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:

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 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



Output:

player 2

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

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