给定两个字符串A和B ，任务是从字典上找到字符串B的最小排列，以使其包含字符串A中的每个子字符串作为其子字符串。如果无法进行有效的排列，请打印“ -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.
天真的方法：最简单的方法是生成字符串B的所有可能置换，然后从所有这些置换中，从字典上找到包含A的所有子串的最小置换。
时间复杂度： O(N！)
辅助空间： O(1)
高效方法：为优化上述方法，主要观察结果是，包含A的所有子字符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。步骤如下：
- 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。
- 为了按字典顺序形成最小的字符串，请初始化一个空字符串结果，然后将其值都小于字符串A的第一个字符的所有剩余字符附加到该结果上。
- 附加所有之前的字符等于第一个字符到结果，检查是否有小于在字符串中的第一个字符的任意字符。如果是这样，则将A附加到第一个结果，然后将所有其余字符等于A的第一个字符，以使重新排序的字符串在字典上最小。
- 否则，附加所有剩余出现的A [0]，然后附加A。
- 最后，将其余字符附加到结果中。
- 完成上述步骤后，输出字符串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 29AjayKumarPython3
# 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 29C#
// 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输出：
aaabbb
时间复杂度： O(N)
辅助空间： O(1)