Day-31: Math & Bit manipulation problems

Author: mandliya

Diff for: README.md

@@ -4,9 +4,9 @@
 
 | Current Status|     Stats     |
 | :------------: | :----------: |
-| Total Problems | 46 |
-| Current Streak | 30 |
-| Longest Streak | 30 ( August 17, 2015 - September 15, 2015 ) |
+| Total Problems | 48 |
+| Current Streak | 31 |
+| Longest Streak | 31 ( August 17, 2015 - September 16, 2015 ) |
 
 </center>
 
@@ -55,6 +55,8 @@ Include contains single header implementation of data structures and some algori
 | Small function to determine position of right most set bit in a given integer.| [right_most_set_bit.cpp](bit_manipulation/right_most_set_bit.cpp)| 
 |Given a vector of numbers, only one number occurs odd number of times, find the number.| [find_odd_one_out.cpp](bit_manipulation/find_odd_one_out.cpp)|
 | Given two integers, determine if their sum would be interger overflow.| [integerOverflow.cpp](bit_manipulation/integerOverflow.cpp)|
+| How many bit flip operation would require to convert number A to B. | [countNumberOfBitFlips.cpp](bit_manipulation/countNumberOfBitFlips.cpp)|
+
 ### Cracking the coding interview problems
 | Problem | Solution |
 | :------------ | :----------: |
@@ -85,6 +87,7 @@ Include contains single header implementation of data structures and some algori
 | Problem | Solution |
 | :------------ | :----------: |
 |  Print all the permutations of a string. Example: Permutations of ABC are ABC, ACB, BCA, BAC, CAB, CBA | [string_permutations.cpp] (math_problems/string_permutations.cpp) |
+| Euclidean algorithm to find greatest common divisor of two numbers. (Iterative and recursive)|[gcd.cpp](math_problems/gcd.cpp)|
 
 ### Stack Problems
 | Problem | Solution |

Diff for: bit_manipulation/countNumberOfBitFlips.cpp

@@ -0,0 +1,35 @@
+/**
+ *  Given two numbers A and B, count number of flips required to convert number A to B
+ *  Approach : Take XOR of A and B, the result will contain set bits only at positions
+ *  where A differed B. Therefore number of set bits in the result of A ^ B is the number
+ *  of flips required to convert A to B
+ */
+
+#include <iostream>
+
+int countSetBits( int x )
+{
+  int count = 0;
+  while( x ) {
+    ++count;
+    x = x & (x - 1);
+  }
+  return count;
+}
+
+int countBitFlips( int a, int b )
+{
+  return countSetBits(a ^ b);
+}
+
+int main()
+{
+  int x, y;
+  std::cout << "Enter number 1 :";
+  std::cin >> x;
+  std::cout << "Enter number 2 :";
+  std::cin >> y;
+  std::cout << "Bit flips required to convert " << x
+            << " to " << y << " is " << countBitFlips(x, y) << std::endl;
+  return 0;
+}

Diff for: math_problems/gcd.cpp

@@ -0,0 +1,60 @@
+/**
+ *  Given two numbers, determine the greatest common divisior of the two numbers.
+ */
+
+#include <iostream>
+
+
+int gcd1( int a, int b ) {
+    while( b > 0 ) {
+        int temp = a % b;
+        a = b;
+        b = temp;
+    }
+    return a;
+}
+
+int gcd2( int a, int b ) {
+    if ( b == 0 ) return a;
+    if ( a == 0 ) return b;
+    return gcd2(b, a % b);
+}
+
+int gcd3( int a, int b ) {
+    if ( a == 0 ) return b;
+    if ( b == 0 ) return a;
+    while ( a != b ) {
+        if ( a > b ) {
+            a = a-b;
+        } else {
+            b = b-a;
+        }
+    }
+    return a;
+}
+
+int gcd4 ( int a, int b ) {
+    if ( a == 0 ) return b;
+    if ( b == 0 ) return a;
+    if ( a == b ) return a;
+    if ( a > b ) return gcd4(b, a-b);
+    else  return gcd4(a, b-a);
+}
+
+int main()
+{
+    int a, b;
+    std::cout << "Enter number 1 : ";
+    std::cin >> a;
+    std::cout << "Enter number 2 : ";
+    std::cin >> b;
+    std::cout << "GCD of " << a << " and "
+              << b << " is " << gcd1(a,b) << std::endl;
+    std::cout << "GCD of " << a << " and "
+              << b << " is " << gcd2(a,b) << std::endl;
+    std::cout << "GCD of " << a << " and "
+              << b << " is " << gcd3(a,b) << std::endl;
+    std::cout << "GCD of " << a << " and "
+              << b << " is " << gcd4(a,b) << std::endl;
+    return 0;
+}