检查日期和月份连接的数字是否可用于形成给定的年份

给定三个字符串D 、 M和Y代表一个日期。任务是检查日期和月份的连接是否产生与年份相同的数字。

例子：

Input: D = 20, M = 12, Y = 2001
Output: Yes
Explanation: Below are the operations performed:
D + M = 2012, Set1 = 0, 1, 2                      
Y = 2001, Set2 = 0, 1, 2
Set1 = Set2, Therefore the answer is Yes. 

Input: D = 26, M = 07, Y = 2001
Output: No

方法：给定的问题可以通过使用散列来解决。请按照以下步骤解决给定的问题。

• 声明两个无序映射，例如s1和s2 ，它们将存储(char, boolean)对。
• 创建将存储D + M的字符串total_concat 。
• 遍历total_concat和Y并将字符分别存储在s1和s2中。
• 运行一个 for 循环并在 set1 中插入所有数量的total_concat ，在 set2 中插入 Y。
• 比较两张地图s1和s2 ：
  • 如果s1 = s2 ，则返回Yes 。
  • 否则返回No 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to return whether the
// date is special or not
bool check(string D, string M, string Y)
{
 
    unordered_map s1;
    unordered_map s2;
 
    // Concatenate date and month
    string total_concat = D + M;
 
    // Insert the elements in set
    for (int i = 0; i < 4; i++) {
        s1[total_concat[i]] = true;
        s2[Y[i]] = true;
    }
 
    // Return 0 if size of both set
    // are unequal
    if (s1.size() != s2.size())
        return 0;
 
    // Return 1 if both set contains
    // same set of numbers otherwise
    // return 0
    else
        return (s1 == s2) ? 1 : 0;
}
 
// Driver Code
int main()
{
    string D = "20";
    string M = "12";
    string Y = "2001";
 
    if (check(D, M, Y))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java implementation for the above approach
import java.util.*;
 
public class GFG
{
    // Function to return whether the
    // date is special or not
    static boolean check(String D, String M, String Y)
    {
 
        HashMap s1 = new HashMap<>();
        HashMap s2 = new HashMap<>();
 
        // Concatenate date and month
        String total_concat = D + M;
 
        // Insert the elements in set
        for (int i = 0; i < 4; i++) {
            s1.put(total_concat.charAt(i), true);
            s2.put(Y.charAt(i), true);
        }
 
        // Return 0 if size of both set
        // are unequal
        if (s1.size() != s2.size())
            return false;
 
        // Return 1 if both set contains
        // same set of numbers otherwise
        // return 0
        else
            return s1.equals(s2);
    }
   
    // Driver Code
    public static void main(String[] args) {
         
        String D = "20";
        String M = "12";
        String Y = "2001";
  
        if (check(D, M, Y))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by Samim Hossain Mondal

Python3

# python program for the above approach
 
# Function to return whether the
# date is special or not
def check(D, M, Y):
 
    s1 = {}
    s2 = {}
 
    # Concatenate date and month
    total_concat = D + M
 
    # Insert the elements in set
    for i in range(0, 4):
        s1[total_concat[i]] = True
        s2[Y[i]] = True
 
    # Return 0 if size of both set
    # are unequal
    if (len(s1) != len(s2)):
        return 0
 
    # Return 1 if both set contains
    # same set of numbers otherwise
    # return 0
    else:
      return s1 == s2
 
# Driver Code
if __name__ == "__main__":
 
    D = "20"
    M = "12"
    Y = "2001"
 
    if (check(D, M, Y)):
        print("Yes")
    else:
        print("No")
 
    # This code is contributed by rakeshsahni

C#

// C# implementation for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
 
public class GFG
{
   
  // Function to return whether the
  // date is special or not
  static bool check(string D, string M, string Y)
  {
 
    Dictionary s1
      = new Dictionary();
    Dictionary s2
      = new Dictionary();
 
    // Concatenate date and month
    string total_concat = D + M;
 
    // Insert the elements in set
    for (int i = 0; i < 4; i++) {
      s1[total_concat[i]] = true;
      s2[Y[i]] = true;
    }
 
    // Return 0 if size of both set
    // are unequal
    if (s1.Count != s2.Count)
      return false;
 
    // Return 1 if both set contains
    // same set of numbers otherwise
    // return 0
    else {
 
      return s1.Count == s2.Count
        && !s1.Except(s2).Any();
    }
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
 
    string D = "20";
    string M = "12";
    string Y = "2001";
 
    if (check(D, M, Y))
      Console.WriteLine("Yes");
    else
      Console.WriteLine("No");
  }
}
 
// This code is contributed by ukasp.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to return whether the
        // date is special or not
        function check(D, M, Y) {
 
            let s1 = new Map();
            let s2 = new Map();
 
            // Concatenate date and month
            let total_concat = D + M;
 
            // Insert the elements in set
            for (let i = 0; i < 4; i++) {
                s1.set(total_concat.charAt(i), true);
                s2.set(Y.charAt(i), true);
            }
 
            // Return 0 if size of both set
            // are unequal
            if (s1.size != s2.size)
                return 0;
 
            // Return 1 if both set contains
            // same set of numbers otherwise
            // return 0
            else
                return (s1.keys == s2.keys) ? 1 : 0;
        }
 
        // Driver Code
        let D = "20";
        let M = "12";
        let Y = "2001";
 
        if (check(D, M, Y))
            document.write("Yes");
        else
            document.write("No");
 
    // This code is contributed by Potta Lokesh
    

输出

Yes

时间复杂度： O(Y)，其中 Y 是年份的大小。

辅助空间： O(1)