Problem
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 * 10^4sandgoalconsist of lowercase letters.
Solution (Java)
class Solution {
public boolean buddyStrings(String s, String goal) {
int first = -1;
int second;
int[] sCounts = new int[26];
if (s.equals(goal)) {
for (int i = 0; i < s.length(); i++) {
sCounts[s.charAt(i) - 'a']++;
if (sCounts[s.charAt(i) - 'a'] > 1) {
return true;
}
}
}
for (int i = 0; i < s.length(); i++) {
char curr = s.charAt(i);
sCounts[curr - 'a']++;
if (curr != goal.charAt(i)) {
if (first == -1) {
first = i;
} else {
second = i;
char[] ss = s.toCharArray();
char temp = ss[first];
ss[first] = ss[second];
ss[second] = temp;
return String.valueOf(ss).equals(goal);
}
}
}
return false;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).