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K-Similar Strings

Hard

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StringsGraphsGreedy AlgorithmsArrays

Strings s1 and s2 are k-similar (for some non-negative integer k) if we can swap the positions of two letters in s1 exactly k times so that the resulting string equals s2.

Given two anagrams s1 and s2, return the smallest k for which s1 and s2 are k-similar.

Example 1:

Input: s1 = "ab", s2 = "ba"
Output: 1
Explanation: The two string are 1-similar because we can use one swap to change s1 to s2: "ab" --> "ba".

Example 2:

Input: s1 = "abc", s2 = "bca"
Output: 2
Explanation: The two strings are 2-similar because we can use two swaps to change s1 to s2: "abc" --> "bac" --> "bca".

Constraints:

1 <= s1.length <= 20
s2.length == s1.length
s1 and s2 contain only lowercase letters from the set {'a', 'b', 'c', 'd', 'e', 'f'}.
s2 is an anagram of s1.

Try coding it in LeetCode

K-Similar Strings

Hard

Asked by:

42 views

Topics:

StringsGraphsGreedy AlgorithmsArrays

Strings s1 and s2 are k-similar (for some non-negative integer k) if we can swap the positions of two letters in s1 exactly k times so that the resulting string equals s2.

Given two anagrams s1 and s2, return the smallest k for which s1 and s2 are k-similar.

Example 1:

Input: s1 = "ab", s2 = "ba"
Output: 1
Explanation: The two string are 1-similar because we can use one swap to change s1 to s2: "ab" --> "ba".

Example 2:

Input: s1 = "abc", s2 = "bca"
Output: 2
Explanation: The two strings are 2-similar because we can use two swaps to change s1 to s2: "abc" --> "bac" --> "bca".

Constraints:

1 <= s1.length <= 20
s2.length == s1.length
s1 and s2 contain only lowercase letters from the set {'a', 'b', 'c', 'd', 'e', 'f'}.
s2 is an anagram of s1.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible lengths of the strings s1 and s2? Are there any constraints on the characters that s1 and s2 can contain (e.g., only lowercase English letters)?
If the strings s1 and s2 are already equal, should I return 0?
Is it guaranteed that a solution always exists, meaning that s1 can always be transformed into s2 with some number of swaps?
Are there any specific performance expectations or constraints I should be aware of beyond general efficiency (e.g., memory usage limitations)?
Could you provide a small example to illustrate the expected input and output?

Brute Force Solution

Approach

The brute force approach for K-Similar Strings is to try every possible pair of swaps between the characters of the first string. After each swap, we check if the swapped string is closer to the target string. We continue swapping and checking until we find the minimum number of swaps needed to make the strings equal.

Here's how the algorithm would work step-by-step:

Consider all possible pairings of positions within the first string.
For each pairing, imagine swapping the characters at those positions.
After each swap, compare the swapped string with the second string to see how similar they are now.
Repeat the swapping process, trying all combinations of pairings.
Keep track of the minimum number of swaps performed to achieve a certain level of similarity or an exact match between the strings.
The final result will be the smallest number of swaps needed to make the two strings equal.

Code Implementation

def k_similar_strings_brute_force(string_one, string_two):
    min_swaps = float('inf')

    def calculate_similarity(current_string):
        similarity_score = 0
        for index in range(len(string_one)):
            if current_string[index] == string_two[index]:
                similarity_score += 1
        return similarity_score

    def swap_and_recurse(current_string, swaps_made):
        nonlocal min_swaps

        # If the strings are equal, update min_swaps.
        if current_string == string_two:
            min_swaps = min(min_swaps, swaps_made)
            return

        # Prune if current swaps already exceed the minimum
        if swaps_made >= min_swaps:
            return

        # Iterate through all possible pairs of indices for swapping
        for first_index in range(len(string_one)):
            for second_index in range(first_index + 1, len(string_one)):
                # Create a new string with the characters swapped
                new_string_list = list(current_string)
                new_string_list[first_index], new_string_list[second_index] = \
                    new_string_list[second_index], new_string_list[first_index]
                new_string = "".join(new_string_list)

                # Explore the possibility of more swaps
                swap_and_recurse(new_string, swaps_made + 1)

    # Initiate the brute force search
    swap_and_recurse(string_one, 0)

    return min_swaps

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach considers all possible pairs of swaps within the first string of length n. In the worst-case scenario, we might have to explore all possible permutations of the string. This involves trying all possible swaps which leads to exploring a tree-like structure where each node represents a possible state after a swap. The number of possible permutations grows factorially with the length of the string. Therefore, the time complexity of this approach is O(n!).

Space Complexity

O(1) – The described brute force approach primarily focuses on swapping characters in place and comparing strings. The steps involve considering pairs of positions and swapping characters, which can be done using a constant amount of extra space for variables to store indices and temporary character values during the swaps. No auxiliary data structures that scale with the input size (N, where N is the length of the strings) are used. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The problem asks for the minimum number of swaps needed to make two strings the same. We use a technique that explores possible swaps in a strategic way, prioritizing swaps that fix mismatches efficiently, avoiding unnecessary detours.

