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K-Similar Strings

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.

Solution Explanation for LeetCode 854: K-Similar Strings

This problem asks for the minimum number of swaps needed to transform string s1 into string s2, given that s1 and s2 are anagrams. This can be efficiently solved using Breadth-First Search (BFS) or a prioritized BFS (using a heap).

Approach 1: Breadth-First Search (BFS)

This approach explores the state space of possible string transformations using BFS. We start with s1 and iteratively swap pairs of characters to generate new strings. The search continues until s2 is found. The distance from s1 to s2 represents the minimum number of swaps.

Algorithm:

  1. Initialization:

    • Create a queue q and add s1 to it.
    • Create a set vis to track visited strings (to avoid cycles). Add s1 to vis.
    • Initialize ans (the minimum swaps) to 0.

  2. BFS Iteration:

    • While the queue is not empty:
      • Process all strings currently in the queue.
      • For each string s:
        • If s is equal to s2, return ans.
        • Generate all possible strings nxt from s by swapping a single pair of characters. Only consider swaps that bring s closer to s2.
        • For each generated nxt:
          • If nxt is not in vis, add it to the queue and vis.
      • Increment ans (we've explored one level in the BFS).

  3. Time Complexity: O(n! * n), where n is the length of the strings. In the worst case, the algorithm might need to explore all possible permutations of the string, which is n!. For each permutation, finding the next states requires O(n) time.

  4. Space Complexity: O(n!), in the worst-case scenario, the visited set and queue could store all possible permutations.

Approach 2: A* Search (Heuristic-Guided BFS)

This approach improves upon the basic BFS by incorporating a heuristic function to guide the search towards s2 more efficiently. The heuristic estimates the remaining number of swaps needed.

Algorithm:

  1. Heuristic Function (f): This function estimates the remaining swaps needed. A simple heuristic is (mismatched_chars + 1) // 2. This is a lower bound, as each swap can fix at most two mismatched characters.

  2. Initialization:

    • Create a priority queue q using a min-heap to store (f(s), s) pairs, ordered by the heuristic. Add (f(s1), s1) to q.
    • Create a dictionary dist to store the distance from s1 to each visited string. Set dist[s1] = 0.

  3. Prioritized BFS:

    • While q is not empty:
      • Pop the element with the lowest heuristic from q.
      • If s is s2, return dist[s].
      • Generate all possible strings nxt from s by swapping a single pair of characters (similar to Approach 1).
      • For each nxt:
        • If nxt is not in dist or dist[nxt] > dist[s] + 1:
          • Update dist[nxt] = dist[s] + 1.
          • Push (dist[nxt] + f(nxt), nxt) into q.

  4. Time Complexity: While the worst-case remains exponential, the heuristic significantly prunes the search space, making it substantially faster in practice than the basic BFS.

  5. Space Complexity: Similar to Approach 1, the space complexity can be O(n!).

Code Examples (Python):

The code examples in various languages are already provided in the original response. The Python code illustrates both approaches. Approach 1 uses basic BFS, and approach 2 uses A* search (though only the first approach is fully complete in all provided languages). Remember to install the heapq module for Approach 2. This approach will be much faster than simple BFS as the test cases get larger.

Note: The complexity analysis provided assumes the worst-case scenario. In practice, the actual runtime can vary depending on the specific input strings. The A* approach (Approach 2) is generally expected to perform much better than the standard BFS (Approach 1), especially for longer strings.