Given two strings, extract indices of all characters from string 1 which are present in the other string, but not in the same index.
Input : test_str1 = ‘pplg’, test_str2 = ‘pineapple’
Output : [0, 1, 2]
Explanation : ppl is found in 2nd string, also not on same index as 1st.Input : test_str1 = ‘pine’, test_str2 = ‘pineapple’
Output : []
Explanation : Found in other string on same index.
Method #1 : Using enumerate() + loop
In this, we employ a nested loop to check for each character its occurrence in 2nd string, and then if it’s in other position, if found the index is appended.
Python3
# Python3 code to demonstrate working of # Extract indices of Present, Non Index matching Strings# using loop + enumerate()# initializing stringstest_str1 = 'apple'test_str2 = 'pineapple'# printing original Stringsprint("The original string 1 is : " + str(test_str1))print("The original string 2 is : " + str(test_str2))# the replaced result res = []for idx, val in enumerate(test_str1): # if present in string 2 if val in test_str2: # if not present at same index if test_str2[idx] != val: res.append(idx)# printing result print("The extracted indices : " + str(res)) |
The original string 1 is : apple The original string 2 is : pineapple The extracted indices : [0, 1, 2, 3, 4]
Method #2 : Using enumerate() + zip() + list comprehension
In this, we perform the task of getting indices using enumerate() and pairing of both strings is done using zip(), the conditional checks occur using list comprehension.
Python3
# Python3 code to demonstrate working of# Extract indices of Present, Non Index matching Strings# using enumerate() + zip() + list comprehension# initializing stringstest_str1 = 'apple'test_str2 = 'pineapple'# printing original Stringsprint("The original string 1 is : " + str(test_str1))print("The original string 2 is : " + str(test_str2))# one-liner to solve this problem.res = [idx for idx, (x, y) in enumerate( zip(test_str1, test_str2)) if x != y and x in test_str2]# printing resultprint("The extracted indices : " + str(res)) |
The original string 1 is : apple The original string 2 is : pineapple The extracted indices : [0, 1, 2, 3, 4]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #3: Use a for loop
- Initialize two empty lists, one for storing indices of present strings and another for storing indices of non-index matching strings.
- Iterate over each character and its index simultaneously using the built-in enumerate function.
- If the character from the first string is in the second string, append its index to the present strings list.
- If the characters at the same index from both strings are different, append the index to the non-index matching strings list.
- Print the final results.
Python3
# Python3 code to demonstrate an alternative approach for# Extracting indices of Present, Non Index matching Strings# initializing stringstest_str1 = 'apple'test_str2 = 'pineapple'# printing original Stringsprint("The original string 1 is : " + str(test_str1))print("The original string 2 is : " + str(test_str2))# initialize two empty listspresent_strs = []non_index_matching_strs = []# iterate over each character and its index using enumeratefor idx, char in enumerate(test_str1): # check if character is present in the second string if char in test_str2: present_strs.append(idx) # check if characters at the same index are different if char != test_str2[idx]: non_index_matching_strs.append(idx)# printing resultsprint("Indices of Present strings : ", present_strs)print("Indices of Non Index matching strings : ", non_index_matching_strs) |
The original string 1 is : apple The original string 2 is : pineapple Indices of Present strings : [0, 1, 2, 3, 4] Indices of Non Index matching strings : [0, 1, 2, 3, 4]
Time complexity: O(n), where n is the length of the first string.
Auxiliary space: O(k), where k is the number of indices that satisfy the conditions.
Method #4: Using a set intersection
- Create two strings test_str1 and test_str2.
- Print the original strings test_str1 and test_str2.
- Find the common characters between the two strings using set intersection and store them in the common_chars set.
- Create a list present_strs which stores the indices of characters in test_str1 that are present in common_chars.
- Create a list of non_index_matching_strs which stores the indices of characters in test_str1 that are not present in test_str2 or are not in the same position in both strings.
- Print the result by displaying the present_strs and non_index_matching_strs
Python3
test_str1 = 'apple'test_str2 = 'pineapple'# printing original Stringsprint("The original string 1 is : " + str(test_str1))print("The original string 2 is : " + str(test_str2))# using set intersection and list comprehensioncommon_chars = set(test_str1) & set(test_str2)present_strs = [idx for idx, char in enumerate(test_str1) if char in common_chars]non_index_matching_strs = [idx for idx, char in enumerate(test_str1) if char not in test_str2 or char != test_str2[idx]]# printing resultsprint("Indices of Present strings : ", present_strs)print("Indices of Non Index matching strings : ", non_index_matching_strs) |
The original string 1 is : apple The original string 2 is : pineapple Indices of Present strings : [0, 1, 2, 3, 4] Indices of Non Index matching strings : [0, 1, 2, 3, 4]
Time complexity: O(n), where n is the length of the first string test_str1.
Auxiliary space: O(n), as the present_strs and non_index_matching_strs lists can potentially store up to n elements each.
Method #5: set comprehension and zip() functions
Python3
# Python3 code to demonstrate another alternative approach for# Extracting indices of Present, Non Index matching Strings# initializing stringstest_str1 = 'apple'test_str2 = 'pineapple'# printing original Stringsprint("The original string 1 is : " + str(test_str1))print("The original string 2 is : " + str(test_str2))# using set comprehension to get the indices of present characterspresent_strs = {idx for idx, char in enumerate(test_str1) if char in test_str2}# using set comprehension to get the indices of non-index matching charactersnon_index_matching_strs = {idx for idx, (char1, char2) in enumerate(zip(test_str1, test_str2)) if char1 != char2}# printing resultsprint("Indices of Present strings : ", present_strs)print("Indices of Non Index matching strings : ", non_index_matching_strs) |
The original string 1 is : apple
The original string 2 is : pineapple
Indices of Present strings : {0, 1, 2, 3, 4}
Indices of Non Index matching strings : {0, 1, 2, 3, 4}
Time complexity: O(n), where n is the length of the input strings.
Auxiliary space: O(n)
