C++程序检查一个字符串是否可以通过最多X个循环顺时针移位从另一个字符串形成

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：

C++

// C++ implementation to check
// that a given string can be
// converted to another string
// by circular clockwise shift
// of each character by atmost
// X times
  
#include 
using namespace std;
  
// Function to check that all
// characters of s1 can be
// converted to s2 by circular
// clockwise shift atmost X times
void isConversionPossible(string s1,
                          string s2, int x)
{
    int diff, 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[i] == s2[i])
            continue;
  
        // Calculate the difference
        // between the ASCII values
        // of the characters
        diff = (int(s2[i] - s1[i])
                + 26)
               % 26;
  
        // If difference exceeds X
        if (diff > x) {
            cout << "NO" << endl;
            return;
        }
    }
  
    cout << "YES" << endl;
}
  
// Driver Code
int main()
{
    string s1 = "you";
    string s2 = "ara";
  
    int x = 6;
  
    // Function call
    isConversionPossible(s1, s2, x);
  
    return 0;
}

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！