尽量减少字符替换为其最接近的字母，以使字符串回文

给定长度为N的小写字母组成的字符串S ，任务是找到将给定字符串转换为回文的最小操作数。在一项操作中，选择任何字符并将其替换为下一个或上一个字母。

Note: The alphabets are cyclic i.e., if z is incremented then it becomes a and if a is decremented then it becomes z.

例子：

Input: S = “arcehesmz”
Output: 16
Explanation:
The possible transformation that given the minimum count of operation is: 

• Decrease 1st character ‘a’ by 1 to get ‘z’. Count of operation = 1
• Decrease 3rd character ‘c’ by 10 to get ‘s’. Count of operations = 1 + 10 = 11
• Increase 8th character ‘m’ by 5  to get ‘r’. Count of operations = 11 + 5 = 16.

Therefore, the total count of operations is 16.

Input: S = “abccdb”
Output: 3
Explanation:
The possible transformation that given the minimum count of operation is: 

• Increase 1st character ‘a’ by 1 to get ‘b’. Count of operation = 1
• Increase 2nd character ‘b’ by 2 to get ‘d’. Count of operations = 1 + 2 = 3.

天真的方法：最简单的方法是生成所有可能的长度为N的字符串。然后检查每个字符串，如果它是回文。如果发现任何字符串都是回文的，则找出将给定字符串转换为该字符串所需的操作成本。对所有生成的字符串重复上述步骤，并打印所有成本中计算出的最低成本。

时间复杂度： O(26 ^N )
辅助空间： O(N)

高效方法：为了优化上述方法，其思想是遍历给定的字符串并独立地找到每个位置的变化。请按照以下步骤解决问题：

1. 在[0，(N / 2)– 1]范围内遍历给定的字符串。
2. 对于索引i处的每个字符，找到索引i与(N – i – 1)处的字符之间的绝对差。
3. 可能存在两个差异，即，当i处的字符递增到索引(N – i – 1)处的字符时，以及该索引处的字符递减时的区别。
4. 取两者中的最小值，并将其添加到结果中。
5. 完成上述步骤后，打印计算出的最低费用。

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 number
// of operations required to convert
// th given string into palindrome
int minOperations(string s)
{
    int len = s.length();
    int result = 0;
 
    // Iterate till half of the string
    for (int i = 0; i < len / 2; i++) {
 
        // Find the absolute difference
        // between the characters
        int D1 = max(s[i], s[len - 1 - i])
                - min(s[i], s[len - 1 - i]);
 
        int D2 = 26 - D1;
 
        // Adding the minimum difference
        // of the two result
        result += min(D1, D2);
    }
 
    // Return the result
    return result;
}
 
// Driver Code
int main()
{
    // Given string
    string s = "abccdb";
 
    // Function Call
    cout << minOperations(s);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
     
// Function to find the minimum number
// of operations required to convert
// th given string into palindrome
public static int minOperations(String s)
{
    int len = s.length();
    int result = 0;
 
    // Iterate till half of the string
    for(int i = 0; i < len / 2; i++)
    {
         
        // Find the absolute difference
        // between the characters
        int D1 = Math.max(s.charAt(i),
                          s.charAt(len - 1 - i)) -
                 Math.min(s.charAt(i),
                          s.charAt(len - 1 - i));
 
        int D2 = 26 - D1;
 
        // Adding the minimum difference
        // of the two result
        result += Math.min(D1, D2);
    }
 
    // Return the result
    return result;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given string
    String s = "abccdb";
 
    // Function call
    System.out.println(minOperations(s));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
 
# Function to find the minimum number
# of operations required to convert
# th given string into palindrome
def minOperations(s):
 
    length = len(s)
    result = 0
 
    # Iterate till half of the string
    for i in range(length // 2):
 
        # Find the absolute difference
        # between the characters
        D1 = (ord(max(s[i], s[length - 1 - i])) -
              ord(min(s[i], s[length - 1 - i])))
         
        D2 = 26 - D1
 
        # Adding the minimum difference
        # of the two result
        result += min(D1, D2)
 
    # Return the result
    return result
 
# Driver Code
 
# Given string
s = "abccdb"
 
# Function call
print(minOperations(s))
 
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the minimum number
// of operations required to convert
// th given string into palindrome
public static int minOperations(String s)
{
    int len = s.Length;
    int result = 0;
 
    // Iterate till half of the string
    for(int i = 0; i < len / 2; i++)
    {
         
        // Find the absolute difference
        // between the characters
        int D1 = Math.Max(s[i],
                          s[len - 1 - i]) -
                 Math.Min(s[i],
                          s[len - 1 - i]);
 
        int D2 = 26 - D1;
 
        // Adding the minimum difference
        // of the two result
        result += Math.Min(D1, D2);
    }
 
    // Return the result
    return result;
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Given string
    String s = "abccdb";
 
    // Function call
    Console.WriteLine(minOperations(s));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

      // JavaScript program for the above approach
 
      // Function to find the minimum number
      // of operations required to convert
      // th given string into palindrome
      function minOperations(s)
      {
        var len = s.length;
        var result = 0;
 
        // Iterate till half of the string
        for (var i = 0; i < len / 2; i++)
        {
         
          // Find the absolute difference
          // between the characters
          var D1 =
            Math.max(s[i].charCodeAt(0), s[len - 1 - i].charCodeAt(0)) -
            Math.min(s[i].charCodeAt(0), s[len - 1 - i].charCodeAt(0));
 
          var D2 = 26 - D1;
 
          // Adding the minimum difference
          // of the two result
          result += Math.min(D1, D2);
        }
 
        // Return the result
        return result;
      }
 
      // Driver Code
       
      // Given string
      var s = "abccdb";
       
      // Function Call
      document.write(minOperations(s));
       
      // This code is contributed by rdtank.
    

输出：

3

时间复杂度： O(N)
辅助空间： O(N)