Here's how the algorithm would work step-by-step:

First, recognize that if the strings are identical, no swaps are needed, and we're done.
Otherwise, find the first place where the strings differ.
Look at the second string and find a character that, if swapped into that first mismatched position, would make the strings match at that position.
Perform that swap. This fixes one mismatch.
Repeat the process: find the next mismatch and look for a 'fix' in the remaining part of the second string.
Keep doing this until the strings are the same. The number of swaps you performed is the answer.
The key is to always prioritize swaps that directly resolve a mismatch. This avoids exploring unnecessary swap combinations and quickly leads to the minimum number of swaps.

Code Implementation

def k_similarity(string_one, string_two):
    if string_one == string_two:
        return 0

    string_one_list = list(string_one)
    string_two_list = list(string_two)
    swap_count = 0

    string_length = len(string_one)

    for index in range(string_length):
        if string_one_list[index] != string_two_list[index]:

            # Find the correct char to swap into current position
            for subsequent_index in range(index + 1, string_length):
                if string_one_list[subsequent_index] == string_two_list[index] and string_one_list[subsequent_index] != string_two_list[subsequent_index]:

                    # Prioritize fixing a mismatch directly.
                    string_one_list[index], string_one_list[subsequent_index] = string_one_list[subsequent_index], string_one_list[index]

                    swap_count += 1

                    break
                
            #If no optimal swap found, find any valid swap to advance
            else:
                for subsequent_index in range(index + 1, string_length):
                    if string_one_list[subsequent_index] == string_two_list[index]:
                        string_one_list[index], string_one_list[subsequent_index] = string_one_list[subsequent_index], string_one_list[index]
                        swap_count += 1
                        break

    return swap_count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the string of length n to find the first mismatched character. For each mismatched character found, it searches the remaining portion of the string (also up to length n) to find a suitable character to swap that will resolve the mismatch. This nested search results in a time complexity proportional to n * n. Therefore, the overall time complexity is O(n²).

Space Complexity

O(1) – The plain English explanation outlines an iterative process of finding mismatches and swapping characters in place. It only requires storing a couple of index variables to track the positions of the mismatches within the strings. Since the amount of extra memory used remains constant regardless of the length of the strings (N), the space complexity is O(1).

Edge Cases

Case	How to Handle
s1 and s2 are empty strings	Return 0 since no swaps are needed.
s1 and s2 are identical strings	Return 0 since no swaps are needed.
s1 and s2 have length 1	Return 0 if s1 == s2, otherwise 1 if they are different.
s1 and s2 are anagrams but require multiple swaps	Breadth-First Search (BFS) should correctly handle multiple swap scenarios to find the minimum swaps.
s1 and s2 contain many repeated characters	The BFS algorithm can handle repeated chars, ensuring the same character at multiple indices are correctly accounted.
Maximum string length (performance considerations)	Optimize BFS by pruning explored states to prevent excessive memory usage and execution time.
Strings s1 and s2 are of considerable length but differ only in one or two positions	BFS should still efficiently determine the minimum number of swaps needed, as it explores only relevant states.
Inputs with very long string lengths potentially cause memory exhaustion issues.	Consider space optimization techniques like using a set to store visited states to prevent revisiting already explored configurations to reduce space complexities.

s1 and s2 are empty strings

How to Handle:

Return 0 since no swaps are needed.

s1 and s2 are identical strings

How to Handle:

Return 0 since no swaps are needed.

s1 and s2 have length 1

How to Handle:

Return 0 if s1 == s2, otherwise 1 if they are different.

s1 and s2 are anagrams but require multiple swaps

How to Handle:

Breadth-First Search (BFS) should correctly handle multiple swap scenarios to find the minimum swaps.

s1 and s2 contain many repeated characters

How to Handle:

The BFS algorithm can handle repeated chars, ensuring the same character at multiple indices are correctly accounted.

Maximum string length (performance considerations)

How to Handle:

Optimize BFS by pruning explored states to prevent excessive memory usage and execution time.

Strings s1 and s2 are of considerable length but differ only in one or two positions

How to Handle:

BFS should still efficiently determine the minimum number of swaps needed, as it explores only relevant states.

Inputs with very long string lengths potentially cause memory exhaustion issues.

How to Handle:

Consider space optimization techniques like using a set to store visited states to prevent revisiting already explored configurations to reduce space complexities.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible lengths of the strings s1 and s2? Are there any constraints on the characters that s1 and s2 can contain (e.g., only lowercase English letters)?
If the strings s1 and s2 are already equal, should I return 0?
Is it guaranteed that a solution always exists, meaning that s1 can always be transformed into s2 with some number of swaps?
Are there any specific performance expectations or constraints I should be aware of beyond general efficiency (e.g., memory usage limitations)?
Could you provide a small example to illustrate the expected input and output?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider all possible pairings of positions within the first string.
For each pairing, imagine swapping the characters at those positions.
After each swap, compare the swapped string with the second string to see how similar they are now.
Repeat the swapping process, trying all combinations of pairings.
Keep track of the minimum number of swaps performed to achieve a certain level of similarity or an exact match between the strings.
The final result will be the smallest number of swaps needed to make the two strings equal.

