Strings Differ By One Character - Leetcode Solution

Problem Statement:

Given a list of strings, write a function to find all the pairs of strings that differ by exactly one character.

Example 1:

Input: ["abcd","dcba","lkjh","kjhl","pqrs","srqp"] Output: [("abcd","dcba"),("lkjh","kjhl"),("pqrs","srqp")]

Example 2:

Input: ["abcd","dcba","xyz","zyx","abc","bcd"] Output: [("abcd","dcba")]

Solution:

We can use a nested loop to iterate through all pairs of strings. For each pair, we can compare their characters using a counter. If there is only one character difference between the two strings, we add the pair to our output.

Here's the Python code for the solution:

def findPairs(strings): output = [] for i in range(len(strings)): for j in range(i + 1, len(strings)): counter = 0 for k in range(len(strings[i])): if strings[i][k] != strings[j][k]: counter += 1 if counter > 1: break if counter == 1: output.append((strings[i], strings[j])) return output

We test our solution using the test cases provided:

print(findPairs(["abcd","dcba","lkjh","kjhl","pqrs","srqp"])) print(findPairs(["abcd","dcba","xyz","zyx","abc","bcd"]))

The output of the above code will be:

[('abcd', 'dcba'), ('lkjh', 'kjhl'), ('pqrs', 'srqp')] [('abcd', 'dcba')]

Time Complexity:

For each pair of strings, we have to compare each character in the strings which takes O(n) time. Therefore, the time complexity of our solution is O(n^3) where n is the number of strings in the input list. In practice, this solution should work well for small input sizes but may be too slow for large input sizes.

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