Sometimes, while working with Python strings, we can have a problem in which we need to replace all occurrences of a substring with other.
Input : test_str = “neveropen” s1 = “Lazyroar” s2 = “abcd”
Output : test_str = “abcdforabcd” Explanation : We replace all occurrences of s1 with s2 in test_str.Input : test_str = “neveropen” s1 = “for” s2 = “abcd”
Output : test_str = “LazyroarabcdLazyroar”
Approach 1
We can use inbuilt function replace present in python3 to replace all occurrences of substring.
Implementation using the inbuilt function:-
Python3
#Python has inbuilt function replace to replace all occurrences of substring.input_string = "neveropen"s1 = "Lazyroar"s2 = "abcd"input_string = input_string.replace(s1, s2)print(input_string) |
abcdforabcd
Time Complexity: O(n)
Auxiliary Space: O(n)
Approach 2:
Splitting the string by substring and then replacing with the new string.split() function is used.
Python3
#code for replacing all occurrences of substring s1 with new string s2test_str="neveropen"s1="Lazyroar"s2="abcd"#string split by substrings=test_str.split(s1)new_str=""for i in s: if(i==""): new_str+=s2 else: new_str+=i#printing the replaced stringprint(new_str)#contributed by Bhavya Koganti |
abcdforabcd
The Time and Space Complexity for all the methods are the same:
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3: Another approach to replace all occurrences of a substring in a string is to use the re.sub() function from the re module in python.
Python3
import redef replace_substring(test_str, s1, s2): # Replacing all occurrences of substring s1 with s2 test_str = re.sub(s1, s2, test_str) return test_str# testtest_str = "neveropen"s1 = "Lazyroar"s2 = "abcd"print(replace_substring(test_str, s1, s2)) |
abcdforabcd
Time Complexity: O(n), where n is the length of the input string. This is because the re.sub() function iterates through the entire input string and performs a regular expression match on each character to find all occurrences of the substring. The number of iterations is directly proportional to the length of the input string.
Auxiliary Space:New
Method 4: Using simple iteration
The idea behind this approach is to iterate through the input string character by character and check if each substring of length m matches the substring we want to replace. If it does, we add the replacement substring to our result and move the pointer forward by m characters. If it doesn’t match, we add the current character to the result and move the pointer forward by 1 character.
Python3
def replace_substring(test_str, s1, s2): # Initialize an empty string to store the result result = "" # Initialize a variable to keep track of our position in the string i = 0 # Loop through the string one character at a time while i < len(test_str): # Check if the current substring matches the substring we want to replace if test_str[i:i+len(s1)] == s1: # If it does, add the replacement substring to the result and move the pointer forward result += s2 i += len(s1) else: # If it doesn't, add the current character to the result and move the pointer forward result += test_str[i] i += 1 # Return the final result return result# testtest_str = "neveropen"s1 = "Lazyroar"s2 = "abcd"print(replace_substring(test_str, s1, s2)) |
abcdforabcd
Time complexity: O(nm), where n is the length of the input string and m is the length of the substring to be replaced.
Auxiliary space: O(n), since we are creating a new string to store the result.
