Software Engineer

Traverse the string , if the (int)s[i] is >= 97 , in that case the character is sure to be a small character as ASCII code of small 'a' is 97 & thus make , s[i]=s[i]-97+65 as 65 is ASCII code of capital 'A' . By this was we'd be able to convert all small letters to capitals.

This is a very basic common question , which involves comparing the string & reverse of the same string. If both turn out to be same , then the string is a palindrome.

Initially , I just told that we'll take a copy of string & reverse the copy & compare both the strings. If both come out to be same , then it is a palindrome , else not.
But since , this solution takes O(n) extra space where n is size of string , the interviewer was not happy and wanted me to solve it with O(1) space.

So , this has a very simple solution that we need to traverse till half of the string & at each step , compare s[i] with s[n-i-1] which would compare the corresponding characters. For eg. First character & last character , 2nd character & 2nd last character & so on. If at any place both comparing characters are not equal , then we need to return false , that the given string is not palindrome.