找到以二进制字符串形式给出分数的游戏的获胜者 |设置 2

给定一个表示排球比赛得分的二进制字符串str 。任务是根据以下条件找到比赛的获胜者：

在排球比赛中，两队互相比赛，先得到15分的球队获胜，除非两队都达到14分。
如果两队都达到 14 分，则保持领先 2 分的球队获胜。

在给定的二进制字符串中， 0表示GEEK 的团队失去一分， 1表示GEEK 的团队赢得一分。任务是找出 GEEK 的团队是赢了还是输了比赛。

例子：

Input: str = “01011111111110110101”
Output: GEEK’S won
Explanation: GEEK wins with score of 15-5

Input: str = “010101010101010101010101010100”
Output: GEEK’s lost
Explanation: The opponent wins with score of 16-14

编程需要懂一点英语

朴素方法：本题的Set-1中提到了朴素方法。

高效方法：无论时间线如何，最后得分的玩家将获胜。原因如下：

The game ends with someone winning the set and meeting the required conditions to win the game.

If the game ends by someone reaching the 15 points first then he will be one to win the last set.
If the game is won by someone maintaining two points lead after reaching 15-14 state, then also the winner will be the one to win the last set maintaining a two point lead.

编程需要懂一点英语

下面是上述方法的实现。

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to find the winner
// from given timeline
string findTheWinner(string str, int N)
{
    // If last point scored is 1 then
    // GEEK is winner
    if (str[N - 1] == '1') {
        return "GEEK's won";
    }
 
    // Else GEEK lost
    return "GEEK's lost";
}
 
// Driver Code
int main()
{
    // Input score timeline
    string str1 = "01011111111110110101";
    int N1 = str1.size();
    string str2 = "010101010101010101010101010100";
    int N2 = str2.size();
 
    // Print the winner
    cout << findTheWinner(str1, N1) << endl;
    cout << findTheWinner(str2, N2) << endl;
    return 0;
}

Java

// C# implementation of the above approach
class GFG {
 
  // Function to find the winner
  // from given timeline
  static String findTheWinner(String str, int N)
  {
 
    // If last point scored is 1 then
    // GEEK is winner
    if (str.charAt(N - 1) == '1')
    {
      return "GEEK's won";
    }
 
    // Else GEEK lost
    return "GEEK's lost";
  }
 
  // Driver Code
  public static void main(String args[])
  {
 
    // Input score timeline
    String str1 = "01011111111110110101";
    int N1 = str1.length();
    String str2 = "010101010101010101010101010100";
    int N2 = str2.length();
 
    // Print the winner
    System.out.println(findTheWinner(str1, N1));
    System.out.println(findTheWinner(str2, N2));
  }
}
 
// This code is contributed by Saurabh Jaiswal

Python3

# Python code for the above approach
 
# Function to find the winner
# from given timeline
def findTheWinner(str, N):
 
    # If last point scored is 1 then
    # GEEK is winner
    if (str[N - 1] == '1'):
        return "GEEK's won"
 
    # Else GEEK lost
    return "GEEK's lost"
 
# Driver Code
 
# Input score timeline
str1 = "01011111111110110101"
N1 = len(str1)
str2 = "010101010101010101010101010100"
N2 = len(str2)
 
# Print the winner
print(findTheWinner(str1, N1))
print(findTheWinner(str2, N2))
 
# This code is contributed by Saurabh Jaiswal

C#

// C# implementation of the above approach
using System;
 
class GFG{
 
// Function to find the winner
// from given timeline
static string findTheWinner(string str, int N)
{
     
    // If last point scored is 1 then
    // GEEK is winner
    if (str[N - 1] == '1')
    {
        return "GEEK's won";
    }
 
    // Else GEEK lost
    return "GEEK's lost";
}
 
// Driver Code
public static void Main()
{
     
    // Input score timeline
    string str1 = "01011111111110110101";
    int N1 = str1.Length;
    string str2 = "010101010101010101010101010100";
    int N2 = str2.Length;
 
    // Print the winner
    Console.WriteLine(findTheWinner(str1, N1));
    Console.WriteLine(findTheWinner(str2, N2));
}
}
 
// This code is contributed by ukasp

Javascript

输出

GEEK's won
GEEK's lost

时间复杂度： O(1)
辅助空间： O(1)