用于检查一个字符串是否可以通过最多 X 次循环顺时针移位从另一个字符串形成的Java程序

给定一个整数X和两个字符串S1和S2 ，任务是通过将字符顺时针循环移动最多 X 次来检查字符串S1是否可以转换为字符串S2 。

Input: S1 = “abcd”, S2 = “dddd”, X = 3 
Output: Yes 
Explanation: 
Given string S1 can be converted to string S2 as- 
Character “a” – Shift 3 times – “d” 
Character “b” – Shift 2 times – “d” 
Character “c” – Shift 1 times – “d” 
Character “d” – Shift 0 times – “d”

Input: S1 = “you”, S2 = “ara”, X = 6 
Output: Yes 
Explanation: 
Given string S1 can be converted to string S2 as – 
Character “y” – Circular Shift 2 times – “a” 
Character “o” – Shift 3 times – “r” 
Character “u” – Circular Shift 6 times – “a”  

方法：想法是遍历字符串和对于每个索引，并找到两个字符串各自索引处字符的ASCII值之间的差异。如果差值小于 0，则对于循环移位，加 26 以获得实际差值。如果对于任何索引，差异超过X ，则S2不能从S1形成，否则可能。
下面是上述方法的实现：

Java

// Java implementation to check 
// that a given string can be 
// converted to another string 
// by circular clockwise shift 
// of each character by atmost 
// X times 
import java.io.*;
import java.util.*;
  
class GFG{
      
// Function to check that all 
// characters of s1 can be 
// converted to s2 by circular 
// clockwise shift atmost X times 
static void isConversionPossible(String s1, 
                                 String s2,
                                 int x) 
{ 
    int diff = 0, n; 
    n = s1.length(); 
      
    // Check for all characters of 
    // the strings whether the 
    // difference between their 
    // ascii values is less than 
    // X or not 
    for(int i = 0; i < n; i++) 
    {
          
       // If both the characters 
       // are same 
       if (s1.charAt(i) == s2.charAt(i))
           continue; 
         
       // Calculate the difference 
       // between the ASCII values 
       // of the characters 
       diff = ((int)(s2.charAt(i) - 
                     s1.charAt(i)) + 26) % 26; 
         
       // If difference exceeds X 
       if (diff > x)
       { 
           System.out.println("NO"); 
           return; 
       } 
    } 
    System.out.println("YES");
} 
      
// Driver Code
public static void main (String[] args)
{
    String s1 = "you"; 
    String s2 = "ara"; 
  
    int x = 6; 
  
    // Function call 
    isConversionPossible(s1, s2, x); 
}
}
  
// This code is contributed by Ganeshchowdharysadanala

输出：

YES

时间复杂度： O(N),N=Length(S1)

辅助空间： O(1)

有关更多详细信息，请参阅有关检查一个字符串是否可以通过最多 X 个圆形顺时针移位从另一个字符串形成的完整文章！