Find the last player to be able to flip a character in a Binary String
Last Updated :
26 Apr, 2021
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 not
void 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 Code
int main()
{
string S = "1100011";
int N = S.length();
findWinner(S, N);
return 0;
}
// Java program for the above approach
import 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 program for the above approach
import sys
# Function to check if player A wins
# the game or not
def 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# program to implement
// the above approach
using 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 Code
public static void Main(String[] args)
{
string S = "1100011";
int N = S.Length;
findWinner(S, N);
}
}
// This code is contributed by splevel62.
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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