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Find the player with least 0s after emptying a Binary String by removing non-empty substrings

  • Last Updated : 03 May, 2021

Given a binary string S, the task is to determine the winner of the game when two players play a game optimally in alternate turns with the given string, as per the following conditions:

  • Player 1 always starts first.
  • In each turn, a player removes a non-empty substring from the given string.
  • After the given string is emptied, the player having the minimum count of 0s will win the game. If both players have an equal count of 0s, then print “Tie”.

Examples:

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Input: S = “00011” 
Output: Player 1 
Explanation: Substrings can be chosen as follows: 
Turn 1: Player 1 removes the substring S[4…5]. Therefore, Player 1 contains “11”. 
Turn 2: Player 2 removes the substring S[0…0]. Therefore, Player 2 contains “0”. 
Turn 3: Player 1 removes the substring S[0…0]. Therefore, Player 1 contains “110”. 
Turn 4: Player 2 removes the substring S[0…0]. Therefore, Player 2 contains “00”. 
Therefore, Player 1 wins the game.



Input: S = “0110011”
Output: Player 2

Approach: The problem can be solved based on the following observations:

  • If the count of 0s in the string is an even number then player 1 and player 2 choose the substring “0” in each turn and no player will win this game.
  • Otherwise, store the count of consecutive 1s in an array and apply the game of nim rule on the array.
  • Nim-Sum: The cumulative XOR value of the number of coins/stones in each pile/heaps(here consecutive 1s) at any point of the game is called Nim-Sum at that point.

Follow the steps below to solve the problem:

  • Initialize a variable, say cntZero, to store the count of 0s in the string.
  • Initialize a variable, say cntConOne, to store the count of consecutive 1s in the string.
  • Initialize a variable, say nimSum, to store the Nim-Sum of consecutive 1s of the given string.
  • Traverse the array and calculate the count of 0s and nimSum.
  • Finally, check if the value of cntZero is an even number or not. If found to be true, then print Tie.
  • Otherwise, check if the value of nimSum is greater than 0 or not. If found to be true, then print Player 1.
  • Otherwise, print player 2.

Below is the implementation of the above approach: 

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the player
// who wins the game
void FindwinnerOfGame(string& S)
{
 
    // Stores total count
    // of 0s in the string
    int cntZero = 0;
 
    // Stores count of
    // consecutive 1s
    int cntConOne = 0;
 
    // Stores Nim-Sum on count
    // of consecutive 1s
    int nimSum = 0;
 
    // Stores length
    // of the string
    int N = S.length();
 
    // Traverse the string
    for (int i = 0; i < N; i++) {
 
        // If the current
        // character is 1
        if (S[i] == '1') {
 
            // Update cntConOne
            cntConOne += 1;
        }
        else {
 
            // Update nimSum
            nimSum ^= cntConOne;
 
            // Update cntConOne
            cntConOne = 0;
 
            // Update cntZero
            cntZero++;
        }
    }
 
    // Update nimSum
    nimSum ^= cntConOne;
 
    // If countZero is
    // an even number
    if (cntZero % 2 == 0) {
        cout << "Tie";
    }
 
    // nimSum is not 0
    else if (nimSum) {
        cout << "player 1";
    }
 
    // If nimSum is zero
    else {
        cout << "player 2";
    }
}
 
// Driver Code
int main()
{
 
    string S = "0110011";
    FindwinnerOfGame(S);
}

Java




// Java program to implement
// the above approach
 
// Function to find the player
// who wins the game
class GFG {
    public static void FindwinnerOfGame(String S)
    {
 
        // Stores total count
        // of 0s in the string
        int cntZero = 0;
 
        // Stores count of
        // consecutive 1s
        int cntConOne = 0;
 
        // Stores Nim-Sum on count
        // of consecutive 1s
        int nimSum = 0;
 
        // Stores length
        // of the string
        int N = S.length();
 
        // Traverse the string
        for (int i = 0; i < N; i++) {
 
            // If the current
            // character is 1
            if (S.charAt(i) == '1') {
 
                // Update cntConOne
                cntConOne += 1;
            }
            else {
 
                // Update nimSum
                nimSum ^= cntConOne;
 
                // Update cntConOne
                cntConOne = 0;
 
                // Update cntZero
                cntZero++;
            }
        }
 
        // Update nimSum
        nimSum ^= cntConOne;
 
        // If countZero is
        // an even number
        if (cntZero % 2 == 0) {
            System.out.print("Tie");
        }
 
        // nimSum is not 0
        else if (nimSum != 0) {
            System.out.print("player 1");
        }
 
        // If nimSum is zero
        else {
            System.out.print("player 2");
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "0110011";
        FindwinnerOfGame(S);
    }
}
 
// This code is contributed by grand_master.

