包含第二个字符串作为子字符串的字符串的字典序最小排列

给定两个字符串str1和str2 ，任务是找到包含str2作为子字符串的str1的字典序最小排列。

注意：假设解总是存在的。

例子：

Input: str1 = “abab”, str2 = “ab”
Output: “aabb”
Explanation: The Lexicographically smallest permutation of string str1 is “aabb”, Since “aabb” contains the string “ab” as a substring, therefore, “aabb” is the required answer.

Input: str1 = “geeksforgeeks”, str2 = “for”
Output: “eeeeforggkkss”

方法：这个问题可以使用频率计数技术的概念来解决。请按照以下步骤解决此问题。

1. 存储字符串str1和str2的所有字符的频率。
2. 用子字符串初始化结果字符串。
3. 从第一个字符串的频率中减去第二个字符串的频率
4. 现在，将按字典顺序小于或等于子字符串第一个字符的剩余字符附加到结果字符串中的子字符串之前。
5. 在结果字符串的子字符串后面按字典顺序附加剩余的字符。
6. 最后，打印结果字符串。

下面是上述方法的实现。

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to print the desired
// lexicographic smaller string
string findSmallestString(string str1,
                          string str2)
{
 
    int freq1[26], freq2[26];
    memset(freq1, 0, sizeof freq1);
    memset(freq2, 0, sizeof freq2);
 
    // Calculate length of the string
    int n1 = str1.length();
    int n2 = str2.length();
 
    // Stores the frequencies of
    // characters of string str1
    for (int i = 0; i < n1; ++i) {
        freq1[str1[i] - 'a']++;
    }
 
    // Stores the frequencies of
    // characters of string str2
    for (int i = 0; i < n2; ++i) {
        freq2[str2[i] - 'a']++;
    }
 
    // Decrease the frequency of
    // second string from that of
    // of the first string
    for (int i = 0; i < 26; ++i) {
        freq1[i] -= freq2[i];
    }
 
    // To store the resultant string
    string res = "";
 
    // To find the index of first
    // character of the string str2
    int minIndex = str2[0] - 'a';
 
    // Append the characters in
    // lexicographical order
    for (int i = 0; i < 26; ++i) {
 
        // Append all the current
        // character (i + 'a')
        for (int j = 0; j < freq1[i]; ++j) {
            res += (char)(i + 'a');
        }
 
        // If we reach first character
        // of string str2 append str2
        if (i == minIndex) {
            res += str2;
        }
    }
 
    // Return the resultant string
    return res;
}
 
// Driver Code
int main()
{
    string str1 = "geeksforgeeksfor";
    string str2 = "for";
    cout << findSmallestString(str1, str2);
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to print the desired
// lexicographic smaller String
static String findSmallestString(String str1,
                                 String str2)
{
    int []freq1 = new int[26];
    int []freq2 = new int[26];
    Arrays.fill(freq1, 0);
    Arrays.fill(freq2, 0);
 
    // Calculate length of the String
    int n1 = str1.length();
    int n2 = str2.length();
 
    // Stores the frequencies of
    // characters of String str1
    for(int i = 0; i < n1; ++i)
    {
        freq1[str1.charAt(i) - 'a']++;
    }
 
    // Stores the frequencies of
    // characters of String str2
    for(int i = 0; i < n2; ++i)
    {
        freq2[str2.charAt(i) - 'a']++;
    }
 
    // Decrease the frequency of
    // second String from that of
    // of the first String
    for(int i = 0; i < 26; ++i)
    {
        freq1[i] -= freq2[i];
    }
 
    // To store the resultant String
    String res = "";
 
    // To find the index of first
    // character of the String str2
    int minIndex = str2.charAt(0) - 'a';
 
    // Append the characters in
    // lexicographical order
    for(int i = 0; i < 26; ++i)
    {
 
        // Append all the current
        // character (i + 'a')
        for(int j = 0; j < freq1[i]; ++j)
        {
            res += (char)(i + 'a');
        }
 
        // If we reach first character
        // of String str2 append str2
        if (i == minIndex)
        {
            res += str2;
        }
    }
 
    // Return the resultant String
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    String str1 = "geeksforgeeksfor";
    String str2 = "for";
     
    System.out.print(findSmallestString(str1, str2));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach
 
# Function to print the desired
# lexicographic smaller string
def findSmallestString(str1, str2):
     
    freq1 = [0] * 26
    freq2 = [0] * 26
     
    # Calculate length of the string
    n1 = len(str1)
    n2 = len(str2)
     
    # Stores the frequencies of
    # characters of string str1
    for i in range(n1):
        freq1[ord(str1[i]) - ord('a')] += 1
         