Code Implementation

def k_similar_strings_brute_force(string_one, string_two):
    min_swaps = float('inf')

    def calculate_similarity(current_string):
        similarity_score = 0
        for index in range(len(string_one)):
            if current_string[index] == string_two[index]:
                similarity_score += 1
        return similarity_score

    def swap_and_recurse(current_string, swaps_made):
        nonlocal min_swaps

        # If the strings are equal, update min_swaps.
        if current_string == string_two:
            min_swaps = min(min_swaps, swaps_made)
            return

        # Prune if current swaps already exceed the minimum
        if swaps_made >= min_swaps:
            return

        # Iterate through all possible pairs of indices for swapping
        for first_index in range(len(string_one)):
            for second_index in range(first_index + 1, len(string_one)):
                # Create a new string with the characters swapped
                new_string_list = list(current_string)
                new_string_list[first_index], new_string_list[second_index] = \
                    new_string_list[second_index], new_string_list[first_index]
                new_string = "".join(new_string_list)

                # Explore the possibility of more swaps
                swap_and_recurse(new_string, swaps_made + 1)

    # Initiate the brute force search
    swap_and_recurse(string_one, 0)

    return min_swaps

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, recognize that if the strings are identical, no swaps are needed, and we're done.
Otherwise, find the first place where the strings differ.
Look at the second string and find a character that, if swapped into that first mismatched position, would make the strings match at that position.
Perform that swap. This fixes one mismatch.
Repeat the process: find the next mismatch and look for a 'fix' in the remaining part of the second string.
Keep doing this until the strings are the same. The number of swaps you performed is the answer.
The key is to always prioritize swaps that directly resolve a mismatch. This avoids exploring unnecessary swap combinations and quickly leads to the minimum number of swaps.

Code Implementation

def k_similarity(string_one, string_two):
    if string_one == string_two:
        return 0

    string_one_list = list(string_one)
    string_two_list = list(string_two)
    swap_count = 0

    string_length = len(string_one)

    for index in range(string_length):
        if string_one_list[index] != string_two_list[index]:

            # Find the correct char to swap into current position
            for subsequent_index in range(index + 1, string_length):
                if string_one_list[subsequent_index] == string_two_list[index] and string_one_list[subsequent_index] != string_two_list[subsequent_index]:

                    # Prioritize fixing a mismatch directly.
                    string_one_list[index], string_one_list[subsequent_index] = string_one_list[subsequent_index], string_one_list[index]

                    swap_count += 1

                    break
                
            #If no optimal swap found, find any valid swap to advance
            else:
                for subsequent_index in range(index + 1, string_length):
                    if string_one_list[subsequent_index] == string_two_list[index]:
                        string_one_list[index], string_one_list[subsequent_index] = string_one_list[subsequent_index], string_one_list[index]
                        swap_count += 1
                        break

    return swap_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
s1 and s2 are empty strings	Return 0 since no swaps are needed.
s1 and s2 are identical strings	Return 0 since no swaps are needed.
s1 and s2 have length 1	Return 0 if s1 == s2, otherwise 1 if they are different.
s1 and s2 are anagrams but require multiple swaps	Breadth-First Search (BFS) should correctly handle multiple swap scenarios to find the minimum swaps.
s1 and s2 contain many repeated characters	The BFS algorithm can handle repeated chars, ensuring the same character at multiple indices are correctly accounted.
Maximum string length (performance considerations)	Optimize BFS by pruning explored states to prevent excessive memory usage and execution time.
Strings s1 and s2 are of considerable length but differ only in one or two positions	BFS should still efficiently determine the minimum number of swaps needed, as it explores only relevant states.
Inputs with very long string lengths potentially cause memory exhaustion issues.	Consider space optimization techniques like using a set to store visited states to prevent revisiting already explored configurations to reduce space complexities.

s1 and s2 are empty strings

How to Handle:

Return 0 since no swaps are needed.

s1 and s2 are identical strings

How to Handle:

Return 0 since no swaps are needed.

s1 and s2 have length 1

How to Handle:

Return 0 if s1 == s2, otherwise 1 if they are different.

s1 and s2 are anagrams but require multiple swaps

How to Handle:

Breadth-First Search (BFS) should correctly handle multiple swap scenarios to find the minimum swaps.

s1 and s2 contain many repeated characters

How to Handle:

The BFS algorithm can handle repeated chars, ensuring the same character at multiple indices are correctly accounted.

Maximum string length (performance considerations)

How to Handle:

Optimize BFS by pruning explored states to prevent excessive memory usage and execution time.

Strings s1 and s2 are of considerable length but differ only in one or two positions

How to Handle:

BFS should still efficiently determine the minimum number of swaps needed, as it explores only relevant states.

Inputs with very long string lengths potentially cause memory exhaustion issues.

How to Handle:

Consider space optimization techniques like using a set to store visited states to prevent revisiting already explored configurations to reduce space complexities.