Day 5: Program to check if a number is power of 4

Author: mandliya

README.md

@@ -6,8 +6,8 @@
 
 | Current Status|     Stats     |
 | :------------: | :----------: |
-| Total Problems | 140 |
-| Current Streak | 4 days |
+| Total Problems | 141 |
+| Current Streak | 5 days |
 | Longest Streak | 91 ( August 17, 2015 - November 15, 2015 ) |
 
 </center>
@@ -72,6 +72,7 @@ Include contains single header implementation of data structures and some algori
 |Louise and Richard play a game. They have a counter set to N. Louise gets the first turn and the turns alternate thereafter. In the game, they perform the following operations: <ul><li>If N is not a power of 2, reduce the counter by the largest power of 2 less than N.</li></ul><ul><li>If N is a power of 2, reduce the counter by half of N.</li></ul> The resultant value is the new N which is again used for subsequent operations.The game ends when the counter reduces to 1, i.e., N == 1, and the last person to make a valid move wins. <ul><li> Given N, your task is to find the winner of the game. If they set counter to 1, Richard wins, because its Louise' turn and she cannot make a move.</li></ul><ul><li>Input Format : -The first line contains an integer T, the number of testcases. T lines follow. Each line contains N, the initial number set in the counter.</ul></li> |[counter_game.cpp](bit_manipulation/counter_game.cpp)|
 |Determine if two integers are of opposite signs.|[check_opposite_signs.cpp](bit_manipulation/check_opposite_signs.cpp)|
 |Swap two bits at position p and q of a given integer.| [swapBits.cpp](bit_manipulation/swapBits.cpp)|
+|Check if a number is power of 4. | [check_if_power_of_4.cpp](bit_manipulation/check_if_power_of_4.cpp)|
 
 
 ### Cracking the coding interview problems

bit_manipulation/check_if_power_of_4.cpp

@@ -0,0 +1,42 @@
+/*
+ * Check if a given number is power of 4 or not.
+ * 
+ * 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)
+ * 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
+ * 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.
+ */
+
+ #include <iostream>
+
+ bool isPowerOf4 (unsigned int n)
+ {
+     // The number should be power of 2, and should have set bit at even position
+     //
+     return (n && !(n & (n-1)) && !(n & 0xAAAAAAAA));
+ }
+
+ int main()
+ {  
+     unsigned int n;
+     std::cout << "Enter a number:";
+     std::cin >> n;
+     if (isPowerOf4(n))
+     {
+         std::cout << n << " is power of 4.\n";
+     }
+     else
+     {
+         std::cout << n << " is not a power of 4.\n";
+     }
+ }