LeetCode 1790. Check if One String Swap Can Make Strings Equal

Description

https://leetcode.com/problems/check-if-one-string-swap-can-make-strings-equal/

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.

Example 4:

Input: s1 = "abcd", s2 = "dcba"
Output: false

Constraints:

  • 1 <= s1.length, s2.length <= 100
  • s1.length == s2.length
  • s1 and s2 consist of only lowercase English letters.

Explanation

Check if letter occurrences are matched and if mismatch counts are no more than 2.

Python Solution

class Solution:
    def areAlmostEqual(self, s1: str, s2: str) -> bool:
        count_s1 = {}
        
        for c in s1:
            count_s1[c] = count_s1.get(c, 0) + 1
            
            
        count_s2 = {}
        for c in s2:
            count_s2[c] = count_s2.get(c, 0) + 1
            
        
        for key, value in count_s1.items():
            if key not in count_s2 or value != count_s2[key]:
                return False
            
        count_mismatch = 0
        
        for c1, c2 in zip(s1, s2):
            if c1 != c2:
                count_mismatch += 1
                
        if count_mismatch > 2:
            return False
            
        return True
        
        
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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