给定两个字符串P和Q，任务是生成满足以下条件的字符串S ：
- 查找S ，以使P重新排列，而Q是其中的子字符串。
- S中Q之前的所有字符均应小于或等于Q中按字母顺序排列的第一个字符。
- 其余字符应按字符顺序排列在Q之后
注意： Q的所有字符始终存在于P中，并且Q的长度始终小于或等于P的长度。
例子：
Input : P = “geeksforgeeksfor” Q = “for”
Output : eeeefforggkkorss
Explanation:
The characters ‘e’ and ‘f’ are the only characters here which are less than or equal to ‘f’ (first character of Q).
So, before “for” the string is lexicographically equal to eeeef.
The rest of the characters in P are greater than ‘f’, so they are placed after “for” in lexicographic order.
Thus, after “for”, the string is ggkkorss.
Therefore the output is eeeefforggkkorss.
Input : P = “lexicographical” Q = “gap”
Output : accegaphiillorx
Explanation:
The string accegaphiillorx satisfies the above conditions for string P and Q.
方法：想法是找到 P和Q中所有字符的频率可以解决该问题。
- 维持P和Q中所有字母的频率数组。
- 找到频率后，根据在Q中的第一个字符P中分离字符，并将它们添加到所得到的字符串。
- 最后返回结果字符串。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to generate a string S from string P
// and Q according to the given conditions
void manipulateStrings(string P, string Q)
{
// Stores the frequencies
int freq[26];
memset(freq, 0, sizeof(freq));
// Counts occurrences of each characters
for(int i = 0; i < P.size(); i++)
{
freq[P[i] - 'a']++;
}
// Reduce the count of the character
// which is also present in Q
for(int i = 0; i < Q.size(); i++)
{
freq[Q[i] - 'a']--;
}
// Stores the resultant string
string sb = "";
// Index in freq[] to segregate the string
int pos = Q[0] - 'a';
for(int i = 0; i <= pos; i++)
{
while (freq[i] > 0)
{
char c = (char)('a' + i);
sb += c;
freq[i]--;
}
}
// Add Q to the resulting string
sb += Q;
for(int i = pos + 1; i < 26; i++)
{
while (freq[i] > 0)
{
char c = (char)('a' + i);
sb += c;
freq[i]--;
}
}
cout << sb << endl;
}
// Driver Code
int main()
{
string P = "geeksforgeeksfor";
string Q = "for";
// Function call
manipulateStrings(P, Q);
}
// This code is contributed by Amit Katiyar Java
// Java Program to implement
// the above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG {
// Function to generate a string S from string P
// and Q according to the given conditions
public static void manipulateStrings(String P,
String Q)
{
// Stores the frequencies
int[] freq = new int[26];
// Counts occurrences of each characters
for (int i = 0; i < P.length(); i++) {
freq[P.charAt(i) - 'a']++;
}
// Reduce the count of the character
// which is also present in Q
for (int i = 0; i < Q.length(); i++) {
freq[Q.charAt(i) - 'a']--;
}
// Stores the resultant string
StringBuilder sb
= new StringBuilder();
// Index in freq[] to segregate the string
int pos = Q.charAt(0) - 'a';
for (int i = 0; i <= pos; i++) {
while (freq[i] > 0) {
char c = (char)('a' + i);
sb.append(c);
freq[i]--;
}
}
// Add Q to the resulting string
sb.append(Q);
for (int i = pos + 1; i < 26; i++) {
while (freq[i] > 0) {
char c = (char)('a' + i);
sb.append(c);
freq[i]--;
}
}
System.out.println(sb);
}
// Driver Code
public static void main(String[] args)
{
String P = "geeksforgeeksfor";
String Q = "for";
// Function call
manipulateStrings(P, Q);
}
}Python3
# Python3 program to implement
# the above approach
# Function to generate a string S
# from string P and Q according to
# the given conditions
def manipulateStrings(P, Q):
# Stores the frequencies
freq = [0 for i in range(26)];
# Counts occurrences of each characters
for i in range(len(P)):
freq[ord(P[i]) - ord('a')] += 1
# Reduce the count of the character
# which is also present in Q
for i in range(len(Q)):
freq[ord(Q[i]) - ord('a')] -= 1
# Stores the resultant string
sb = ""
# Index in freq[] to segregate the string
pos = ord(Q[0]) - ord('a')
for i in range(pos + 1):
while freq[i] > 0:
sb += chr(ord('a') + i)
freq[i] -= 1
# Add Q to the resulting string
sb += Q
for i in range(pos + 1, 26):
while freq[i] > 0:
sb += chr(ord('a') + i)
freq[i] -= 1
print(sb)
# Driver Code
if __name__=="__main__":
P = "geeksforgeeksfor";
Q = "for";
# Function call
manipulateStrings(P, Q);
# This code is contributed by rutvik_56C#
// C# Program to implement
// the above approach
using System;
class GFG{
// Function to generate a string S from string P
// and Q according to the given conditions
public static void manipulateStrings(String P,
String Q)
{
// Stores the frequencies
int[] freq = new int[26];
// Counts occurrences of each characters
for (int i = 0; i < P.Length; i++)
{
freq[P[i] - 'a']++;
}
// Reduce the count of the character
// which is also present in Q
for (int i = 0; i < Q.Length; i++)
{
freq[Q[i] - 'a']--;
}
// Stores the resultant string
String sb = "";
// Index in []freq to segregate the string
int pos = Q[0] - 'a';
for (int i = 0; i <= pos; i++)
{
while (freq[i] > 0)
{
char c = (char)('a' + i);
sb += c;
freq[i]--;
}
}
// Add Q to the resulting string
sb += Q;
for (int i = pos + 1; i < 26; i++)
{
while (freq[i] > 0)
{
char c = (char)('a' + i);
sb += c;
freq[i]--;
}
}
Console.WriteLine(sb);
}
// Driver Code
public static void Main(String[] args)
{
String P = "geeksforgeeksfor";
String Q = "for";
// Function call
manipulateStrings(P, Q);
}
}
// This code is contributed by Rohit_ranjan输出：
eeeefforggkkorss
时间复杂度： O(N + M)，其中N和M分别是P和Q的长度。
辅助空间： O(1)