Description
You are given two strings s1 and s2 of equal length. A string swap is an operation where you choose two indices in a string (not necessarily different) and swap the characters at these indices.
Return true if it is possible to make both strings equal by performing at most one string swap on exactly one of the strings. Otherwise, return false.
Example 1:
Input: s1 = "bank", s2 = "kanb" Output: true Explanation: For example, swap the first character with the last character of s2 to make "bank".
Example 2:
Input: s1 = "attack", s2 = "defend" Output: false Explanation: It is impossible to make them equal with one string swap.
Example 3:
Input: s1 = "kelb", s2 = "kelb" Output: true Explanation: The two strings are already equal, so no string swap operation is required.
Constraints:
1 <= s1.length, s2.length <= 100s1.length == s2.lengths1ands2consist of only lowercase English letters.
Solutions
Solution 1: Counting
We use a variable cnt to record the number of characters at the same position in the two strings that are different. If the two strings meet the requirements of the problem, then cnt must be 0 or 2. We also use two character variables c1 and c2 to record the characters that are different at the same position in the two strings.
While traversing the two strings simultaneously, for two characters a and b at the same position, if a \ne b, then cnt is incremented by 1. If at this time cnt is greater than 2, or cnt is 2 and a \ne c2 or b \ne c1, then we directly return false. Note to record c1 and c2.
At the end of the traversal, if cnt ≠ 1, return true.
The time complexity is O(n), where n is the length of the string. The space complexity is O(1).
class Solution: def areAlmostEqual(self, s1: str, s2: str) -> bool: cnt = 0 c1 = c2 = None for a, b in zip(s1, s2): if a != b: cnt += 1 if cnt > 2 or (cnt == 2 and (a != c2 or b != c1)): return False c1, c2 = a, b return cnt != 1(code-box)
