Length of the longest substring without repeating characters
Method 1 (Brute Force Method)
We can consider all substrings.
One by one and check for each substring whether it contains all unique characters or not.
There will be n*(n+1)/2 substrings.
Whether a substirng contains all unique characters or not can be checked in linear time by scanning it from left to right and keeping a map of visited characters.
Time complexity of this solution would be O(n^3).
Method 2 (Linear Time)
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.
Explanation
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 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)
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);
for (i = 0; i < NO_OF_CHARS; i++)
visited[i] = -1;
visited[str[0]] = 0;
for (i = 1; i < n; i++)
{
prev_index = visited[str[i]];
if (prev_index == -1 || i - cur_len > prev_index)
cur_len++;
else
{
if (cur_len > max_len)
max_len = cur_len;
cur_len = i - prev_index;
}
visited[str[i]] = i; // update the index of current character
}
if (cur_len > max_len)
max_len = cur_len;
free(visited); // free memory allocated for visited
return max_len;
}