Find the minimum no of operations to convert a string into a palindrome?

Question

For eg: z is changed to y & not y to z ....the value of alphabets are changed from right to left ( eg.in the word apple ,the value of 'e' is reduced to 'd' until it becomes a palindrome or when the value becomes 'a'.

Comments

Can you add some example i.e. what is the input and what is the expected output.
what I can get is one operation is from e to a (for the last character) and another is p to l (for the second character) as l to p is not allowed so number of operation is two only. But I am not clear about my understanding.

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Eg: if the input is "abcd" ...we have to change  d->c then the presently changed c is changed from c->b & then again fromb->a ....then the 2nd alphabet from right is coconsidered(c) ...c->b ...so the word becones abba (palindrome) ...no of operations is 4

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The value of the alphabets can be reduced by 1unit @ a time ( from e to d  ,and then d can be changed to c & so on until the alphabet is 'a' or the word is a palindrome.

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Got it, first we need to find the characters which to be changed and take the sum of the diff.

Let me work on the code by tomorrow :)

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:-) thank u so much Sir :-)

Answer 1

main()
{
    char *input="apple";
    int length=strlen(input);
    int i,j;

    int cost = 0;

    for(i=0;i<length/2;i++)
    {
       if (input[i] != input[length-1-i])
       {
           if (input[i] < input[length-1-i])
           {
             printf("%c to be changed %c\n", input[length-1-i], input[i]);
             cost += input[length-1-i] - input[i];
           }
           else
           {
             printf("%c to be changed %c\n", input[i], input[length-1-i]);
             cost += input[i] - input[length-1-i];         
           }
       }
    }
    printf("Total cost to make it palindrome - %d\n", cost);
}

Answer 2

main()
{
           char c[ ]="palindrom";
           int i=0,l,sum=0;
           l=strlen(c);
           l--;
           while(i<=l)
           {
                 if(a[i]<a[l]) {
                        printf("Total operation to convert %c to %c is %d \n",a[l],a[i],a[l]-a[i]);
                        sum+=(a[l--]-a[i++]);
                  }
                  else {
                        printf("Total operation to convert %c to %c is %d \n",a[i],a[l],a[i]-a[l]);
                        sum+=(a[i++]-a[l--]);
                   }
          }
         printf("\nTotal No. of operation= %d",sum);
}