Length of the longest substring without repeating characters

Source

Given a string, find the length of the longest substring without repeating characters. For example, the longest substrings without repeating characters for “ABDEFGABEF” are “BDEFGA” and “DEFGAB”, with length 6. For “BBBB” the longest substring is “B”, with length 1. For “GEEKSFORGEEKS”, there are two longest substrings shown in the below diagrams, with length 7.



The desired time complexity is O(n) where n is the length of the string.
Method 1 (Brute Force)
Let us cover the Brute-Force solution as the first method, no one would like this method though. The Brute-Force solution is to generate all possible substrings, find substrings with all unique characters and return the length of longest such substring. Time complexity of such a solution will be exponential which is not practical.
Method 2 (Linear Time)
Let us talk about the linear time solution now. This solution uses extra space to store the last indexes of already visited characters. The idea is to scan the string from left to right, keep track of the maximum length Non-Repeating Character Substring (NRCS) seen so far. Let the maximum length be max_len. When we traverse the string, we also keep track of length of the current NRCS using cur_len variable. For every new character, we look for it in already processed part of the string (A temp array called visited[] is used for this purpose). If it is not present, then we increase the cur_len by 1. If present, then there are two cases:
a) The previous instance of character is not part of current NRCS (The NRCS which is under process). In this case, we need to simply increase cur_len by 1.
b) If the previous instance is part of the current NRCS, then our current NRCS changes. It becomes the substring staring from the next character of previous instance to currently scanned character. We also need to compare cur_len and max_len, before changing current NRCS (or changing cur_len).
Implementation

#include<stdlib.h> #include<stdio.h> #define NO_OF_CHARS 256   int min(int a, int b);   int longestUniqueSubsttr(char *str) {     int n = strlen(str);     int cur_len = 1;  // To store the lenght of current substring     int max_len = 1;  // To store the result     int prev_index;  // To store the previous index     int i;     int *visited = (int *)malloc(sizeof(int)*NO_OF_CHARS);       /* Initialize the visited array as -1, -1 is used to indicate that        character has not been visited yet. */     for (i = 0; i < NO_OF_CHARS;  i++)         visited[i] = -1;       /* Mark first character as visited by storing the index of first        character in visited array. */     visited[str[0]] = 0;       /* Start from the second character. First character is already processed        (cur_len and max_len are initialized as 1, and visited[str[0]] is set */     for (i = 1; i < n; i++)     {         prev_index =  visited[str[i]];           /* If the currentt character is not present in the already processed            substring or it is not part of the current NRCS, then do cur_len++ */         if (prev_index == -1 || i - cur_len > prev_index)             cur_len++;           /* If the current character is present in currently considered NRCS,            then update NRCS to start from the next character of previous instance. */         else         {             /* Also, when we are changing the NRCS, we should also check whether               length of the previous NRCS was greater than max_len or not.*/             if (cur_len > max_len)                 max_len = cur_len;               cur_len = i - prev_index;         }           visited[str[i]] = i; // update the index of current character     }       // Compare the length of last NRCS with max_len and update max_len if needed     if (cur_len > max_len)         max_len = cur_len;       free(visited); // free memory allocated for visited       return max_len; }   /* A utility function to get the minimum of two integers */ int min(int a, int b) {     return (a>b)?b:a; }   /* Driver program to test above function */ int main() {     char str[] = "ABDEFGABEF";     printf("The input string is %s \n", str);     int len =  longestUniqueSubsttr(str);     printf("The length of the longest non-repeating character substring is %d", len);       getchar();     return 0; }

Output

The input string is ABDEFGABEF
  The length of the longest non-repeating character substring is 6

Time Complexity: O(n + d) where n is length of the input string and d is number of characters in input string alphabet. For example, if string consists of lowercase English characters then value of d is 26.
Auxiliary Space: O(d)
Algorithmic Paradigm: Dynamic Programming