Given a binary string S of length N, the task is to find the winner of the game if two players A and B plays optimally as per the following rules:
- Player A always starts the game.
- In a player’s first turn, he can move to any index (1-based indexing) consisting of ‘0’ and make it ‘1’.
- For the subsequent turns, if any player is at index i, then he can move to one of it’s adjacent indice, if it contains 0, and convert it to ‘1’ after moving.
- If any player is unable to move to any position during his turn, then the player loses the game.
The task is to find the winner of the game.
Examples:
Input: S = “1100011”
Output: Player A
Explanation:
The indices 3, 4 and 5 consists of 0s and indices 1, 2, 6 and 7 consists of 1s.
A starts by flipping the character at index 4..
B flips either the index 3 or 5.
A is now left with only one index adjacent to 4, which B did not pick. After A flips the character at that index, B does not have any character to flip. Since B has no moves, A wins.
Hence, print “Player A”.Input: S = “11111”
Output: Player B
Approach: The idea is to store the length of all the substrings consisting only of 0s from the given array arr[] in another array, say V[]. Now, the following cases arise:
- If the size of V is 0: In this case, the array does not contain any 0s. Therefore, Player A can’t make any move and loses the game. Hence, print Player B.
- If the size of V is 1: In this case, there is 1 substring consisting only of 0s, say of length L. If the value of L is odd, then Player A wins the game. Otherwise, Player B wins the game.
- In all other cases: Store the length of the largest and the second-largest consecutive segment of 0s in first and second respectively. Player A can win the game if and only if the value of first is odd and (first + 1)/2 > second. Otherwise, Player B wins the game.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if player A wins// the game or notvoid findWinner(string a, int n){ // Stores size of the groups of 0s vector<int> v; // Stores size of the group of 0s int c = 0; // Traverse the array for (int i = 0; i < n; i++) { // Increment c by 1 if a[i] is 0 if (a[i] == '0') { c++; } // Otherwise, push the size // in array and reset c to 0 else { if (c != 0) v.push_back(c); c = 0; } } if (c != 0) v.push_back(c); // If there is no substring of // odd length consisting only of 0s if (v.size() == 0) { cout << "Player B"; return; } // If there is only 1 substring of // odd length consisting only of 0s if (v.size() == 1) { if (v[0] & 1) cout << "Player A"; // Otherwise else cout << "Player B"; return; } // Stores the size of the largest // and second largest substrings of 0s int first = INT_MIN; int second = INT_MIN; // Traverse the array v[] for (int i = 0; i < v.size(); i++) { // If current element is greater // than first, then update both // first and second if (a[i] > first) { second = first; first = a[i]; } // If arr[i] is in between // first and second, then // update second else if (a[i] > second && a[i] != first) second = a[i]; } // If the condition is satisfied if ((first & 1) && (first + 1) / 2 > second) cout << "Player A"; else cout << "Player B";}// Driver Codeint main(){ string S = "1100011"; int N = S.length(); findWinner(S, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;public class GFG{ // Function to check if player A wins // the game or not static void findWinner(String a, int n) { // Stores size of the groups of 0s Vector<Integer> v = new Vector<Integer>(); // Stores size of the group of 0s int c = 0; // Traverse the array for (int i = 0; i < n; i++) { // Increment c by 1 if a[i] is 0 if (a.charAt(i) == '0') { c++; } // Otherwise, push the size // in array and reset c to 0 else { if (c != 0) v.add(c); c = 0; } } if (c != 0) v.add(c); // If there is no substring of // odd length consisting only of 0s if (v.size() == 0) { System.out.print("Player B"); return; } // If there is only 1 substring of // odd length consisting only of 0s if (v.size() == 1) { if ((v.get(0) & 1) != 0) System.out.print("Player A"); // Otherwise else System.out.print("Player B"); return; } // Stores the size of the largest // and second largest substrings of 0s int first = Integer.MIN_VALUE; int second = Integer.MIN_VALUE; // Traverse the array v[] for (int i = 0; i < v.size(); i++) { // If current element is greater // than first, then update both // first and second if (a.charAt(i) > first) { second = first; first = a.charAt(i); } // If arr[i] is in between // first and second, then // update second else if (a.charAt(i) > second && a.charAt(i) != first) second = a.charAt(i); } // If the condition is satisfied if ((first & 1) != 0 && (first + 1) / 2 > second) System.out.print("Player A"); else System.out.print("Player B"); } // Driver code public static void main(String[] args) { String S = "1100011"; int N = S.length(); findWinner(S, N); }}// This code is contributed by divyeshrabadiya07. |
