Find the last player to be able to flip a character in a Binary String
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++
#include <bits/stdc++.h>
using namespace std;
void findWinner(string a, int n)
{
vector< int > v;
int c = 0;
for ( int i = 0; i < n; i++) {
if (a[i] == '0' ) {
c++;
}
else {
if (c != 0)
v.push_back(c);
c = 0;
}
}
if (c != 0)
v.push_back(c);
if (v.size() == 0) {
cout << "Player B" ;
return ;
}
if (v.size() == 1) {
if (v[0] & 1)
cout << "Player A" ;
else
cout << "Player B" ;
return ;
}
int first = INT_MIN;
int second = INT_MIN;
for ( int i = 0; i < v.size(); i++) {
if (a[i] > first) {
second = first;
first = a[i];
}
else if (a[i] > second
&& a[i] != first)
second = a[i];
}
if ((first & 1)
&& (first + 1) / 2 > second)
cout << "Player A" ;
else
cout << "Player B" ;
}
int main()
{
string S = "1100011" ;
int N = S.length();
findWinner(S, N);
return 0;
}
|
Java
import java.util.*;
public class GFG
{
static void findWinner(String a, int n)
{
Vector<Integer> v = new Vector<Integer>();
int c = 0 ;
for ( int i = 0 ; i < n; i++)
{
if (a.charAt(i) == '0' )
{
c++;
}
else
{
if (c != 0 )
v.add(c);
c = 0 ;
}
}
if (c != 0 )
v.add(c);
if (v.size() == 0 )
{
System.out.print( "Player B" );
return ;
}
if (v.size() == 1 )
{
if ((v.get( 0 ) & 1 ) != 0 )
System.out.print( "Player A" );
else
System.out.print( "Player B" );
return ;
}
int first = Integer.MIN_VALUE;
int second = Integer.MIN_VALUE;
for ( int i = 0 ; i < v.size(); i++)
{
if (a.charAt(i) > first) {
second = first;
first = a.charAt(i);
}
else if (a.charAt(i) > second
&& a.charAt(i) != first)
second = a.charAt(i);
}
if ((first & 1 ) != 0
&& (first + 1 ) / 2 > second)
System.out.print( "Player A" );
else
System.out.print( "Player B" );
}
public static void main(String[] args)
{
String S = "1100011" ;
int N = S.length();
findWinner(S, N);
}
}
|
Python3
import sys
def findWinner(a, n) :
v = []
c = 0
for i in range ( 0 , n) :
if (a[i] = = '0' ) :
c + = 1
else :
if (c ! = 0 ) :
v.append(c)
c = 0
if (c ! = 0 ) :
v.append(c)
if ( len (v) = = 0 ) :
print ( "Player B" , end = "")
return
if ( len (v) = = 1 ) :
if ((v[ 0 ] & 1 ) ! = 0 ) :
print ( "Player A" , end = "")
else :
print ( "Player B" , end = "")
return
first = sys.minsize
second = sys.minsize
for i in range ( len (v)) :
if (a[i] > first) :
second = first
first = a[i]
elif (a[i] > second and a[i] ! = first) :
second = a[i]
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)
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
class GFG{
static void findWinner( string a, int n)
{
List< int > v = new List< int >();
int c = 0;
for ( int i = 0; i < n; i++)
{
if (a[i] == '0' )
{
c++;
}
else
{
if (c != 0)
v.Add(c);
c = 0;
}
}
if (c != 0)
v.Add(c);
if (v.Count == 0)
{
Console.Write( "Player B" );
return ;
}
if (v.Count == 1)
{
if ((v[0] & 1) != 0)
Console.Write( "Player A" );
else
Console.Write( "Player B" );
return ;
}
int first = Int32.MinValue;
int second = Int32.MinValue;
for ( int i = 0; i < v.Count; i++)
{
if (a[i] > first) {
second = first;
first = a[i];
}
else if (a[i] > second
&& a[i] != first)
second = a[i];
}
if ((first & 1) != 0
&& (first + 1) / 2 > second)
Console.Write( "Player A" );
else
Console.Write( "Player B" );
}
public static void Main(String[] args)
{
string S = "1100011" ;
int N = S.Length;
findWinner(S, N);
}
}
|
Javascript
<script>
function findWinner(a, n)
{
let v = [];
let c = 0;
for (let i = 0; i < n; i++)
{
if (a[i] == '0' )
{
c++;
}
else
{
if (c != 0)
v.push(c);
c = 0;
}
}
if (c != 0)
v.push(c);
if (v.length == 0)
{
document.write( "Player B" );
return ;
}
if (v.length == 1)
{
if ((v[0] & 1) != 0)
document.write( "Player A" );
else
document.write( "Player B" );
return ;
}
let first = Number.MIN_VALUE;
let second = Number.MIN_VALUE;
for (let i = 0; i < v.length; i++)
{
if (a[i] > first) {
second = first;
first = a[i];
}
else if (a[i] > second && a[i] != first)
second = a[i];
}
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);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Last Updated :
26 Apr, 2021
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