包含另一个字符串的所有子字符串的字典序排列最小

给定两个字符串A和B ，任务是按字典顺序查找字符串B的最小排列，使其包含字符串A中的每个子字符串作为其子字符串。如果不可能有这样的有效安排，则打印“ -1” 。
例子：

Input: A = “aa”, B = “ababab” 
Output: aaabbb 
Explanation: 
All possible substrings of A are (‘a’, ‘a’, ‘aa’) 
Rearrange string B to “aaabb”. 
Now “aaabb” is the lexicographically the smallest arrangement of B which contains all the substrings of A.

Input: A = “aaa”, B = “ramialsadaka” 
Output: aaaaadiklmrs 
Explanation: 
All possible substrings of A are (‘a’, ‘aa’, ‘aaa’) 
Rearrange string B to “aaaaadiklmrs”. 
Now “aaaaadiklmrs” is the lexicographically smallest arrangement of B which contains all the substrings of A.

朴素方法：最简单的方法是生成字符串B 的所有可能排列，然后从所有这些排列中，按字典顺序找到包含A 的所有子串的最小排列。

时间复杂度： O(N!)
辅助空间： O(1)

高效方法：为了优化上述方法，主要观察是包含A 的所有子串的最小字符串是字符串A本身。因此，对于要重新排序的字符串B并包含A 的所有子字符串，它必须包含A作为子字符串。仅当字符串B中每个字符的频率大于或等于其在A 中的频率时，重新排序的字符串B 才能包含A作为其子字符串。以下是步骤：

1. 计算数组freq[]中字符串B中每个字符的频率，然后从中减去字符串A 中相应字符的频率。
2. 为了形成字典序最小的字符串，初始化一个空字符串result ，然后附加所有剩余的字符，这些字符的值小于字符串A的第一个字符。
3. 在将等于第一个字符A 的所有字符附加到结果之前，请检查是否有任何字符小于字符串A 中的第一个字符。如果是，则首先将 A 附加到 result ，然后将所有剩余的字符附加到 A 的第一个字符，以使重新排序的字符串字典顺序最小。
4. 否则，追加所有剩余的 A[0]，然后追加 A。
5. 最后，将剩余的字符附加到结果中。
6. 完成上述步骤后，打印字符串result 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to reorder the string B
// to contain all the substrings of A
string reorderString(string A, string B)
{
    // Find length of strings
    int size_a = A.length();
    int size_b = B.length();
 
    // Initialize array to count the
    // frequencies of the character
    int freq[300] = { 0 };
 
    // Counting frequencies of
    // character in B
    for (int i = 0; i < size_b; i++)
        freq[B[i]]++;
 
    // Find remaining character in B
    for (int i = 0; i < size_a; i++)
        freq[A[i]]--;
 
    for (int j = 'a'; j <= 'z'; j++) {
        if (freq[j] < 0)
            return "-1";
    }
 
    // Declare the reordered string
    string answer;
 
    for (int j = 'a'; j < A[0]; j++)
 
        // Loop until freq[j] > 0
        while (freq[j] > 0) {
            answer.push_back(j);
 
            // Decrement the value
            // from freq array
            freq[j]--;
        }
 
    int first = A[0];
 
    for (int j = 0; j < size_a; j++) {
 
        // Check if A[j] > A[0]
        if (A[j] > A[0])
            break;
 
        // Check if A[j] < A[0]
        if (A[j] < A[0]) {
            answer += A;
            A.clear();
            break;
        }
    }
 
    // Append the remaining characters
    // to the end of the result
    while (freq[first] > 0) {
        answer.push_back(first);
        --freq[first];
    }
 
    answer += A;
 
    for (int j = 'a'; j <= 'z'; j++)
 
        // Push all the values from
        // frequency array in the answer
        while (freq[j]--)
            answer.push_back(j);
 
    // Return the answer
    return answer;
}
 
// Driver Code
int main()
{
    // Given strings A and B
    string A = "aa";
    string B = "ababab";
 
