Python Program To Find Length Of The Longest Substring Without Repeating Characters - GeeksforGeeks

20 Dec, 2021
Given a string str, find the length of the longest substring without repeating characters. 
For “ABDEFGABEF”, the longest substring 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 (Simple : O(n3)): 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 substring 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).
Python3
# Python3 program to find the length# of the longest substring without# repeating characters  # This functionr eturns true if all# characters in strr[i..j] are # distinct, otherwise returns falsedef areDistinct(strr, i, j):      # Note : Default values in visited are false    visited = [0] * (26)      for k in range(i, j + 1):        if (visited[ord(strr[k]) -                    ord('a')] == True):            return False                      visited[ord(strr[k]) -                ord('a')] = True      return True  # Returns length of the longest substring# with all distinct characters.def longestUniqueSubsttr(strr):          n = len(strr)          # Result    res = 0           for i in range(n):        for j in range(i, n):            if (areDistinct(strr, i, j)):                res = max(res, j - i + 1)                      return res  # Driver codeif __name__ == '__main__':          strr = "geeksforgeeks"    print("The input is ", strr)          len = longestUniqueSubsttr(strr)    print("The length of the longest "          "non-repeating character substring is ", len)  # This code is contributed by mohit kumar 29
The input string is geeksforgeeks
The length of the longest non-repeating character substring is 7
Method 2 (Better : O(n2)) The idea is to use window sliding. Whenever we see repetition, we remove the previous occurrence and slide the window.
Python3
# Python3 program to find the # length of the longest substring# without repeating charactersdef longestUniqueSubsttr(str):          n = len(str)          # Result    res = 0        for i in range(n):                   # Note : Default values in         # visited are false        visited = [0] * 256              for j in range(i, n):               # If current character is visited            # Break the loop            if (visited[ord(str[j])] == True):                break               # Else update the result if            # this window is larger, and mark            # current character as visited.            else:                res = max(res, j - i + 1)                visited[ord(str[j])] = True                      # Remove the first character of previous        # window        visited[ord(str[i])] = False          return res  # Driver codestr = "geeksforgeeks"print("The input is ", str)  len = longestUniqueSubsttr(str)print("The length of the longest "       "non-repeating character substring is ", len)  # This code is contributed by sanjoy_62
The input string is geeksforgeeks
The length of the longest non-repeating character substring is 7
Method 4 (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 seen so far in res. When we traverse the string, to know the length of current window we need two indexes. 1) Ending index ( j ) : We consider current index as ending index. 2) Starting index ( i ) : It is same as previous window if current character was not present in the previous window. To check if the current character was present in the previous window or not, we store last index of every character in an array lasIndex[]. If lastIndex[str[j]] + 1 is more than previous start, then we updated the start index i. Else we keep same i.  
Below is the implementation of the above approach :
Python3
# Python3 program to find the length# of the longest substring# without repeating charactersdef longestUniqueSubsttr(string):      # last index of every character    last_idx = {}    max_len = 0      # starting index of current     # window to calculate max_len    start_idx = 0      for i in range(0, len(string)):                # Find the last index of str[i]        # Update start_idx (starting index of current window)        # as maximum of current value of start_idx and last        # index plus 1        if string[i] in last_idx:            start_idx = max(start_idx, last_idx[string[i]] + 1)          # Update result if we get a larger window        max_len = max(max_len, i-start_idx + 1)          # Update last index of current char.        last_idx[string[i]] = i      return max_len    # Driver program to test the above functionstring = "geeksforgeeks"print("The input string is " + string)length = longestUniqueSubsttr(string)print("The length of the longest non-repeating character" +      " substring is " + str(length))
The input string is geeksforgeeks
The length of the longest non-repeating character substring is 7
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) 
Alternate Implementation : 
Python
# Here, we are planning to implement a simple sliding window methodology   def longestUniqueSubsttr(string):           # Creating a set to store the last positions of occurrence    seen = {}    maximum_length = 0       # starting the initial point of window to index 0    start = 0            for end in range(len(string)):           # Checking if we have already seen the element or not        if string[end] in seen:              # If we have seen the number, move the start pointer            # to position after the last occurrence            start = max(start, seen[string[end]] + 1)           # Updating the last seen value of the character        seen[string[end]] = end        maximum_length = max(maximum_length, end-start + 1)    return maximum_length   # Driver Codestring = "geeksforgeeks"print("The input string is", string)length = longestUniqueSubsttr(string)print("The length of the longest non-repeating character substring is", length)
The input String is geeksforgeeks
The length of the longest non-repeating character substring is 7
As an exercise, try the modified version of the above problem where you need to print the maximum length NRCS also (the above program only prints the length of it).
Please refer complete article on Length of the longest substring without repeating characters for more details!
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