Make Number of Distinct Characters Equal - Try Each Pair of Characters [JS]

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Description 

Solution: Try Each Pair of Characters

  1. Count the occurances of each character in word1 and word2.
  2. Try each pair of characters (26 * 26).
    We can figure out the number of distinct characters in word1 and word2 after we swap character i from word1 with character j from word2.
    • New distinct character count in word1:
      • If character j does not exist in word1, +1
      • If character i has only one occurance in word1, -1
    • New distinct character count in word2:
      • If character i does not exist in word2, +1
      • If character j has only one occurance in word2, -1
    • Note: When i and j are equal, we return true if the original distinct character counts are equal.

n = length of word1,m = length of word2
Time Complexity: O(n + m + 26*26)
Space Complexity: O(1)

var isItPossible = function(word1, word2) {
  let count1 = Array(26).fill(0), distinctCount1 = 0;  
  let count2 = Array(26).fill(0), distinctCount2 = 0;  
  for (let i = 0; i < word1.length; i++) {
    count1[word1.charCodeAt(i) - 97]++;
    if (count1[word1.charCodeAt(i) - 97] === 1) distinctCount1++;
  }
  for (let i = 0; i < word2.length; i++) {
    count2[word2.charCodeAt(i) - 97]++;
    if (count2[word2.charCodeAt(i) - 97] === 1) distinctCount2++;
  }
  
  for (let i = 0; i < 26; i++) {
    for (let j = 0; j < 26; j++) {
      if (count1[i] === 0 || count2[j] === 0) continue;
      if (i === j) {
        if (distinctCount1 === distinctCount2) return true;
        continue;
      }
      let newCount1 = distinctCount1 + (count1[j] === 0 ? 1 : 0) + (count1[i] === 1 ? -1 : 0);
      let newCount2 = distinctCount2 + (count2[i] === 0 ? 1 : 0) + (count2[j] === 1 ? -1 : 0);
      if (newCount1 === newCount2) return true;
    }
  }
  return false;
};
Location: Auckland, New Zealand

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