Let's say that you want to add a value to a list that's already sorted. Using push() will insert the value at the end of the list, and this isn't what you want. The insert() method can put the value at a particular index—you just have to figure out which index to use to keep the list sorted:

const sortedInsert = (item, list) => {
  let low = 0;
  let high = list.count();

  while (low < high) {
    const mid = (low + high) >>> 1;
    if (item > list.get(mid)) {
      low = mid + 1;
    } else {
      high = mid;
    }
  }
  return list.insert(low, item);
};

The sortedIndex() function takes a value and a list into which to insert the value as arguments. It then does a simple binary search to find the correct index that will maintain the sort order of the collection. Let's see how this function works in practice:

const myList = List.of(1, 3, 5);
const withTwo = sortedInsert(2, myList);
const withFour = sortedInsert(4, withTwo);

console.log('myList', myList.toJS());
// -> myList [ 1, 3, 5 ]
console.log('withTwo', withTwo.toJS());
// -> withTwo [ 1, 2, 3, 5 ]
console.log('withFour', withFour.toJS());
// -> withFour [ 1, 2, 3, 4, 5 ]

The withTwo list has the value 2 inserted in the correct position, and withFour has 4 in the correct position. The main advantage of implementing this approach is that you can avoid having to re-sort the entire collection after pushing a new value. Finding the correct insertion index to maintain the sort order isn't without it's downsides, though.