When given two sorted arrays, nums1 and nums2, the solution should merge them into one by modifying the first array. nums1 has just enough space to store all elements from nums1 and nums2.

Since both arrays are already sorted, you can merge two arrays by comparing the heads of each array.

If an element from the array gets filled(nums1) is less than the other array, let that element be as is because it is already sorted.

Otherwise, you need to swap the heads of nums1 and nums2. Then, place the head of nums2 in the correct(sorted) position so that nums2 can keep its order.

/**
 * @param {number[]} nums1
 * @param {number} m
 * @param {number[]} nums2
 * @param {number} n
 * @return {void} Do not return anything, modify nums1 in-place instead.
 */
var merge = function (
  nums1 = [1, 2, 3, 0, 0, 0],
  m = 3,
  nums2 = [2, 5, 6],
  n = 3
) {
  let nums1Idx = 0;

  while (nums1Idx < m + n || nums2.lenth > 0) {
    if (nums1Idx === m) {
            // nums1 is empty, push all nums2 element into nums1
      while (nums1Idx < m + n && nums2.length) {
        nums1[nums1Idx] = nums2.shift();
        nums1Idx++;
      }
      break;
    }

    if (nums2[0] < nums1[nums1Idx]) {
            // Place the smaller element on the left side.
      swap(nums1, nums2, nums1Idx, 0);

      let swappedNums2Index = 0;
            // If the element from nums1 is smaller than the head of nums2
            // Move the new element to the right position.
      while (
        nums2[swappedNums2Index] > nums2[swappedNums2Index + 1] &&
        swappedNums2Index < nums2.length - 1
      ) {
        swap(nums2, nums2, swappedNums2Index, swappedNums2Index + 1);
        swappedNums2Index++;
      }
    }

    nums1Idx++;
  }
};

function swap (a, b, idxA, idxB) {
  const temp = a[idxA];
  a[idxA] = b[idxB];
  b[idxB] = temp;
};