Day-52 Heap Sort 2

README.md

@@ -4,9 +4,9 @@
 
 | Current Status|     Stats     |
 | :------------: | :----------: |
-| Total Problems | 72 |
-| Current Streak | 52 |
-| Longest Streak | 52 ( August 17, 2015 - October 7, 2015 ) |
+| Total Problems | 73 |
+| Current Streak | 53 |
+| Longest Streak | 53 ( August 17, 2015 - October 8, 2015 ) |
 
 </center>
 
@@ -64,6 +64,21 @@ Include contains single header implementation of data structures and some algori
 | How many bit flip operation would require to convert number A to B. | [countNumberOfBitFlips.cpp](bit_manipulation/countNumberOfBitFlips.cpp)|
 | Given a number x and two positions (from right side) in binary representation of x, write a function that swaps n right bits at given two positions and returns the result. It is also given that the two sets of bits do not overlap.|[swapSetOfBits.cpp](bit_manipulation/swapSetOfBits.cpp)|
 | Add two numbers without using any arithmetic operators | [addition_without_operators.cpp](bit_manipulation/addition_without_operators.cpp)
+| 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.
+
+If N is not a power of 2, reduce the counter by the largest power of 2 less than N.
+If N is a power of 2, reduce the counter by half of N.
+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.
+
+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.
+
+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.
+|[counter_game.cpp](bit_manipulation/counter_game.cpp)| 
 
 ### Cracking the coding interview problems
 | Problem | Solution |

bit_manipulation/counter_game.cpp

@@ -0,0 +1,82 @@
+
+/*
+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.
+
+If N is not a power of 2, reduce the counter by the largest power of 2 less than N.
+If N is a power of 2, reduce the counter by half of N.
+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.
+
+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.
+
+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.
+
+*/
+
+
+
+#include <cmath>
+#include <cstdio>
+#include <vector>
+#include <iostream>
+#include <algorithm>
+using namespace std;
+
+bool isPowerOf2(unsigned long long int x) {
+    return ( x && ((x &(x-1)) == 0));
+}
+
+unsigned long long int lowerPowerof2( unsigned long long int x) {
+     if (x == 0) {
+        return 0;
+    }
+    x--;
+    x |= (x >> 1);
+    x |= (x >> 2);
+    x |= (x >> 4);
+    x |= (x >> 8);
+    x |= (x >> 16);
+    x |= (x >> 32);
+    return x - (x >> 1);
+}
+
+std::string winner( bool win ) {
+    if (win) {
+        return std::string("Louise");
+    } else {
+        return std::string("Richard");
+    }
+}
+
+int main() {
+
+    int T;
+    unsigned long long int N;
+    cin >> T;
+
+    while(T) {
+        cin >> N;
+        if (N == 1) {
+            std::cout << "Richard\n";
+            continue;
+        }
+        bool win = false;
+        while(N > 1) {
+            if (isPowerOf2(N)) {
+                N = N/2;
+            } else {
+                N = N - lowerPowerof2(N);
+            }
+            win = !win;
+        }
+        std::cout << winner(win) << std::endl;
+        --T;
+    }
+
+    return 0;
+}
+
+