Update check_if_power_of_4.cpp

fixed decimal to binary conversions in notes, and improved formatting. Also a suggestion on a slightly more straightforward solution.

bit_manipulation/check_if_power_of_4.cpp

@@ -4,15 +4,15 @@
  * Approach: If a value is power of 4, it has to be power of 2 and
  * it will have set bits at even position (as the number is power of 2, it will have
  * only one set bit.) For example:
- * 4 (10)
- * 16 (10000)
- * 64 (1000000)
+ * 4 (0100)
+ * 16 (0001 0000)
+ * 64 (0100 0000)
  * If a number is power of 2 then it would have only one set bit and that set
  * bit would be reset by the expression (n & (n-1)), thus if (n & (n-1)) == 0,
  * means n is power of 2.
  * Now, since we have one set bit and if it is at even position it would be
  * power of 4, to check if set bit is at even position, we should AND it with
- * expression 10101010101010101010101010101010 i.e. 0xAAAAAAAA. Notice we have
+ * expression 1010 1010 1010 1010 1010 1010 1010 1010 i.e. 0xAAAAAAAA. Notice we have
  * set bits at all odd positions (it is 0 indexed), thus if expression
  * (n & 0xAAAAAAAA) == 0, then we have that set bit at even position.
  */
@@ -24,6 +24,8 @@
      // The number should be power of 2, and should have set bit at even position
      //
      return (n && !(n & (n-1)) && !(n & 0xAAAAAAAA));
+     //note an alternative that avoids the last negation could be 
+     //return (n && !(n & (n-1)) && (n & 0x55555555));
  }
 
  int main()
@@ -33,10 +35,10 @@
      std::cin >> n;
      if (isPowerOf4(n))
      {
-         std::cout << n << " is power of 4.\n";
+         std::cout << n << " is a power of 4.\n";
      }
      else
      {
          std::cout << n << " is not a power of 4.\n";
      }
- }
+ }