Python3
# Python3 program for the above approachimport sys# Function to check if player A wins# the game or notdef findWinner(a, n) : # Stores size of the groups of 0s v = [] # Stores size of the group of 0s c = 0 # Traverse the array for i in range(0, n) : # Increment c by 1 if a[i] is 0 if (a[i] == '0') : c += 1 # Otherwise, push the size # in array and reset c to 0 else : if (c != 0) : v.append(c) c = 0 if (c != 0) : v.append(c) # If there is no substring of # odd length consisting only of 0s if (len(v) == 0) : print("Player B", end = "") return # If there is only 1 substring of # odd length consisting only of 0s if (len(v) == 1) : if ((v[0] & 1) != 0) : print("Player A", end = "") # Otherwise else : print("Player B", end = "") return # Stores the size of the largest # and second largest substrings of 0s first = sys.minsize second = sys.minsize # Traverse the array v[] for i in range(len(v)) : # If current element is greater # than first, then update both # first and second if (a[i] > first) : second = first first = a[i] # If arr[i] is in between # first and second, then # update second elif (a[i] > second and a[i] != first) : second = a[i] # If the condition is satisfied if (((first & 1) != 0) and (first + 1) // 2 > second) : print("Player A", end = "") else : print("Player B", end = "")S = "1100011"N = len(S)findWinner(S, N)# This code is contributed by divyesh072019. |
C#
// C# program to implement// the above approachusing System;using System.Collections.Generic; using System.Linq;class GFG{ // Function to check if player A wins // the game or not static void findWinner(string a, int n) { // Stores size of the groups of 0s List<int> v = new List<int>(); // Stores size of the group of 0s int c = 0; // Traverse the array for (int i = 0; i < n; i++) { // Increment c by 1 if a[i] is 0 if (a[i] == '0') { c++; } // Otherwise, push the size // in array and reset c to 0 else { if (c != 0) v.Add(c); c = 0; } } if (c != 0) v.Add(c); // If there is no substring of // odd length consisting only of 0s if (v.Count == 0) { Console.Write("Player B"); return; } // If there is only 1 substring of // odd length consisting only of 0s if (v.Count == 1) { if ((v[0] & 1) != 0) Console.Write("Player A"); // Otherwise else Console.Write("Player B"); return; } // Stores the size of the largest // and second largest substrings of 0s int first = Int32.MinValue; int second = Int32.MinValue; // Traverse the array v[] for (int i = 0; i < v.Count; i++) { // If current element is greater // than first, then update both // first and second if (a[i] > first) { second = first; first = a[i]; } // If arr[i] is in between // first and second, then // update second else if (a[i] > second && a[i] != first) second = a[i]; } // If the condition is satisfied if ((first & 1) != 0 && (first + 1) / 2 > second) Console.Write("Player A"); else Console.Write("Player B"); }// Driver Codepublic static void Main(String[] args){ string S = "1100011"; int N = S.Length; findWinner(S, N);}}// This code is contributed by splevel62. |
Javascript
<script> // Javascript program to implement the above approach // Function to check if player A wins // the game or not function findWinner(a, n) { // Stores size of the groups of 0s let v = []; // Stores size of the group of 0s let c = 0; // Traverse the array for (let i = 0; i < n; i++) { // Increment c by 1 if a[i] is 0 if (a[i] == '0') { c++; } // Otherwise, push the size // in array and reset c to 0 else { if (c != 0) v.push(c); c = 0; } } if (c != 0) v.push(c); // If there is no substring of // odd length consisting only of 0s if (v.length == 0) { document.write("Player B"); return; } // If there is only 1 substring of // odd length consisting only of 0s if (v.length == 1) { if ((v[0] & 1) != 0) document.write("Player A"); // Otherwise else document.write("Player B"); return; } // Stores the size of the largest // and second largest substrings of 0s let first = Number.MIN_VALUE; let second = Number.MIN_VALUE; // Traverse the array v[] for (let i = 0; i < v.length; i++) { // If current element is greater // than first, then update both // first and second if (a[i] > first) { second = first; first = a[i]; } // If arr[i] is in between // first and second, then // update second else if (a[i] > second && a[i] != first) second = a[i]; } // If the condition is satisfied if ((first & 1) != 0 && parseInt((first + 1) / 2, 10) > second) document.write("Player A"); else document.write("Player B"); } let S = "1100011"; let N = S.length; findWinner(S, N); // This code is contributed by suresh07.</script> |
Output:
Player A
Time Complexity: O(N)
Auxiliary Space: O(N)
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