    // Function Call
    cout << reorderString(A, B);
 
    return 0;
}

Java

// Java program for
// the above approach
class GFG{
 
// Function to reorder the String B
// to contain all the subStrings of A
static String reorderString(char []A,
                            char []B)
{
  // Find length of Strings
  int size_a = A.length;
  int size_b = B.length;
 
  // Initialize array to count the
  // frequencies of the character
  int freq[] = new int[300];
 
  // Counting frequencies of
  // character in B
  for (int i = 0; i < size_b; i++)
    freq[B[i]]++;
 
  // Find remaining character in B
  for (int i = 0; i < size_a; i++)
    freq[A[i]]--;
 
  for (int j = 'a'; j <= 'z'; j++)
  {
    if (freq[j] < 0)
      return "-1";
  }
 
  // Declare the reordered String
  String answer = "";
 
  for (int j = 'a'; j < A[0]; j++)
 
    // Loop until freq[j] > 0
    while (freq[j] > 0)
    {
      answer+=j;
 
      // Decrement the value
      // from freq array
      freq[j]--;
    }
 
  int first = A[0];
 
  for (int j = 0; j < size_a; j++)
  {
    // Check if A[j] > A[0]
    if (A[j] > A[0])
      break;
 
    // Check if A[j] < A[0]
    if (A[j] < A[0])
    {
      answer += String.valueOf(A);
      A = new char[A.length];
      break;
    }
  }
 
  // Append the remaining characters
  // to the end of the result
  while (freq[first] > 0)
  {
    answer += String.valueOf((char)first);
    --freq[first];
  }
 
  answer += String.valueOf(A);
 
  for (int j = 'a'; j <= 'z'; j++)
 
    // Push all the values from
    // frequency array in the answer
    while (freq[j]-- > 0)
      answer += ((char)j);
 
  // Return the answer
  return answer;
}
 
// Driver Code
public static void main(String[] args)
{
  // Given Strings A and B
  String A = "aa";
  String B = "ababab";
 
  // Function Call
  System.out.print(reorderString(A.toCharArray(),
                                 B.toCharArray()));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Function to reorder the B
# to contain all the substrings of A
def reorderString(A, B):
     
    # Find length of strings
    size_a = len(A)
    size_b = len(B)
 
    # Initialize array to count the
    # frequencies of the character
    freq = [0] * 300
 
    # Counting frequencies of
    # character in B
    for i in range(size_b):
        freq[ord(B[i])] += 1
 
    # Find remaining character in B
    for i in range(size_a):
        freq[ord(A[i])] -= 1
 
    for j in range(ord('a'), ord('z') + 1):
        if (freq[j] < 0):
            return "-1"
 
    # Declare the reordered string
    answer = []
 
    for j in range(ord('a'), ord(A[0])):
 
        # Loop until freq[j] > 0
        while (freq[j] > 0):
            answer.append(j)
 
            # Decrement the value
            # from freq array
            freq[j] -= 1
 
    first = A[0]
 
    for j in range(size_a):
 
        # Check if A[j] > A[0]
        if (A[j] > A[0]):
            break
 
        # Check if A[j] < A[0]
        if (A[j] < A[0]):
            answer += A
            A = ""
            break
 
    # Append the remaining characters
    # to the end of the result
    while (freq[ord(first)] > 0):
        answer.append(first)
        freq[ord(first)] -= 1
 
    answer += A
 
    for j in range(ord('a'), ord('z') + 1):
 
        # Push all the values from
        # frequency array in the answer
        while (freq[j]):
            answer.append(chr(j))
            freq[j] -= 1
 
    # Return the answer
    return "".join(answer)
 
# Driver Code
if __name__ == '__main__':
     
    # Given strings A and B
    A = "aa"
    B = "ababab"
 
    # Function call
    print(reorderString(A, B))
 
# This code is contributed by mohit kumar 29

C#

// C# program for
// the above approach
using System;
class GFG{
 
// Function to reorder the String B
// to contain all the subStrings of A
static String reorderString(char []A,
                            char []B)
{
  // Find length of Strings
  int size_a = A.Length;
  int size_b = B.Length;
 
