Count Number of Distinct Integers After Reverse Operations - Hashset [JS]

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Description 

Solution: Hashset

Add all numbers and reversed numbers to a hashset and return the size.
Reversing a number costs O(log(n)).

Time Complexity: O(n log(n)) 243ms
Space Complexity: O(n) 78.3MB

var countDistinctIntegers = function(nums) {
  let set = new Set(nums);
  for (let num of nums) {
    set.add(reverse(num));
  }
  return set.size;
};

function reverse(num) {
  let reversed = 0;
  while (num > 0) {
    let digit = num % 10;
    reversed = reversed * 10 + digit;
    num = Math.floor(num / 10);
  }
  return reversed;
}
Location: Auckland, New Zealand

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