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

## 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 <= 100
• s1.length == s2.length
• s1 and s2 consist 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 $cnt \neq 1$, return true.

The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.

• class Solution {
public boolean areAlmostEqual(String s1, String s2) {
int cnt = 0;
char c1 = 0, c2 = 0;
for (int i = 0; i < s1.length(); ++i) {
char a = s1.charAt(i), b = s2.charAt(i);
if (a != b) {
if (++cnt > 2 || (cnt == 2 && (a != c2 || b != c1))) {
return false;
}
c1 = a;
c2 = b;
}
}
return cnt != 1;
}
}

• class Solution {
public:
bool areAlmostEqual(string s1, string s2) {
int cnt = 0;
char c1 = 0, c2 = 0;
for (int i = 0; i < s1.size(); ++i) {
char a = s1[i], b = s2[i];
if (a != b) {
if (++cnt > 2 || (cnt == 2 && (a != c2 || b != c1))) {
return false;
}
c1 = a, c2 = b;
}
}
return cnt != 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


• func areAlmostEqual(s1 string, s2 string) bool {
cnt := 0
var c1, c2 byte
for i := range s1 {
a, b := s1[i], s2[i]
if a != b {
cnt++
if cnt > 2 || (cnt == 2 && (a != c2 || b != c1)) {
return false
}
c1, c2 = a, b
}
}
return cnt != 1
}

• function areAlmostEqual(s1: string, s2: string): boolean {
let c1, c2;
let cnt = 0;
for (let i = 0; i < s1.length; ++i) {
const a = s1.charAt(i);
const b = s2.charAt(i);
if (a != b) {
if (++cnt > 2 || (cnt == 2 && (a != c2 || b != c1))) {
return false;
}
c1 = a;
c2 = b;
}
}
return cnt != 1;
}


• impl Solution {
pub fn are_almost_equal(s1: String, s2: String) -> bool {
if s1 == s2 {
return true;
}
let (s1, s2) = (s1.as_bytes(), s2.as_bytes());
let mut idxs = vec![];
for i in 0..s1.len() {
if s1[i] != s2[i] {
idxs.push(i);
}
}
if idxs.len() != 2 {
return false;
}
s1[idxs[0]] == s2[idxs[1]] && s2[idxs[0]] == s1[idxs[1]]
}
}