chore(Python): add remove duplicate and first non-repeating character string (#312)

algorithms/Python/README.md

@@ -34,3 +34,5 @@
 1. [Is Good Str](strings/is_good_str.py)
 2. [Palindrome](strings/palindrome.py)
 3. [Word Count](strings/word_count.py)
+4. [Remove Duplicates from a String](strings/remove_duplicates_from_a_string.py)
+5. [First Non Repeating Character](strings/first_non_repeating_character.py)

algorithms/Python/strings/first_non_repeating_character.py

@@ -0,0 +1,33 @@
+"""
+Output the first character which occurred once in the string
+I/P: "abcbadabfaa"
+O/P: "c" (-1 if no such character is there)
+
+Time Complexity: O(n)
+Space Complexity: O(26) => O(1)
+"""
+string = "abcbadabfaa"  # Constraint: only smallcase letters are there
+
+def first_non_repeating_character(string):
+    """
+    One way to do this is to have two dictionaries one which keeps the count of character,
+    other dictionary will keep the first appearance of the character (index).
+    After a traversal. Look in first dictionary for characters occurring once and output the one which
+    has the first appearance as per second dictionary.
+
+    Since in this case we have special constraint in mind. We will go with a cleverer way by reducing more stack space.
+    """
+    n = len(string)
+    occurrence = [n]*26
+    for i in range(len(string)):
+        index = ord(string[i])-ord('a')
+        if index < 0 or index > 25:
+            return ("Invalid")
+        if occurrence[index] == n:
+            occurrence[index] = i
+        else:
+            occurrence[index] += n
+    return string[min(occurrence)] if min(occurrence) < n else -1
+
+if __name__ == "__main__":
+    print(first_non_repeating_character(string))

algorithms/Python/strings/remove_duplicates_from_a_string.py

@@ -0,0 +1,31 @@
+"""
+Remove all duplicates from a string
+
+I/P: "bbaaccdefabafaz"
+O/P: "bacdefz"
+Time Complexity: O(n)
+Space Complexity: O(26) (constant)
+"""
+
+string = "bbaaccdefabafaz"  # Constraint: only smallcase letters are there
+
+def remove_duplicate(string):
+    alphabet = "0" * 26 # instead of going for classic use of sets, dictionaries we can use string containing bits. Either way is fine.
+    ans_string = ""
+    for letter in string:
+        index = ord(letter)-ord('a')
+        if index < 0 or index > 25:
+            return ("Invalid")
+        if alphabet[index] == "1":
+            pass
+        else:
+            index = ord(letter)-ord('a')
+            if index == 0:
+                alphabet = "1" + alphabet[1:]
+            else:
+                alphabet = alphabet[0:index] + "1" + alphabet[index+1:]
+            ans_string += letter
+    return ans_string
+
+if __name__ == "__main__":
+    print(remove_duplicate(string))