Javascript 程序 tp 检查一个字符串是否可以通过最多 X 次循环顺时针移位由另一个字符串组成

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：

Javascript

// Javascript implementation to check
// that a given string can be
// converted to another string
// by circular clockwise shift
// of each character by atmost
// X times
  
  
// Function to check that all
// characters of s1 can be
// converted to s2 by circular
// clockwise shift atmost X times
function isConversionPossible(s1,s2,x)
{
    let 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(let i = 0; i < n; i++)
    {
           
       // If both the characters
       // are same
       if (s1[i] == s2[i])
           continue;
          
       // Calculate the difference
       // between the ASCII values
       // of the characters
       diff = ((s2[i].charCodeAt(0) -
                     s1[i].charCodeAt(0)) + 26) % 26;
          
       // If difference exceeds X
       if (diff > x)
       {
           document.write("NO<br>");
           return;
       }
    }
    document.write("YES<br>");
}
  
// Driver Code
let s1 = "you";
let s2 = "ara";
let x = 6;
  
// Function call
isConversionPossible(s1, s2, x);
  
// This code is contributed by avanitrachhadiya2155

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！