Python3




# Python 3 program to implement
# the above approach
 
# Function to find the player
# who wins the game
def FindwinnerOfGame(S):
   
    # Stores total count
    # of 0s in the string
    cntZero = 0
 
    # Stores count of
    # consecutive 1s
    cntConOne = 0
 
    # Stores Nim-Sum on count
    # of consecutive 1s
    nimSum = 0
 
    # Stores length
    # of the string
    N = len(S)
 
    # Traverse the string
    for i in range(N):
       
        # If the current
        # character is 1
        if (S[i] == '1'):
           
            # Update cntConOne
            cntConOne += 1
        else:
           
            # Update nimSum
            nimSum ^= cntConOne
 
            # Update cntConOne
            cntConOne = 0
 
            # Update cntZero
            cntZero += 1
 
    # Update nimSum
    nimSum ^= cntConOne
 
    # If countZero is
    # an even number
    if (cntZero % 2 == 0):
        print("Tie")
 
    # nimSum is not 0
    elif(nimSum):
        print("player 1")
 
    # If nimSum is zero
    else:
        print("player 2")
 
# Driver Code
if __name__ == '__main__':
    S = "0110011"
    FindwinnerOfGame(S)
 
    # this code is contributed by SURENDRA_GANGWAR.

C#




// C# program to implement
// the above approach
using System;
 
// Function to find the player
// who wins the game
class GFG {
  public static void FindwinnerOfGame(string S)
  {
 
    // Stores total count
    // of 0s in the string
    int cntZero = 0;
 
    // Stores count of
    // consecutive 1s
    int cntConOne = 0;
 
    // Stores Nim-Sum on count
    // of consecutive 1s
    int nimSum = 0;
 
    // Stores length
    // of the string
    int N = S.Length;
 
    // Traverse the string
    for (int i = 0; i < N; i++) {
 
      // If the current
      // character is 1
      if (S[i] == '1') {
 
        // Update cntConOne
        cntConOne += 1;
      }
      else {
 
        // Update nimSum
        nimSum ^= cntConOne;
 
        // Update cntConOne
        cntConOne = 0;
 
        // Update cntZero
        cntZero++;
      }
    }
 
    // Update nimSum
    nimSum ^= cntConOne;
 
    // If countZero is
    // an even number
    if (cntZero % 2 == 0) {
      Console.Write("Tie");
    }
 
    // nimSum is not 0
    else if (nimSum != 0) {
      Console.Write("player 1");
    }
 
    // If nimSum is zero
    else {
      Console.Write("player 2");
    }
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    string S = "0110011";
    FindwinnerOfGame(S);
  }
}
 
// This code is contributed by ukasp.

Javascript




<script>
// javascript program of the above approach
 
    // Function to find the player
// who wins the game
    function FindwinnerOfGame(S)
    {
 
        // Stores total count
        // of 0s in the string
        let cntZero = 0;
 
        // Stores count of
        // consecutive 1s
        let cntConOne = 0;
 
        // Stores Nim-Sum on count
        // of consecutive 1s
        let nimSum = 0;
 
        // Stores length
        // of the string
        let N = S.length;
 
        // Traverse the string
        for (let i = 0; i < N; i++) {
 
            // If the current
            // character is 1
            if (S[i] == '1') {
 
                // Update cntConOne
                cntConOne += 1;
            }
            else {
 
                // Update nimSum
                nimSum ^= cntConOne;
 
                // Update cntConOne
                cntConOne = 0;
 
                // Update cntZero
                cntZero++;
            }
        }
 
        // Update nimSum
        nimSum ^= cntConOne;
 
        // If countZero is
        // an even number
        if (cntZero % 2 == 0) {
            document.write("Tie");
        }
 
        // nimSum is not 0
        else if (nimSum != 0) {
            document.write("player 1");
        }
 
        // If nimSum is zero
        else {
            document.write("player 2");
        }
    }
 
    // Driver Code
     
           let S = "0110011";
        FindwinnerOfGame(S);
 
</script>
Output: 
player 2

 

Time Complexity: O(N), where N is the length of the string
Auxiliary Space: O(1)




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