  // Initialize array to count the
  // frequencies of the character
  int []freq = new int[300];
 
  // Counting frequencies of
  // character in B
  for (int i = 0; i < size_b; i++)
    freq[B[i]]++;
 
  // Find remaining character in B
  for (int i = 0; i < size_a; i++)
    freq[A[i]]--;
 
  for (int j = 'a'; j <= 'z'; j++)
  {
    if (freq[j] < 0)
      return "-1";
  }
 
  // Declare the reordered String
  String answer = "";
 
  for (int j = 'a'; j < A[0]; j++)
 
    // Loop until freq[j] > 0
    while (freq[j] > 0)
    {
      answer+=j;
 
      // Decrement the value
      // from freq array
      freq[j]--;
    }
 
  int first = A[0];
 
  for (int j = 0; j < size_a; j++)
  {
    // Check if A[j] > A[0]
    if (A[j] > A[0])
      break;
 
    // Check if A[j] < A[0]
    if (A[j] < A[0])
    {
      answer += String.Join("", A);
      A = new char[A.Length];
      break;
    }
  }
 
  // Append the remaining characters
  // to the end of the result
  while (freq[first] > 0)
  {
    answer += String.Join("", (char)first);
    --freq[first];
  }
 
  answer += String.Join("", A);
 
  for (int j = 'a'; j <= 'z'; j++)
 
    // Push all the values from
    // frequency array in the answer
    while (freq[j]-- > 0)
      answer += ((char)j);
 
  // Return the answer
  return answer;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given Strings A and B
  String A = "aa";
  String B = "ababab";
 
  // Function Call
  Console.Write(reorderString(A.ToCharArray(),
                              B.ToCharArray()));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// Javascript program for
// the above approach
 
// Function to reorder the String B
// to contain all the subStrings of A
function reorderString(A,B)
{
    // Find length of Strings
  let size_a = A.length;
  let size_b = B.length;
  
  // Initialize array to count the
  // frequencies of the character
  let freq = new Array(300);
  for(let i=0;i<300;i++)
  {
      freq[i]=0;
  }
  
  // Counting frequencies of
  // character in B
  for (let i = 0; i < size_b; i++)
    freq[B[i].charCodeAt(0)]++;
  
  // Find remaining character in B
  for (let i = 0; i < size_a; i++)
    freq[A[i].charCodeAt(0)]--;
  
  for (let j = 'a'.charCodeAt(0); j <= 'z'.charCodeAt(0); j++)
  {
    if (freq[j] < 0)
      return "-1";
  }
  
  // Declare the reordered String
  let answer = "";
  
  for (let j = 'a'.charCodeAt(0); j < A[0].charCodeAt(0); j++)
  
    // Loop until freq[j] > 0
    while (freq[j] > 0)
    {
      answer+=j;
  
      // Decrement the value
      // from freq array
      freq[j]--;
    }
  
  let first = A[0];
  
  for (let j = 0; j < size_a; j++)
  {
    // Check if A[j] > A[0]
    if (A[j] > A[0])
      break;
  
    // Check if A[j] < A[0]
    if (A[j] < A[0])
    {
      answer += (A).join("");
      A = new Array(A.length);
      break;
    }
  }
  
  // Append the remaining characters
  // to the end of the result
  while (freq[first] > 0)
  {
    answer += (String.fromCharCode(first));
    --freq[first];
  }
  
  answer += (A).join("");
  
  for (let j = 'a'.charCodeAt(0); j <= 'z'.charCodeAt(0); j++)
  
    // Push all the values from
    // frequency array in the answer
    while (freq[j]-- > 0)
      answer += String.fromCharCode(j);
  
  // Return the answer
  return answer;
}
 
// Driver Code
// Given Strings A and B
let A = "aa";
let B = "ababab";
 
// Function Call
document.write(reorderString(A.split(""),
                               B.split("")));
 
 
// This code is contributed by patel2127

输出：

aaabbb

时间复杂度： O(N)
辅助空间： O(1)

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