最长的公共子序列，允许排列

给定两个字符串小写，找到最长的字符串，其排列是给定两个字符串。输出的最长字符串必须排序。

例子：

Input  :  str1 = "pink", str2 = "kite"
Output : "ik" 
The string "ik" is the longest sorted string 
whose one permutation "ik" is subsequence of
"pink" and another permutation "ki" is 
subsequence of "kite". 

Input  : str1 = "working", str2 = "women"
Output : "now"

Input  : str1 = "geeks" , str2 = "cake"
Output : "ek"

Input  : str1 = "aaaa" , str2 = "baba"
Output : "aa"

我们的想法是在计算两个字符串的字符。

1. 计算每个字符串的字符频率，并将它们存储在各自的计数数组中，例如str1的count1 []和str2的count2 []。
2. 现在我们有26个字符的计数数组。因此遍历count1 []，对于任何索引’i’，在结果字符串’result’中将字符(’a’+ i)追加min(count1 [i]，count2 [i])次。
3. 由于我们以升序遍历计数数组，因此最终的字符串字符将按排序顺序。

   C++

   // C++ program to find LCS with permutations allowed
   #include
   using namespace std;
     
   // Function to calculate longest string
   // str1     --> first string
   // str2     --> second string
   // count1[]  --> hash array to calculate frequency
   //              of characters in str1
   // count[2]  --> hash array to calculate frequency
   //              of characters in str2
   // result   --> resultant longest string whose
   // permutations are sub-sequence of given two strings
   void longestString(string str1, string str2)
   {
       int count1[26] = {0}, count2[26]= {0};
     
       // calculate frequency  of characters
       for (int i=0; i

   ---

   Java

   //Java program to find LCS with permutations allowed
     
   class GFG {
     
   // Function to calculate longest String 
   // str1     --> first String 
   // str2     --> second String 
   // count1[] --> hash array to calculate frequency 
   //             of characters in str1 
   // count[2] --> hash array to calculate frequency 
   //             of characters in str2 
   // result --> resultant longest String whose 
   // permutations are sub-sequence of given two strings 
       static void longestString(String str1, String str2) {
           int count1[] = new int[26], count2[] = new int[26];
     
           // calculate frequency of characters 
           for (int i = 0; i < str1.length(); i++) {
               count1[str1.charAt(i) - 'a']++;
           }
           for (int i = 0; i < str2.length(); i++) {
               count2[str2.charAt(i) - 'a']++;
           }
     
           // Now traverse hash array 
           String result = "";
           for (int i = 0; i < 26; i++) // append character ('a'+i) in resultant 
           // String 'result' by min(count1[i],count2i]) 
           // times 
           {
               for (int j = 1; j <= Math.min(count1[i], count2[i]); j++) {
                   result += (char)('a' + i);
               }
           }
     
           System.out.println(result);
       }
     
   // Driver program to run the case 
       public static void main(String[] args) {
           String str1 = "geeks", str2 = "cake";
           longestString(str1, str2);
     
       }
   }
   /* This java code is contributed by 29AjayKumar*/

   Python 3

   # Python 3 program to find LCS
   # with permutations allowed
     
   # Function to calculate longest string
   # str1     --> first string
   # str2     --> second string
   # count1[] --> hash array to calculate frequency
   #               of characters in str1
   # count[2] --> hash array to calculate frequency
   #               of characters in str2
   # result --> resultant longest string whose
   # permutations are sub-sequence 
   # of given two strings
   def longestString(str1, str2):
     
       count1 = [0] * 26
       count2 = [0] * 26
     
       # calculate frequency of characters
       for i in range( len(str1)):
           count1[ord(str1[i]) - ord('a')] += 1
       for i in range(len(str2)):
           count2[ord(str2[i]) - ord('a')] += 1
     
       # Now traverse hash array
       result = ""
       for i in range(26):
     
           # append character ('a'+i) in 
           # resultant string 'result' by 
           # min(count1[i],count2i]) times
           for j in range(1, min(count1[i],
                                 count2[i]) + 1):
               result = result + chr(ord('a') + i)
     
       print(result)
     
   # Driver Code
   if __name__ == "__main__":
         
       str1 = "geeks"
       str2 = "cake"
       longestString(str1, str2)
     
   # This code is contributed by ita_c

   C#

   // C# program to find LCS with 
   // permutations allowed
   using System;
     
   class GFG
   {
     
   // Function to calculate longest String 
   // str1 --> first String 
   // str2 --> second String 
   // count1[] --> hash array to calculate  
   //        frequency of characters in str1 
   // count[2] --> hash array to calculate 
   //         frequency of characters in str2 
   // result --> resultant longest String whose 
   // permutations are sub-sequence of
   // given two strings 
   static void longestString(String str1, 
                             String str2) 
   {
       int []count1 = new int[26];
       int []count2 = new int[26];
     
       // calculate frequency of characters 
       for (int i = 0; i < str1.Length; i++)
       {
           count1[str1[i] - 'a']++;
       }
       for (int i = 0; i < str2.Length; i++) 
       {
           count2[str2[i] - 'a']++;
       }
     
       // Now traverse hash array 
       String result = "";
       for (int i = 0; i < 26; i++) 
         
       // append character ('a'+i) in resultant 
       // String 'result' by min(count1[i],count2i]) 
       // times 
       {
           for (int j = 1; 
                    j <= Math.Min(count1[i], 
                                  count2[i]); j++)
           {
               result += (char)('a' + i);
           }
       }
     
   Console.Write(result);
   }
     
   // Driver Code
   public static void Main()
   {
       String str1 = "geeks", str2 = "cake";
       longestString(str1, str2);
   }
   }
     
   // This code is contributed
   // by PrinciRaj1992

   PHP

    first string
   // str2     --> second string
   // count1[] --> hash array to calculate frequency
   //                of characters in str1
   // count[2] --> hash array to calculate frequency
   //                of characters in str2
   // result --> resultant longest string whose
   // permutations are sub-sequence of given two strings
   function longestString($str1, $str2)
   {
       $count1 = array_fill(0, 26, NULL);
       $count2 = array_fill(0, 26, NULL);
     
       // calculate frequency of characters
       for ($i = 0; $i < strlen($str1); $i++)
           $count1[ord($str1[$i]) - ord('a')]++;
       for ($i = 0; $i < strlen($str2); $i++)
           $count2[ord($str2[$i]) - ord('a')]++;
     
       // Now traverse hash array
       $result = "";
       for ($i = 0; $i < 26; $i++)
     
           // append character ('a'+i) in resultant
           // string 'result' by min(count1[$i],
           // count2[$i]) times
           for ($j = 1; $j <= min($count1[$i], 
                                  $count2[$i]); $j++)
               $result = $result.chr(ord('a') + $i);
     
       echo $result;
   }
     
   // Driver Code
   $str1 = "geeks";
   $str2 = "cake";
   longestString($str1, $str2);
     
   // This code is contributed by ita_c
   ?>

   输出：

   ek
   

   时间复杂度： O(m + n)，其中m和n是输入字符串的长度。
   辅助空间： O(1)如果您有其他解决此问题的方法，请分享。