You are given an array nums of n positive integers.

You can perform two types of operations on any element of the array any number of times:

• If the element is even, divide it by 2.
  • For example, if the array is [1,2,3,4], then you can do this operation on the last element, and the array will be [1,2,3,2].
• If the element is odd, multiply it by 2.
  • For example, if the array is [1,2,3,4], then you can do this operation on the first element, and the array will be [2,2,3,4].

The deviation of the array is the maximum difference between any two elements in the array.

Return the minimum deviation the array can have after performing some number of operations.

Example

Input: nums = [1,2,3,4]
Output: 1
Explanation: You can transform the array to [1,2,3,2], then to [2,2,3,2], then the deviation will be 3 - 2 = 1.

Solution

Great solution here>

/**
 * @param {number[]} nums
 * @return {number}
 */
var minimumDeviation = function (nums) {
  // https://github.com/datastructures-js/priority-queue
  const mpq = new MaxPriorityQueue();

  // Convert all the numbers to even and enqueue them
  nums.forEach((num) => {
    if (num % 2 !== 0) {
      const value = num * 2;

      mpq.enqueue(value, value);
    } else {
      mpq.enqueue(num, num);
    }
  });

  // Get difference between max and min values
  let deviation = mpq.front().element - mpq.back().element;

  // Loop until we have any max even number left in the queue
  while (mpq.front().element % 2 === 0) {
    // Get max even value
    const { element } = mpq.dequeue();

    // Convert it to odd number and enqueue again
    mpq.enqueue(element / 2, element / 2);

    // Get minimum between previous deviation and after above conversion
    deviation = Math.min(deviation, mpq.front().element - mpq.back().element);
  }

  return deviation;
};