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Solution: Buddy Strings
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Problem Statement
Examples
Solution
Step-by-step Algorithm
Code
Complexity Analysis
Problem Statement
Given two strings a and b, return true if they can be considered "buddy strings". Otherwise, return false.
Two strings are buddy strings if you can swap any two letters in one string to make it equal to the other string.
Examples
-
Example 1:
- Input:
a = "xy", b = "yx" - Expected Output: true
- Justification: Swapping 'x' and 'y' in string
aresults in "yx", which matches stringb.
- Input:
-
Example 2:
- Input:
a = "ab", b = "cd" - Expected Output: false
- Justification: There are differences at every position, and swapping any characters in
acannot make it equal tob.
- Input:
-
Example 3:
- Input:
a = "abcde", b = "acbde" - Expected Output: true
- Justification: Swapping 'b' and 'c' in string
aresults in "acbde", which matches stringb. So, both strings are buddy strings.
- Input:
Solution
To solve this problem, we start by comparing the lengths of the two strings. If their lengths differ, they cannot be buddy strings. For strings of equal length, we track the positions where they differ. If there are no differences and the first string has at least one repeating character, it means a swap could occur within the string to make it its own buddy string. If there are exactly two differences, we check if swapping the characters at these positions in one string makes the two strings equal. This approach ensures we accurately identify buddy strings by addressing both scenarios where strings are initially equal or need a specific swap to become equal.
Our approach is effective because it minimizes the number of checks needed to determine if two strings are buddy strings. By first checking the length, then the positions of difference, and finally the conditions for being buddy strings, we efficiently narrow down the possibilities with minimal computational overhead. This method balances thoroughness with efficiency, making it suitable for a wide range of inputs.
Step-by-step Algorithm
-
Initial Length Check:
- If
a.lengthis not equal tob.length, return false immediately, as strings of different lengths cannot be buddy strings.
- If
-
Handle Identical Strings:
- If the strings
aandbare identical, then for them to be buddy strings, there must be at least one character inathat appears more than once. Use a frequency count array or a set to check for repeating characters. If a repeat is found, return true.
- If the strings
-
Identify and Record Differences:
- If
aandbare not identical, iterate through both strings simultaneously to compare characters at each position. - Record the indices where
aandbdiffer. Use two variables,firstandsecond, initialized to -1, to store these indices. - If more than two differences are found, return false as a swap cannot resolve more than two differences.
- If
-
Check for Exactly Two Differences:
- After identifying the indices of differences, check if swapping the characters at these positions in
awould make it equal tob. This means that the character atfirstindex inashould match the character atsecondindex inband vice versa.
- After identifying the indices of differences, check if swapping the characters at these positions in
-
Final Verification:
- If the conditions in step 4 are satisfied, return true, indicating that the strings are buddy strings by a single swap.
- Otherwise, return false.
a = "abcde" and b = "acbde" have the same length, so a swap is possible in principle.
1 of 4
Code
Complexity Analysis
- Time Complexity: O(n) where n is the length of the strings, as the algorithm iterates over the strings at most once.
- Space Complexity: O(1) for checking differences and O(n) in the worst case when the strings are identical to store character frequencies. However, because the character set is limited (assuming ASCII), this can also be considered O(1) in practical scenarios where the alphabet size is fixed.
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On This Page
Problem Statement
Examples
Solution
Step-by-step Algorithm
Code
Complexity Analysis