Count of total Heads and Tails after N flips in a coin

14 Jun, 2021
Given character C and an integer N representing N number of coins in C position where C can be either head or tail. We can flip the coins N times, wherein the ith round the player will flip the face of all the coins whose number is less than or equal to i. The task is to determine the total number of head and tail after flipping N possible times.
Examples:
Input: C = ‘H’, N = 5 Output: Head = 2, Tail = 3 Explanation: H means initially all the coins are facing in head direction, N means the total number of coins. So initially for i = 0, we have: H H H H H After the first round that is i = 1: T H H H H After the second round that is i = 2: H T H H H After the third round that is i = 3: T H T H H After the fourth round that is i = 4: H T H T H After the fifth round that is i = 5: T H T H T Hence the total count of the head is 2 and tail is 3.
Input: C = ‘T’, N = 7 Output: Head = 4, Tail = 3 Explanation: After all the possible flips the head and tail count is 4 and 3.  
Approach: To solve the problem mentioned above we have to follow the steps given below:  
In the question above if we observe then there is a pattern that if initially, all the coins are facing towards head direction then the total number of heads after N rounds will be the floor value of (n / 2) and tails will be the cell value of (n / 2).
Otherwise, if all the coins are facing towards tail direction then a total number of tails after N rounds will be a floor value of (n / 2) and heads will be ceil value of (n / 2).
Below is the implementation: 
C++
Java
Python3
C#
Javascript
// C++ program to count total heads// and tails after N flips in a coin#include<bits/stdc++.h>using namespace std; // Function to find count of head and tailpair<int, int> count_ht(char s, int N){         // Check if initially all the    // coins are facing towards head    pair<int, int>p;    if(s == 'H')    {        p.first = floor(N / 2.0);        p.second = ceil(N / 2.0);    }         // Check if initially all the coins    // are facing towards tail    else if(s == 'T')    {        p.first = ceil(N / 2.0);        p.second = floor(N / 2.0);    }         return p;} // Driver codeint main(){    char C = 'H';    int N = 5;         pair<int, int> p = count_ht(C, N);         cout << "Head = " << (p.first) << "\n";    cout << "Tail = " << (p.second) << "\n";} // This code is contributed by virusbuddah_
// Java program to count// total heads and tails// after N flips in a coinimport javafx.util.Pair;public class Main{// Function to find count of head and tail public static Pair <Integer,                    Integer> count_ht(char s,                                      int N){                // Check if initially all the   // coins are facing towards head   Pair <Integer,        Integer> p = new Pair <Integer,                               Integer> (0, 0);  if(s == 'H')  {     p = new Pair <Integer,                   Integer> ((int)Math.floor(N / 2.0),                             (int)Math.ceil(N / 2.0));  }   // Check if initially all the coins   // are facing towards tail   else if(s == 'T')  {     p = new Pair <Integer,                   Integer> ((int)Math.ceil(N / 2.0),                             (int)Math.floor(N / 2.0));  }  return p;} public static void main(String[] args){  char C = 'H';  int N = 5;  Pair <Integer,        Integer> p = count_ht(C, N);  System.out.println("Head = " + p.getKey());  System.out.println("Tail = " + p.getValue());}} // This code is contributed by divyeshrabadiya07
# Python3 program to Count total heads# and tails after N flips in a coin # Function to find count of head and tailimport mathdef count_ht( s, N ):         # Check if initially all the    # coins are facing towards head    if s == "H":        h = math.floor( N / 2 )        t = math.ceil( N / 2 )             # Check if initially all the coins    # are facing towards tail    elif s == "T":        h = math.ceil( N / 2 )        t = math.floor( N / 2 )             return [h, t] # Driver Codeif __name__ == "__main__":    C = "H"    N = 5    l = count_ht(C, n)    print("Head = ", l[0])    print("Tail = ", l[1])
// C# program to count total heads// and tails after N flips in a coinusing System; class GFG{     // Function to find count of head and tail public static Tuple<int, int> count_ht(char s,                                       int N){          // Check if initially all the     // coins are facing towards head     Tuple<int, int> p = Tuple.Create(0, 0);         if (s == 'H')    {        p = Tuple.Create((int)Math.Floor(N / 2.0),                         (int)Math.Ceiling(N / 2.0));    }         // Check if initially all the coins     // are facing towards tail     else if (s == 'T')    {        p = Tuple.Create((int)Math.Ceiling(N / 2.0),                         (int)Math.Floor(N / 2.0));    }    return p;} // Driver Codestatic void Main(){    char C = 'H';    int N = 5;    Tuple<int, int> p = count_ht(C, N);         Console.WriteLine("Head = " + p.Item1);    Console.WriteLine("Tail = " + p.Item2);}} // This code is contributed by divyesh072019
<script> // JavaScript program to count total heads// and tails after N flips in a coin // Function to find count of head and tailfunction count_ht(s, N){         // Check if initially all the    // coins are facing towards head    var p = [0,0];    if(s == 'H')    {        p[0] = Math.floor(N / 2.0);        p[1] = Math.ceil(N / 2.0);    }         // Check if initially all the coins    // are facing towards tail    else if(s == 'T')    {        p[0] = Math.ceil(N / 2.0);        p[1] = Math.floor(N / 2.0);    }         return p;} // Driver codevar C = 'H';var N = 5; var p = count_ht(C, N); document.write( "Head = " + (p[0]) + "<br>");document.write( "Tail = " + (p[1]) + "<br>"); </script>
Head =  2
Tail =  3
 
Time complexity: O(1)
virusbuddha
divyeshrabadiya07
divyesh072019
rutvik_56
Mathematical
Pattern Searching
Mathematical
Pattern Searching
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