1 - cnt2 bits left to make result small.
                cnt1 -= 1
            if cnt1 < cnt2 and (1 << i) & num1 == 0: # In this case, there are still left bits to set, try to set them from the least significant bit until x have cnt2 bits

                res ^= 1 << i
                cnt1 += 1
        return res

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