1114MediumHash TableStringSortingCounting

Are the Two Words Kin

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Trains the technique from

LeetCode 1657Determine if Two Strings Are Close

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Two words first and second hold only lowercase letters. Two moves are allowed on a word, each usable as often as you like:

Swap the letters sitting at any two positions.
Pick two letters that both already appear somewhere in the word and exchange them throughout: every copy of one becomes the other and every copy of the other becomes the first.

Call the words kin when some run of moves turns first into second.

Return true when they are kin.

Examples

Example 1

Input: first = "aaabb", second = "aabbb"
Output: true

Exchanging every a with every b turns the first word into three b and two a, and rearranging those gives the second word.

Example 2

Input: first = "aaab", second = "aabb"
Output: false

Both words use only a and b, but the first holds three a and one b while the second holds two of each, and no move can change a tally's value.

Example 3

Input: first = "aab", second = "bbc"
Output: false

The first word uses an a and the second does not, and no move can bring in a letter that is not already present.

Constraints

1 <= first.length <= 10^5
1 <= second.length <= 10^5
first and second hold only lowercase English letters

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def close_strings(first: str, second: str) -> bool:

Java

public boolean closeStrings(String first, String second)

The intended approach

Time O(n) to tally plus the cost of ordering the tallies.Space O(1), since only twenty-six letters exist.

Neither move can bring in a letter that is not already there, and neither can drop one entirely: the exchange move insists both letters already appear, so it trades their tallies rather than wiping one out. So the set of letters used is fixed for good. The first move rearranges freely without touching any tally, and the second move swaps two tallies. Swaps of pairs can build any reordering, so the tallies can be shuffled among the letters however you like but never changed in value. What survives every move is therefore exactly two things: which letters appear, and the collection of tallies with no attachment to letters. That gives the test. Build a tally per letter for both words, insist the letter sets match, and insist the tallies match once put in order. Matching tallies force equal lengths, and both conditions are needed: two words can share their letters with different tallies, or share tallies while using different letters.

Pythonentry point: close_strings

def close_strings(first, second):
    if len(first) != len(second):
        return False
    left = {}
    right = {}
    for ch in first:
        left[ch] = left.get(ch, 0) + 1
    for ch in second:
        right[ch] = right.get(ch, 0) + 1
    if set(left) != set(right):
        return False
    return sorted(left.values()) == sorted(right.values())

The obvious version

Search outwards from the first word. From any word reached, list every word one move away, both by swapping two positions holding different letters and by exchanging two letters that appear in it, and keep walking until the second word turns up or the reachable words run out. Costs the alphabet to the power of the word length in the worst case.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_close_strings(first, second):
    if len(first) != len(second):
        return False
    n = len(first)
    seen = set([first])
    stack = [first]
    while stack:
        word = stack.pop()
        if word == second:
            return True
        moves = []
        for i in range(n):
            for j in range(i + 1, n):
                if word[i] != word[j]:
                    body = list(word)
                    body[i], body[j] = body[j], body[i]
                    moves.append("".join(body))
        pool = sorted(set(word))
        for a in range(len(pool)):
            for b in range(a + 1, len(pool)):
                one = pool[a]
                other = pool[b]
                body = []
                for ch in word:
                    if ch == one:
                        body.append(other)
                    elif ch == other:
                        body.append(one)
                    else:
                        body.append(ch)
                moves.append("".join(body))
        for other in moves:
            if other not in seen:
                seen.add(other)
                stack.append(other)
    return False

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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