给定字符串S和两个正整数X和Y ，任务是找到获得原始字符串所需的最少操作数。在每一个操作中，来自字符串到所述字符串的分别在每次操作中的前部的端部追加的X或Y的字符。
例子：
Input: S = “GeeksforGeeks”, X = 5, Y = 3
Output: 3
Explanation:
Below are the operations performed:
Operation 1: Append 5 characters from the back of the string S to the front of the string S. Now the updated string is “GeeksGeeksfor”.
Operation 2: Append 3 characters from the back of the string S to the front of the string S. Now the updated string is “forGeeksGeeks”.
Operation 3: Append 5 characters from the back of the string S to the front of the string S. Now the updated string is “GeeksforGeeks”.
Therefore, the minimum count of operations required is 3.
Input: S = “AbcDef”, X = 1, Y = 2
Output: 4
Explanation:
Below are the operations performed:
Operation 1: Append 1 characters from the back of the string S to the front of the string S. Now the updated string is “fAbcDe”.
Operation 2: Append 2 characters from the back of the string S to the front of the string S. Now the updated string is “fDeAbc”.
Operation 3: Append 1 characters from the back of the string S to the front of the string S. Now the updated string is “cfDeAb”.
Operation 4: Append 2 characters from the back of the string S to the front of the string S. Now the updated string is “AbcDef”.
Therefore, the minimum count of operations required is 4.
方法：想法是将给定字符串中的X和Y字符附加到字符串的开头，并在执行操作时跟踪操作计数。步骤如下：
- 将计数初始化为0 ，以存储最少的操作。
- 将给定的字符串S存储在需要执行操作的另一个字符串(例如newString )中。
- 现在，在newString上执行以下操作，直到它等于S为止：
- 从字符串newString的末尾追加X字符，并将计数增加1 。
- 如果newString与字符串S相同。
- 从字符串newString的末尾追加Y字符，并将计数增加1 。
- 如果newString与字符串S相同。
- 完成上述步骤后，将计数值打印为所需的最少操作数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the minimum operations
// required to get the given string after
// appending m or n characters from the end
// to the front of the string in each operation
int minimumOperations(string orig_str, int m, int n)
{
// Store the original string
string orig = orig_str;
// Stores the count of operations
int turn = 1;
int j = 1;
// Traverse the string
for(auto i : orig_str)
{
// Cut m letters from end
string m_cut = orig_str.substr(
orig_str.length() - m);
orig_str.erase(orig_str.length() - m);
// Add cut m letters to beginning
orig_str = m_cut + orig_str;
// Update j
j = j + 1;
// Check if strings are the same
if (orig != orig_str)
{
turn = turn + 1;
// Cut n letters from end
string n_cut = orig_str.substr(
orig_str.length() - n);
orig_str.erase(orig_str.length() - n);
// Add cut n letters to beginning
orig_str = n_cut + orig_str;
// Update j
j = j + 1;
}
// Check if strings are the same
if (orig == orig_str)
{
break;
}
// Update the turn
turn = turn + 1;
}
cout << turn;
}
// Driver Code
int main()
{
// Given string S
string S = "GeeksforGeeks";
int X = 5, Y = 3;
// Function Call
minimumOperations(S, X, Y);
return 0;
}
// This code is contributed by akhilsaini Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to find the minimum operations
// required to get the given string after
// appending m or n characters from the end
// to the front of the string in each operation
static void minimumOperations(String orig_str,
int m, int n)
{
// Store the original string
String orig = orig_str;
// Stores the count of operations
int turn = 1;
int j = 1;
// Traverse the string
for(int i = 0; i < orig_str.length(); i++)
{
// Cut m letters from end
String m_cut = orig_str.substring(
orig_str.length() - m);
orig_str = orig_str.substring(
0, orig_str.length() - m);
// Add cut m letters to beginning
orig_str = m_cut + orig_str;
// Update j
j = j + 1;
// Check if strings are the same
if (!orig.equals(orig_str))
{
turn = turn + 1;
// Cut n letters from end
String n_cut = orig_str.substring(
orig_str.length() - n);
orig_str = orig_str.substring(
0, orig_str.length() - n);
// Add cut n letters to beginning
orig_str = n_cut + orig_str;
// Update j
j = j + 1;
}
// Check if strings are the same
if (orig.equals(orig_str))
{
break;
}
// Update the turn
turn = turn + 1;
}
System.out.println( turn );
}
// Driver Code
public static void main(String[] args)
{
// Given string S
String S = "GeeksforGeeks";
int X = 5, Y = 3;
// Function Call
minimumOperations(S, X, Y);
}
}
// This code is contributed by akhilsainiPython
# Python program for the above approach
# Function to find the minimum operations
# required to get the given string after
# appending m or n characters from the end
# to the front of the string in each operation
def minimumOperations(orig_str, m, n):
# Store the original string
orig = orig_str
# Stores the count of operations
turn = 1
j = 1
# Traverse the string
for i in orig_str:
# Cut m letters from end
m_cut = orig_str[-m:]
orig_str = orig_str.replace(' ', '')[:-m]
# Add cut m letters to beginning
orig_str = m_cut + orig_str
# Update j
j = j + 1
# Check if strings are the same
if orig != orig_str:
turn = turn + 1
# Cut n letters from end
n_cut = orig_str[-n:]
orig_str = orig_str.replace(' ', '')[:-n]
# Add cut n letters to beginning
orig_str = n_cut + orig_str
# Update j
j = j + 1
# Check if strings are the same
if orig == orig_str:
break
# Update the turn
turn = turn + 1
print(turn)
# Driver Code
# Given string S
S = "GeeksforGeeks"
X = 5
Y = 3
# Function Call
minimumOperations(S, X, Y)C#
// C# program for the above approach
using System;
class GFG{
// Function to find the minimum operations
// required to get the given string after
// appending m or n characters from the end
// to the front of the string in each operation
static void minimumOperations(string orig_str, int m,
int n)
{
// Store the original string
string orig = orig_str;
// Stores the count of operations
int turn = 1;
int j = 1;
// Traverse the string
for(int i = 0; i < orig_str.Length; i++)
{
// Cut m letters from end
string m_cut = orig_str.Substring(
orig_str.Length - m);
orig_str = orig_str.Substring(
0, orig_str.Length - m);
// Add cut m letters to beginning
orig_str = m_cut + orig_str;
// Update j
j = j + 1;
// Check if strings are the same
if (!orig.Equals(orig_str))
{
turn = turn + 1;
// Cut n letters from end
String n_cut = orig_str.Substring(
orig_str.Length - n);
orig_str = orig_str.Substring(
0, orig_str.Length - n);
// Add cut n letters to beginning
orig_str = n_cut + orig_str;
// Update j
j = j + 1;
}
// Check if strings are the same
if (orig.Equals(orig_str))
{
break;
}
// Update the turn
turn = turn + 1;
}
Console.WriteLine(turn);
}
// Driver Code
public static void Main()
{
// Given string S
string S = "GeeksforGeeks";
int X = 5, Y = 3;
// Function Call
minimumOperations(S, X, Y);
}
}
// This code is contributed by akhilsaini3
时间复杂度： O(N)
辅助空间： O(1)