Problem Description

The problem introduces a concept called "close" strings. Two strings are "close" if one can be obtained from the other by performing any of the following operations, any number of times:

• Operation 1: Swap any two existing characters in the string. For example, changing "abce" to "aebc" by swapping 'b' with 'e'.
• Operation 2: Transform every occurrence of one existing character into another existing character and vice versa. For instance, changing "aacabb" to "bbcbaa" by swapping all 'a' characters with 'b' characters, and all 'b' characters with 'a' characters.

The goal is to determine if two given strings, word1 and word2, can be considered "close" by applying these operations.

Intuition

To solve this problem, we need to understand that the operations allowed don't change the frequency of characters, only their positions or representations. Therefore, two "close" strings must have the same set of characters and the same frequency of each character, although the characters themselves can be different.

The crucial realization is that operation 1 allows us to reorder characters in any fashion, making the relative order of characters inconsequential. Operation 2 allows us to transform characters into each other, given that both characters exist in both strings. The consequence of this is:

• Both word1 and word2 must contain the same unique characters - they must have the same set of keys in their character counts (Counter).
• Both word1 and word2 must have the same character frequencies, which implies, after sorting their frequency counts, these should match.

With these constraints in mind, the solution approach is to count the frequency of each character in word1 and word2 using Python's Counter, then compare the sorted values of the counters to check for matching frequencies, as well as compare the sets of keys to ensure that both words contain the same unique characters.

If both the sorted values of the counts and the sets of unique characters match, we return true. Otherwise, we return false.

Learn more about Sorting patterns.

Solution Approach

The solution takes the following approach:

1. Use Python's Counter from the collections module to count the frequency of each character in both word1 and word2. Counter is a dictionary subclass that's designed to count hashable objects. It's a collection where elements are stored as dictionary keys and their counts are stored as dictionary values.

2. Check if the set of keys from both Counters is the same—set(cnt1.keys()) == set(cnt2.keys()). This checks whether word1 and word2 contain exactly the same unique characters. This is a requirement because operation 2 can only be performed if a character exists in both strings.

3. Compare the sorted values of both counters—sorted(cnt1.values()) == sorted(cnt2.values()). This step checks if word1 and word2 have the same frequency of characters (the same count of each character). Sorting is crucial here since we need to compare frequencies regardless of the character they are associated with.

If both conditions are met, then word1 and word2 are "close", and the function returns True. If any of the conditions is not met, the function returns False. These comparisons effectively implement the constraints of the two operations without having to manually perform the operations themselves.

Here's a step-by-step breakdown of how the provided solution achieves this:

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        # Count the frequency of each character in both words
        cnt1, cnt2 = Counter(word1), Counter(word2)
      
        # If the sets of unique characters (keys of the counters) are the same and
        # the sorted lists of character counts (values of the counters) are the same,
        # then the strings are "close"
        return sorted(cnt1.values()) == sorted(cnt2.values()) and set(cnt1.keys()) == set(cnt2.keys())

The algorithmic complexity of this solution is mainly dependent on the sorting of the counted values, which is O(n log n) where n is the number of unique characters. Comparing the sets of keys essentially happens in linear time, O(m), where m is the length of the string.

Example Walkthrough

Let's take an example to illustrate the solution approach with two strings, word1 = "baba" and word2 = "abab".

1. We first count the frequency of each character in both words. The Counter for word1 is { 'b': 2, 'a': 2 } and for word2 is { 'a': 2, 'b': 2 }. This shows that both 'a' and 'b' appear twice in both strings.

2. We then compare the sets of keys from both dictionaries—set(cnt1.keys()) == set(cnt2.keys()). For our example, this means comparing { 'b', 'a' } == { 'a', 'b' }, which evaluates to True because the set operation takes care of the order, and both sets contain the same elements.

3. For the final step, we compare the sorted values of both counters—sorted(cnt1.values()) == sorted(cnt2.values()). In this case, sorted([2, 2]) == sorted([2, 2]). The sorted list of character counts for both word1 and word2 is [2, 2], which means that both strings have the same frequency for their characters.

Since both conditions are satisfied, we conclude that word1 and word2 are indeed "close" as per the problem's definition. The function would return True.

In a case where word1 = "baba" and word2 = "abbc", this is how the approach would result in a False:

1. Counter(word1) gives { 'b': 2, 'a': 2 } and Counter(word2) gives { 'a': 1, 'b': 2, 'c': 1 }.

2. When we compare the sets of the keys { 'b', 'a' } and { 'a', 'b', 'c' }, we get False because word2 contains an extra character 'c'.

Since the sets of keys don't match, we don't even need to compare the values. The function would return False, indicating word1 and word2 are not "close".

Solution Implementation

Time and Space Complexity

The given Python function closeStrings determines if two words can be made equal by swapping the positions of the letters. This is done by checking two conditions: the sorted values of the character counters for both words are the same and the sets of characters (keys of the counters) in both words are the same.

Time Complexity:

• Creating the counters for word1 and word2 has a time complexity of O(N + M), where N is the length of word1 and M is the length of word2. This is because each word is traversed once to create the counts of characters.
• Sorting the values of the counters has a time complexity of O(NlogN + MlogM) in the worst case, because the time complexity of sorting n elements is O(nlogn), and we are sorting the frequency counts of both words.
• Comparing two sorted lists of values has a time complexity of O(N + M) in the worst case, assuming N and M are the numbers of unique characters in word1 and word2, respectively.
• Comparing two sets of keys has a time complexity of O(N + M), because sets are typically implemented as hash tables, and comparing two sets involves checking that every element of one set is in the other (and vice versa), which takes O(N + M) time in this case.

Considering the most expensive operations, the overall time complexity is O(NlogN + MlogM), since the logarithmic factors in sorting dominates the linear factors from other operations.

Space Complexity:

• The counters for word1 and word2 consume O(N + M) space, as they store counts for the characters present in both word1 and word2.
• Sorting the values of the counters requires O(N + M) space to hold the sorted lists.
• The sets of keys take O(N + M) space as well.

Hence, the overall space complexity is O(N + M), which accounts for the storage of the character counts and the unique characters.

Learn more about how to find time and space complexity quickly using problem constraints.