Algorithm
Problem Name: 859. Buddy Strings
Problem Link: https://leetcode.com/problems/buddy-strings/
Given two strings s and goal, return true if you can swap two letters in s so the result is equal to goal, otherwise, return false.
Swapping letters is defined as taking two indices i and j (0-indexed) such that i != j and swapping the characters at s[i] and s[j].
- For example, swapping at indices
0and2in"abcd"results in"cbad".
Example 1:
Input: s = "ab", goal = "ba" Output: true Explanation: You can swap s[0] = 'a' and s[1] = 'b' to get "ba", which is equal to goal.
Example 2:
Input: s = "ab", goal = "ab" Output: false Explanation: The only letters you can swap are s[0] = 'a' and s[1] = 'b', which results in "ba" != goal.
Example 3:
Input: s = "aa", goal = "aa" Output: true Explanation: You can swap s[0] = 'a' and s[1] = 'a' to get "aa", which is equal to goal.
Constraints:
1 <= s.length, goal.length <= 2 * 104sandgoalconsist of lowercase letters.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
bool buddyStrings(string A, string B) {
if(A.size() != B.size()) return false;
int n = A.size(), pos = -1;
vector<int>count(26);
bool repeat = false, swaped = false;
for(int i = 0; i < n; i++){
if(A[i] != B[i]){
if(pos == -1) pos = i;
else if(swaped || A[pos] != B[i] || A[i] != B[pos]> return false;
else swaped = true;
}
if(++count[A[i] - 'a'] > 1) repeat = true;
}
return swaped || repeat;
}
};
Input
s = "ab", goal = "ba"
Output
true
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public boolean buddyStrings(String s, String goal) {
if (s.length() != goal.length()) {
return false;
}
if (s.equals(goal)) {
Set < Character> set = new HashSet<>();
for (char c : s.toCharArray()) {
if (set.contains(c)) {
return true;
}
set.add(c);
}
return false;
}
char[] mismatch = {'-', '-'};
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != goal.charAt(i)) {
if (mismatch[0] == '|') {
return false;
}
if (mismatch[0] == '-') {
mismatch[0] = s.charAt(i);
mismatch[1] = goal.charAt(i);
} else {
if (goal.charAt(i) == mismatch[0] && s.charAt(i) == mismatch[1]) {
mismatch[0] = '|';
continue;
}
return false;
}
}
}
return mismatch[0] == '|';
}
}
Input
s = "ab", goal = "ba"
Output
true
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const buddyStrings = function(A, B) {
if(A.length !== B.length) return false
const aCode = ('a').charCodeAt(0)
if(A === B) {
const count = new Array(26).fill(0)
for(let i = 0; i < A.length; i++) {
count[A.charCodeAt(i) - aCode]++
}
for(let el of count> {
if(el > 1) return true
}
return false
} else {
const arr = []
for(let i = 0; i < A.length; i++) {
if(A[i] !== B[i]) {
arr.push(i>
if(arr.length > 2) return false
}
}
if(arr.length !== 2) return false
return A[arr[0]] === B[arr[1]] && A[arr[1]] === B[arr[0]]
}
};
Input
s = "ab", goal = "ab"
Output
false
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def buddyStrings(self, A, B):
if len(A) != len(B):
return False
dif, dup = [[s1, B[i]] for i, s1 in enumerate(A) if s1 != B[i]], len(A) != len(set(A))
return len(dif) == 2 and dif[0] == dif[1][::-1] or (not dif and dup)
Input
s = "ab", goal = "ab"
Output
false
#5 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0859_BuddyStrings
{
public bool BuddyStrings(string A, string B)
{
if (A.Length != B.Length) return false;
if (A == B)
{
int[] counts = new int[26];
for (int i = 0; i < A.Length; ++i)
counts[A[i] - 'a']++;
foreach (int c in counts)
if (c > 1) return true;
return false;
}
int first = -1, second = -1;
for (int i = 0; i < A.Length; ++i)
{
if (A[i] != B[i])
{
if (first == -1)
first = i;
else if (second == -1)
second = i;
else
return false;
}
}
return second != -1 && A[first] == B[second] && A[second] == B[first];
}
}
}
Input
s = "aa", goal = "aa"
Output
true