Leetcode #854: K-Similar Strings
In this guide, we solve Leetcode #854 K-Similar Strings in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Breadth-First Search, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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".
Python Solution
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
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.