给定分别为N和M的两个字符串数组arr []和brr [] ，任务是当两个玩家按照以下规则最佳玩游戏时，找到游戏的赢家：
- 玩家1开始游戏。
- 播放器1删除该数组ARR一个字符串，如果[]它尚未从阵列BRR []删除。
- 播放机2删除该数组BRR一个字符串，如果[]它尚未从阵列ARR []删除。
- 无法从数组中删除字符串的玩家将失去游戏。
例子：
Input: arr[] = { “geeks”, “geek” }, brr[] = { “geeks”, “geeksforgeeks” }
Output: Player 1
Explanation:
Turn 1: Player 1 removed “geeks” from arr[].
Turn 2: Player 2 removed “geeksforgeeks” from brr[]
Turn 3: Player 1 removed “geek” from brr[].
Now, player 2 cannot remove any string.
Therefore, the required output is Player 1.
Input: arr[] = { “a”, “b” }, brr[] = { “a”, “b” }
Output: Player 2
Explanation:
Turn 1: Player 1 removed “a” from arr[].
Turn 2: Player 2 removed “b” from brr[].
Therefore, the required output is Player 2
方法：基于以下事实的想法：两个数组中的公共字符串只能从一个数组中删除。请按照以下步骤解决问题：
- 如果来自两个数组的公共字符串的计数均为奇数，则从玩家brr []中删除一个字符串，因为玩家1开始游戏，并且玩家1删除了第一个公共字符串。
- 如果通过从两个数组中删除公共字符串， arr []中的字符串数大于brr []中的字符串，则打印“ Player 1” 。
- 否则，打印“ Player 2” 。
下面是上述方法的实现：
C++
// C++ Program for the above approach
#include
using namespace std;
// Function to find last player to be
// able to remove a string from one array
// which has not been removed from the other array
void lastPlayer(int n, int m, vector arr,
vector brr)
{
// Stores common strings
// from both the array
set common;
for (int i = 0; i < arr.size(); i++)
{
for (int j = 0; j < brr.size(); j++)
{
if (arr[i] == brr[j])
{
// add common elements
common.insert(arr[i]);
break;
}
}
}
// Removing common strings from arr[]
set a;
bool flag;
for (int i = 0; i < arr.size(); i++)
{
flag = false;
for (auto value : common)
{
if (value == arr[i])
{
// add common elements
flag = true;
break;
}
}
if (flag)
a.insert(arr[i]);
}
// Removing common elements from B
set b;
for (int i = 0; i < brr.size(); i++)
{
flag = false;
for (auto value : common)
{
if (value == brr[i])
{
// add common elements
flag = true;
break;
}
}
if (flag)
b.insert(brr[i]);
}
// Stores strings in brr[] which
// is not common in arr[]
int LenBrr = b.size();
if ((common.size()) % 2 == 1)
{
// Update LenBrr
LenBrr -= 1;
}
if (a.size() > LenBrr)
{
cout<<("Player 1")< arr{ "geeks", "geek" };
// Set of strings for player B
vector brr{ "geeks", "geeksforgeeks" };
int n = arr.size();
int m = brr.size();
lastPlayer(n, m, arr, brr);
}
// This code is contributed by SURENDRA_GANGWAR. Java
// Java Program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to find last player to be
// able to remove a string from one array
// which has not been removed from the other array
static void lastPlayer(int n, int m, String[] arr,
String[] brr)
{
// Stores common strings
// from both the array
Set common = new HashSet<>();
for (int i = 0; i < arr.length; i++)
{
for (int j = 0; j < brr.length; j++)
{
if (arr[i] == brr[j])
{
// add common elements
common.add(arr[i]);
break;
}
}
}
// Removing common strings from arr[]
Set a = new HashSet<>();
boolean flag;
for (int i = 0; i < arr.length; i++)
{
flag = false;
for (String value : common)
{
if (value == arr[i])
{
// add common elements
flag = true;
break;
}
}
if (flag)
a.add(arr[i]);
}
// Removing common elements from B
Set b = new HashSet<>();
for (int i = 0; i < brr.length; i++)
{
flag = false;
for (String value : common)
{
if (value == brr[i])
{
// add common elements
flag = true;
break;
}
}
if (flag)
b.add(brr[i]);
}
// Stores strings in brr[] which
// is not common in arr[]
int LenBrr = b.size();
if ((common.size()) % 2 == 1)
{
// Update LenBrr
LenBrr -= 1;
}
if (a.size() > LenBrr)
{
System.out.print("Player 1");
}
else
{
System.out.print("Player 2");
}
}
// Driver Code
public static void main(String[] args)
{
// Set of strings for player A
String[] arr = { "geeks", "geek" };
// Set of strings for player B
String[] brr = { "geeks", "geeksforgeeks" };
int n = arr.length;
int m = brr.length;
lastPlayer(n, m, arr, brr);
}
}
// This code is contributed by Dharanendra L V. Python
# Python Program for the above approach
# Function to find last player to be
# able to remove a string from one array
# which has not been removed from the other array
def lastPlayer(n, m, arr, brr):
# Stores common strings
# from both the array
common = list(set(arr) & set(brr))
# Removing common strings from arr[]
a = list(set(arr) ^ set(common))
# Removing common elements from B
b = list(set(brr) ^ set(common))
# Stores strings in brr[] which
# is not common in arr[]
LenBrr = len(b)
if len(common) % 2 == 1:
# Update LenBrr
LenBrr -= 1
if len(a) > LenBrr:
print("Player 1")
else:
print("Player 2")
# Driver Code
if __name__ == '__main__':
# Set of strings for player A
arr = ["geeks", "geek"]
# Set of strings for player B
brr = ["geeks", "geeksforgeeks"]
n = len(arr)
m = len(brr)
lastPlayer(n, m, arr, brr)C#
// C# Program for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to find last player to be
// able to remove a string from one array
// which has not been removed from the other array
static void lastPlayer(int n, int m, String[] arr,
String[] brr)
{
// Stores common strings
// from both the array
HashSet common = new HashSet();
for (int i = 0; i < arr.Length; i++)
{
for (int j = 0; j < brr.Length; j++)
{
if (arr[i] == brr[j])
{
// add common elements
common.Add(arr[i]);
break;
}
}
}
// Removing common strings from []arr
HashSet a = new HashSet();
bool flag;
for (int i = 0; i < arr.Length; i++)
{
flag = false;
foreach (String value in common)
{
if (value == arr[i])
{
// add common elements
flag = true;
break;
}
}
if (flag)
a.Add(arr[i]);
}
// Removing common elements from B
HashSet b = new HashSet();
for (int i = 0; i < brr.Length; i++)
{
flag = false;
foreach (String value in common)
{
if (value == brr[i])
{
// add common elements
flag = true;
break;
}
}
if (flag)
b.Add(brr[i]);
}
// Stores strings in brr[] which
// is not common in []arr
int LenBrr = b.Count;
if ((common.Count) % 2 == 1)
{
// Update LenBrr
LenBrr -= 1;
}
if (a.Count > LenBrr)
{
Console.Write("Player 1");
}
else
{
Console.Write("Player 2");
}
}
// Driver Code
public static void Main(String[] args)
{
// Set of strings for player A
String[] arr = { "geeks", "geek" };
// Set of strings for player B
String[] brr = { "geeks", "geeksforgeeks" };
int n = arr.Length;
int m = brr.Length;
lastPlayer(n, m, arr, brr);
}
}
// This code is contributed by 29AjayKumar 输出：
Player 1时间复杂度： O(N + M)
辅助空间： O(N + M)