#1657 - Determine if Two Strings Are Close

MediumHash Map / Set

Two strings are considered close if you can attain one from the other using these operations any number of times: (1) swap any two existing characters, (2) transform every occurrence of one existing character into another existing character, and simultaneously do the reverse. Return true if word1 and word2 are close, or false otherwise.

Input	Output
word1 = "abc", word2 = "bca"	true
word1 = "a", word2 = "aa"	false
word1 = "cabbba", word2 = "abbccc"	true

Solution Space

Try it

/**
 * Determines whether two strings are "close" — transformable into each
 * other via character swaps and pairwise frequency transforms.
 *
 * @param word1 First string of lowercase English letters
 * @param word2 Second string of lowercase English letters
 * @returns Whether the two strings are close
 */
export function closeStrings(word1: string, word2: string): boolean {
  if (word1.length !== word2.length) return false;

  const freq1 = new Array(26).fill(0);
  const freq2 = new Array(26).fill(0);

  for (let i = 0; i < word1.length; i++) {
    freq1[word1.charCodeAt(i) - 97]++;
    freq2[word2.charCodeAt(i) - 97]++;
  }

  // Both strings must use the exact same set of characters
  for (let i = 0; i < 26; i++) {
    if ((freq1[i] === 0) !== (freq2[i] === 0)) return false;
  }

  // Sorted frequency distributions must match (operation 2 lets us
  // reassign which character carries which count)
  freq1.sort((a, b) => a - b);
  freq2.sort((a, b) => a - b);

  for (let i = 0; i < 26; i++) {
    if (freq1[i] !== freq2[i]) return false;
  }

  return true;
}

Approach: frequency analysis – two strings are close iff they share the same character set **and** the same multiset of character frequencies (sorted frequency arrays must match).

Time: O(n + k log k) where k = 26 (alphabet size) – one pass for frequencies, sorting a fixed-size array.

Space: O(k) = O(1) – two arrays of size 26.

#1207 Unique Number of Occurrences

#2215 Find the Difference of Two Arrays