    # Stores the frequencies of
    # characters of string str2
    for i in range(n2):
        freq2[ord(str2[i]) - ord('a')] += 1
         
    # Decrease the frequency of
    # second string from that of
    # of the first string
    for i in range(26):
        freq1[i] -= freq2[i]
         
    # To store the resultant string
    res = ""
     
    # To find the index of first
    # character of the string str2
    minIndex = ord(str2[0]) - ord('a')
     
    # Append the characters in
    # lexicographical order
    for i in range(26):
         
        # Append all the current
        # character (i + 'a')
        for j in range(freq1[i]):
            res += chr(i+ ord('a'))
             
        # If we reach first character
        # of string str2 append str2
        if i == minIndex:
            res += str2
             
    # Return the resultant string
    return res
 
# Driver code
str1 = "geeksforgeeksfor"
str2 = "for"
 
print(findSmallestString(str1, str2))
 
# This code is contributed by Stuti Pathak

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
    // Function to print the desired
    // lexicographic smaller String
    static String findSmallestString(String str1,
                                     String str2)
    {
        int[] freq1 = new int[26];
        int[] freq2 = new int[26];
 
        // Calculate length of the String
        int n1 = str1.Length;
        int n2 = str2.Length;
 
        // Stores the frequencies of
        // characters of String str1
        for (int i = 0; i < n1; ++i)
        {
            freq1[str1[i] - 'a']++;
        }
 
        // Stores the frequencies of
        // characters of String str2
        for (int i = 0; i < n2; ++i)
        {
            freq2[str2[i] - 'a']++;
        }
 
        // Decrease the frequency of
        // second String from that of
        // of the first String
        for (int i = 0; i < 26; ++i)
        {
            freq1[i] -= freq2[i];
        }
 
        // To store the resultant String
        String res = "";
 
        // To find the index of first
        // character of the String str2
        int minIndex = str2[0] - 'a';
 
        // Append the characters in
        // lexicographical order
        for (int i = 0; i < 26; ++i)
        {
 
            // Append all the current
            // character (i + 'a')
            for (int j = 0; j < freq1[i]; ++j)
            {
                res += (char)(i + 'a');
            }
 
            // If we reach first character
            // of String str2 append str2
            if (i == minIndex)
            {
                res += str2;
            }
        }
 
        // Return the resultant String
        return res;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        String str1 = "geeksforgeeksfor";
        String str2 = "for";
          Console.Write(findSmallestString(str1, str2));
    }
}
 
// This code is contributed by shikhasingrajput

Javascript

 
    // Javascript program to implement
    // the above approach
 
    // Function to prlet the desired
    // lexicographic smaller String
      function findSmallestString(str1, str2)
    {
        let freq1 = Array.from({length: 26}, (_, i) => 0);
        let freq2 = Array.from({length: 26}, (_, i) => 0);
  
        // Calculate length of the String
        let n1 = str1.length;
        let n2 = str2.length;
  
        // Stores the frequencies of
        // characters of String str1
        for (let i = 0; i < n1; ++i)
        {
            freq1[str1[i].charCodeAt() - 'a'.charCodeAt()]++;
        }
  
        // Stores the frequencies of
        // characters of String str2
        for (let i = 0; i < n2; ++i)
        {
            freq2[str2[i].charCodeAt() - 'a'.charCodeAt()]++;
        }
  
        // Decrease the frequency of
        // second String from that of
        // of the first String
        for (let i = 0; i < 26; ++i)
        {
            freq1[i] -= freq2[i];
        }
  
        // To store the resultant String
        let res = "";
  
        // To find the index of first
        // character of the String str2
        let minIndex = str2[0].charCodeAt() - 'a'.charCodeAt();
  
        // Append the characters in
        // lexicographical order
        for (let i = 0; i < 26; ++i)
        {
  
            // Append all the current
            // character (i + 'a')
            for (let j = 0; j < freq1[i]; ++j)
            {
                res += String.fromCharCode(i + 'a'.charCodeAt());
            }
  
            // If we reach first character
            // of String str2 append str2
            if (i == minIndex)
            {
                res += str2;
            }
        }
  
        // Return the resultant String
        return res;
    }
 
 
    // Driver code
 
    let str1 = "geeksforgeeksfor";
    let str2 = "for";
    document.write(findSmallestString(str1, str2));
 

输出：

eeeefforggkkorss

时间复杂度： O(N)
辅助空间： O(1)

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