Original solution makes 4 passes:

• find min element
• find index of min element
• find max element
• find index of max element

these can be combined into a single pass.

class Solution {
  /**
   * Time: O(n) - one pass over each value to find min & max
   * Space: O(1) - variables does not scale with the input size
   * @param nums
   */
  minMoves(nums: number[]) {
    let minIdx = -1;
    let min = Infinity;
    let maxIdx = -1;
    let max = -Infinity;

    for (let i = 0; i < nums.length; i++) { // Time: O(n)
      if (nums[i] < min) {
        min = nums[i];
        minIdx = i;
      }
      if (nums[i] > max) {
        max = nums[i];
        maxIdx = i;
      }
    }
    // removing from both sides
    // (smallest index + 1) => total removals from the left
    // (length - largest index) => total removals from the right
    const both = Math.min(minIdx, maxIdx) + 1 + (nums.length - Math.max(minIdx, maxIdx)); // Time: O(1)
    const left = Math.max(minIdx, maxIdx) + 1; // Time: O(1)
    const right = nums.length - Math.min(minIdx, maxIdx); // Time: O(1);
    return Math.min(both, left, right); // Time: O(1)
  }
}