一个数与原数之和等于另一个给定数的排列

给定两个整数A和C，任务是检查是否存在数量的排列，使得数量之和其排列是等于C。

例子：

Input: A = 133, C = 446
Output: Yes
Explanation: One of the permutation of A is 313. Therefore, sum = 133 + 313 = 446, which is equal to C.

Input: A = 200, C = 201
Output: No

朴素方法：解决问题的最简单方法是生成数字A 的所有排列并将其与A的原始值相加。现在，检查它们的总和是否等于C。

时间复杂度： O((log ₁₀ A)！)
辅助空间： O(1)

高效方法：为了优化上述方法，这个想法是减去A的从C值，并检查是否存在A的置换，它等于C和A或不之间的差异。请按照以下步骤解决问题：

• 从C 中减去A并检查C是否小于0 。如果发现是真的，则打印“否”。
• 否则，请执行以下步骤：
  • 将A转换为其等效的字符串表示形式并将其存储在一个变量中，例如S 。
  • 将C转换为其等效的字符串表示形式并将其存储在一个变量中，例如K 。
  • 对新生成的字符串S和K 进行排序。
  • 如果S等于K ，则打印“是”以及排列。否则，打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if there
// exists a permutation of a number
// A whose sum with A results to C
void checkPermutation(int a, int c)
{
    // Subtract a from c
    c -= a;
 
    // Check if c is less than 0
    if (c <= 0) {
 
        // If true, print "No"
        cout << "No";
        return;
    }
 
    // Otherwise, convert a to its
    // equivalent string
    string s = to_string(a);
 
    // convert c to its
    // equivalent string
    string k = to_string(c);
 
    // Sort string s
    sort(s.begin(), s.end());
 
    // Sort string k
    sort(k.begin(), k.end());
 
    // If both strings are equal,
    // then print "Yes"
    if (k == s) {
        cout << "Yes\n"
             << c;
    }
 
    // Otherwise, print "No"
    else {
        cout << "No";
    }
}
 
// Driver Code
int main()
{
    int A = 133, C = 446;
    checkPermutation(A, C);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to check if there
// exists a permutation of a number
// A whose sum with A results to C
static void checkPermutation(int a, int c)
{
     
    // Subtract a from c
    c -= a;
 
    // Check if c is less than 0
    if (c <= 0)
    {
         
        // If true, print "No"
        System.out.println("No");
        return;
    }
 
    // Otherwise, convert a to its
    // equivalent string
    String s = sortString(Integer.toString(a));
 
    // Convert c to its
    // equivalent string
    String k = sortString(Integer.toString(c));
 
    // If both strings are equal,
    // then print "Yes"
    if (k.equals(s))
    {
        System.out.println("Yes");
        System.out.println(c);
    }
 
    // Otherwise, print "No"
    else
    {
        System.out.println("No");
    }
}
 
// Method to sort a string alphabetically
public static String sortString(String inputString)
{
     
    // Convert input string to char array
    char tempArray[] = inputString.toCharArray();
       
    // sort tempArray
    Arrays.sort(tempArray);
       
    // Return new sorted string
    return new String(tempArray);
}
 
// Driver code
public static void main(String[] args)
{
    int A = 133, C = 446;
     
    checkPermutation(A, C);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
 
# Function to check if there
# exists a permutation of a number
# A whose sum with A results to C
def checkPermutation(a, c):
    # Subtract a from c
    c -= a
 
    # Check if c is less than 0
    if (c <= 0):
        # If true, print "No"
        print("No")
        return
 
    # Otherwise, convert a to its
    # equivalent string
    s = str(a)
 
    # convert c to its
    # equivalent string
    k = str(c)
 
    res = ''.join(sorted(s))
 
    # Sort string s
    s = str(res)
 
    # Sort string k
    res = ''.join(sorted(k))
    k = str(res)
 
    # If both strings are equal,
    # then print "Yes"
    if (k == s):
        print("Yes")
        print(c)
 
    # Otherwise, print "No"
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
    A = 133
    C = 446
    checkPermutation(A, C)
     
    # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;
class GFG
{
   
  // Function to check if there
  // exists a permutation of a number
  // A whose sum with A results to C
  static void checkPermutation(int a, int c)
  {
     
    // Subtract a from c
    c -= a;
 
    // Check if c is less than 0
    if (c <= 0) {
 
      // If true, print "No"
      Console.Write("No");
      return;
    }
 
    // Otherwise, convert a to its
    // equivalent string
    string s = a.ToString();
 
    // convert c to its
    // equivalent string
    string k = c.ToString();
 
    // Sort string s
    char[] arr = s.ToCharArray();
    Array.Sort(arr);
 
    // Sort string k
    char[] brr = k.ToCharArray();
    Array.Sort(brr);
 
    // If both strings are equal,
    // then print "Yes"
    if (String.Join("", brr) == String.Join("", arr)) {
      Console.Write("Yes\n" + c);
    }
 
    // Otherwise, print "No"
    else {
      Console.WriteLine("No");
    }
  }
 
  // Driver Code
  public static void Main()
  {
    int A = 133, C = 446;
    checkPermutation(A, C);
  }
}
 
// This code is contributed by ukasp.

Javascript

 
// javascript program for the above approach
 
// Function to check if there
// exists a permutation of a number
// A whose sum with A results to C
function checkPermutation(a, c)
{
    // Subtract a from c
    c -= a;
 
    // Check if c is less than 0
    if (c <= 0) {
 
        // If true, print "No"
        print("No");
        return;
    }
 
    // Otherwise, convert a to its
    // equivalent string
    var s = a.toString();
 
    // convert c to its
    // equivalent string
    var k = c.toString();
 
    // Sort string s
    s= s.split('').sort((a, b) => a.localeCompare(b)).join('');
   // sort(s);
 
    // Sort string k
    k = k.split('').sort((a, b) => a.localeCompare(b)).join('');
    //sort(k);
 
    // If both strings are equal,
    // then print "Yes"
    if (k == s) {
        document.write("Yes" + "<br>" + c);
    }
 
    // Otherwise, print "No"
    else {
        document.write("No");
    }
}
 
// Driver Code
    var A = 133, C = 446;
    checkPermutation(A, C);
 

输出：

Yes
313

时间复杂度： O((log ₁₀ A)*log (log ₁₀ A))
辅助空间： O(log ₁₀ A)