#854

Hard

LeetCode Problem Workspace

K-Similar Strings

K-Similar Strings involves finding the minimal number of swaps to transform one string into another through character swaps.

hash table string breadth first search

Problem Statement

Given two anagrams s1 and s2, the task is to find the smallest integer k such that s1 can be transformed into s2 using exactly k swaps of characters in s1. These swaps can only change the positions of two letters at a time, and the result must match the target string s2.

The challenge lies in efficiently determining the minimum number of swaps, leveraging string manipulation techniques along with hash table strategies. Since s1 and s2 are anagrams, they have the same characters, but in different orders. The objective is to find the smallest k to perform the necessary transformations.

Examples

Example 1

Input: s1 = "ab", s2 = "ba"

Output: 1

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

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.

Solution Approach

Breadth-First Search (BFS)

Breadth-First Search (BFS) is a good fit for finding the minimum swap count because BFS explores all possible transformations level by level. Starting from s1, we can use BFS to simulate each swap, progressively transforming s1 toward s2, ensuring the minimal swap count is found as soon as s2 is reached.

Hash Table for Optimizing State Transitions

A hash table can be used to store previously seen states of s1, preventing redundant calculations. This will help track the transformations already explored during the BFS, thus ensuring that the solution remains efficient by reducing unnecessary recalculations.

String Comparison and Swap Optimization

Comparing strings at each level of the BFS allows for immediate identification of the target string s2. The optimization comes from targeting only those swaps that will bring s1 closer to s2, minimizing redundant moves and ensuring that we don't perform unnecessary swaps.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach used for BFS traversal and how efficiently states are stored and processed. In the worst case, the space complexity is proportional to the number of unique permutations of the string, while the time complexity also grows depending on the number of states processed during BFS traversal.

What Interviewers Usually Probe

Candidate demonstrates an understanding of BFS for finding the shortest path in a graph.
Candidate applies optimization techniques using hash tables to avoid redundant state processing.
Candidate effectively uses string manipulation techniques to reduce unnecessary swaps.

Common Pitfalls or Variants

Common pitfalls

Overlooking the importance of BFS for level-based state exploration could lead to an inefficient solution.
Failing to use a hash table to track visited states will result in redundant calculations and higher time complexity.
Not considering early termination when the target string is reached can lead to unnecessary calculations and slow performance.

Follow-up variants

What if the two strings are already equal? In this case, the solution should immediately return 0 without further computation.
Consider the case where only one swap is needed: this will test if the algorithm handles minimal cases effectively.
What if there are repeated characters? Ensure that the algorithm handles string transformations correctly, even when characters appear multiple times.

FAQ

What is the minimum number of swaps for K-Similar Strings?

The minimum number of swaps required is the smallest integer k such that s1 can be transformed into s2 using exactly k swaps of characters.

How does the BFS approach work for K-Similar Strings?

BFS explores all possible transformations level by level, ensuring the minimal swap count is found when s2 is reached.

Can K-Similar Strings be solved without BFS?

While BFS is the most efficient way to find the minimal number of swaps, other techniques like greedy methods or recursive backtracking could also be used, but they may not be as optimal.

What role does the hash table play in the solution?

The hash table stores previously seen states of s1, helping to avoid redundant calculations and ensuring efficient traversal of the string transformations.

What are the constraints of the K-Similar Strings problem?

The input strings s1 and s2 are anagrams of each other, with lengths between 1 and 20, and consist of lowercase letters from {'a', 'b', 'c', 'd', 'e', 'f'}.

terminal

Solution

Solution 1

#### Python3

class Solution:
    def kSimilarity(self, s1: str, s2: str) -> int:
        def next(s):
            i = 0
            while s[i] == s2[i]:
                i += 1
            res = []
            for j in range(i + 1, n):
                if s[j] == s2[i] and s[j] != s2[j]:
                    res.append(s2[: i + 1] + s[i + 1 : j] + s[i] + s[j + 1 :])
            return res

        q = deque([s1])
        vis = {s1}
        ans, n = 0, len(s1)
        while 1:
            for _ in range(len(q)):
                s = q.popleft()
                if s == s2:
                    return ans
                for nxt in next(s):
                    if nxt not in vis:
                        vis.add(nxt)
                        q.append(nxt)
            ans += 1

Solution 2

#### Python3

class Solution:
    def kSimilarity(self, s1: str, s2: str) -> int:
        def next(s):
            i = 0
            while s[i] == s2[i]:
                i += 1
            res = []
            for j in range(i + 1, n):
                if s[j] == s2[i] and s[j] != s2[j]:
                    res.append(s2[: i + 1] + s[i + 1 : j] + s[i] + s[j + 1 :])
            return res

        q = deque([s1])
        vis = {s1}
        ans, n = 0, len(s1)
        while 1:
            for _ in range(len(q)):
                s = q.popleft()
                if s == s2:
                    return ans
                for nxt in next(s):
                    if nxt not in vis:
                        vis.add(nxt)
                        q.append(nxt)
            ans += 1

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