<p>Given 3 positives numbers <code>a</code>, <code>b</code> and <code>c</code>. Return the minimum flips required in some bits of <code>a</code> and <code>b</code> to make (&nbsp;<code>a</code> OR <code>b</code> == <code>c</code>&nbsp;). (bitwise OR operation).<br />
Flip operation&nbsp;consists of change&nbsp;<strong>any</strong>&nbsp;single bit 1 to 0 or change the bit 0 to 1&nbsp;in their binary representation.</p>

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

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/01/06/sample_3_1676.png" style="width: 260px; height: 87px;" /></p>

<pre>
<strong>Input:</strong> a = 2, b = 6, c = 5
<strong>Output:</strong> 3
<strong>Explanation: </strong>After flips a = 1 , b = 4 , c = 5 such that (<code>a</code> OR <code>b</code> == <code>c</code>)</pre>

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

<pre>
<strong>Input:</strong> a = 4, b = 2, c = 7
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> a = 1, b = 2, c = 3
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= a &lt;= 10^9</code></li>
	<li><code>1 &lt;= b&nbsp;&lt;= 10^9</code></li>
	<li><code>1 &lt;= c&nbsp;&lt;= 10^9</code></li>
</ul>
class Solution:
    def minFlips(self, a: int, b: int, c: int) -> int:
        ans = 0
        while a or b or c:
            c_digit = c % 2
            a_digit = a % 2
            b_digit = b % 2
            if c_digit == 0:
                if a_digit == 1:
                    ans += 1
                if b_digit == 1:
                    ans += 1
            else:
                if a_digit == 0 and b_digit == 0:
                    ans += 1

            a //= 2
            b //= 2
            c //= 2
        return ans
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