Sync LeetCode submission Runtime - 3 ms (18.40%), Memory - 17.7 MB (41.60%)

Author: jimit105

2509-minimize-xor/README.md

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+<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>

2509-minimize-xor/solution.py

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+# Approach 2: Building the Answer
+
+# Time: O(log n)
+# Space: O(1)
+
+class Solution:
+    def minimizeXor(self, num1: int, num2: int) -> int:
+        result = 0
+
+        target_set_bits_count = bin(num2).count('1')
+        set_bits_count = 0
+        current_bit = 31 # Start from most significant bit
+
+        while set_bits_count < target_set_bits_count:
+            # If the current bit of num1 is set or we must set all remaining bits in result
+            if self._is_set(num1, current_bit) or (target_set_bits_count - set_bits_count > current_bit):
+                result = self._set_bit(result, current_bit)
+                set_bits_count += 1
+            current_bit -= 1
+
+        return result
+
+
+    # Helper function to check if the given bit position in result is set (1).
+    def _is_set(self, x, bit):
+        return (x & (1 << bit)) != 0
+
+    # Helper function to set the given bit position in result to 1.
+    def _set_bit(self, x, bit):
+        return x | (1 << bit)
+