String

Searching a string inside another string is a very easy task. Given a string A and a string B, start a loop over A from left to right and, for every letter, start an internal loop to see if there is a match with the letters of B. A ball of string Pseudo code would look something like this: 1 2 3 4 5 6 7 function NaiveSearch(string s[1..n], string sub[1..m]) for i from 1 to n-m+1 for j from 1 to m if s[i+j-1] ≠ sub[j] jump to next iteration of outer loop return i return not found Time Complexity? O(n * m), where n is the size of A and m is the size of B. ...

My colleague and friend Luca (@lucabox) described a better solution to the problem of "Finding the first non repearing char in a string in O(n) time and O(1) space". It uses smartly the space, making the solution nicer and slicker. Or we are just 2 geeks that need to give more attention to their girlfriends :P Luca’s solution description The logic of this solution is based on the usage of an array of unsigned chars. Every char (assumed to be lowecase) has an associated small byte (1 char = 8 bits), where the bits 0x1 and 0x2 (the 2 least significant) represents, respectively, “present once in the input string” and “present multiple times in the input string”. After the input is “scanned” once, and every letter is marked with the correspondent “presence bit” (once, multiple or none), it get’s scanned a second time to find the first char of the input which has the bit “present once” set to “1”. ...