Python program to count the pairs of reverse strings
Last Updated :
18 May, 2023
Given the String list, write a Python program to count pairs of reverse strings.
Examples:
Input : test_list = [“geeks”, “best”, “tseb”, “for”, “skeeg”]
Output : 2
Explanation : geeks, skeeg and best, tseb are 2 pairs of reverse strings available.
Input : test_list = [“geeks”, “best”, “for”, “skeeg”]
Output : 1
Explanation : geeks, skeeg is 1 pair of reverse strings available.
Method #1 : Using reversed() + loop
In this, we perform task of comparing strings to reversed versions of it using reversed() and conditional statements. In this, loop is used for task of comparing with each string for pairing.
Python3
test_list = ["geeks", "best", "tseb", "for", "skeeg"]
print("The original list is : " + str(test_list))
res = 0
for idx in range(0, len(test_list)):
for idxn in range(idx, len(test_list)):
if test_list[idxn] == str(''.join(list(reversed(test_list[idx])))):
res += 1
print("Reversed Pairs : " + str(res))
|
Output
The original list is : ['geeks', 'best', 'tseb', 'for', 'skeeg']
Reversed Pairs : 2
Output:
The original list is : [‘geeks’, ‘best’, ‘tseb’, ‘for’, ‘skeeg’] Reversed Pairs : 2
Time Complexity: O(n*n)
Auxiliary Space: O(n)
Method #2 : Using list comprehension + sum()
In this, list comprehension handles nested loop and sum() handles part of increment counter for counting pairs.
Python3
test_list = ["geeks", "best", "tseb", "for", "skeeg"]
print("The original list is : " + str(test_list))
res = sum([1 for idx in range(0, len(test_list)) for idxn in range(idx, len(
test_list)) if test_list[idxn] == str(''.join(list(reversed(test_list[idx]))))])
print("Reversed Pairs : " + str(res))
|
Output
The original list is : ['geeks', 'best', 'tseb', 'for', 'skeeg']
Reversed Pairs : 2
Output:
The original list is : [‘geeks’, ‘best’, ‘tseb’, ‘for’, ‘skeeg’] Reversed Pairs : 2
Time Complexity: O(n2)
Auxiliary Space: O(n)
Approach 3: Using hashmap (dictionary)
The given code is using a hashmap (dictionary) to count the pairs of reverse strings in a given list of strings. The approach works as follows:
- Initialize an empty hashmap (dictionary) and a variable to count the number of pairs.
- Loop through each word in the list of strings.
- Reverse the current word.
- Check if the reversed word already exists in the hashmap.
- If it exists, increment the count of pairs.
- Add the current word to the hashmap.
- Print the count of pairs.
Python3
test_list = ["geeks", "best", "tseb", "for", "skeeg"]
print("The original list is : " + str(test_list))
hashmap = {}
res = 0
cnt = 0
for word in test_list:
rev_word = word[::-1]
if rev_word in hashmap:
cnt+=1
if not word in hashmap:
hashmap[word] = 1
print("Reversed Pairs : " + str(cnt))
|
Output
The original list is : ['geeks', 'best', 'tseb', 'for', 'skeeg']
Reversed Pairs : 2
Time complexity: O(n), where n is the number of words in the list. This is because the loop iterates through each word once, and each dictionary operation (inserting a key-value pair or checking if a key exists) takes constant time on average. The space complexity of this approach is also O(n), since the hashmap can contain at most n key-value pairs.
The approach is justified because it is simple and efficient for the problem at hand. It avoids the need for nested loops or sorting the list, both of which would result in higher time complexity. The use of a hashmap also allows for constant-time lookup of keys, making it a good choice for this problem.
Approach 5: Using sets
Stepwise Approach:
- Initialize a variable count to 0 to count the reversed pairs.
- Initialize an empty set.
- Loop through each word in the list and for each word:
a. Get the reversed version of the current word.
b. Check if the reversed version is in the set.
c. If the reversed version is in the set, increment the count variable.
d. Add the current word to the set.
- Print the count variable as the result.
Python3
test_list = ["geeks", "best", "tseb", "for", "skeeg"]
print("The original list is : " + str(test_list))
seen = set()
cnt = 0
for word in test_list:
rev_word = word[::-1]
if rev_word in seen:
cnt += 1
seen.add(word)
print("Reversed Pairs : " + str(cnt))
|
Output
The original list is : ['geeks', 'best', 'tseb', 'for', 'skeeg']
Reversed Pairs : 2
Time complexity: O(n), where n is the length of the input list since we loop through the list once.
Auxiliary space: O(n), where n is the length of the input list since we need to store each word in the set.
Method 5: using a two-pointer approach
Python3
test_list = ["geeks", "best", "tseb", "for", "skeeg"]
print("The original list is: " + str(test_list))
cnt = 0
for i in range(len(test_list)):
word = test_list[i]
rev_word = word[::-1]
for j in range(i+1, len(test_list)):
if rev_word == test_list[j]:
cnt += 1
print("Reversed Pairs: " + str(cnt))
|
Output
The original list is: ['geeks', 'best', 'tseb', 'for', 'skeeg']
Reversed Pairs: 2
Time complexity: O(n^2), where n is the number of words in the list.
Auxiliary space: O(1) because it uses a constant amount of extra space.
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