Requirement: sort the first list and reorder the second list to keep corresponding values.

The solution is identical to question 1. Pair each element of the first list with its counterpart in the second list, sort the pairs by the first element, then separate them back into two lists. Complexity remains O(n log n) time and O(n) space. Example in JavaScript:

const list1 = [3, 1, 2];
const list2 = ['c', 'a', 'b'];
const paired = list1.map((v, i) => [v, list2[i]]);
paired.sort((a, b) => a[0] - b[0]);
const sortedList1 = paired.map(p => p[0]);
const sortedList2 = paired.map(p => p[1]);
console.log(sortedList1); // [1, 2, 3]
console.log(sortedList2); // ['a', 'b', 'c']

This keeps the relationship between the two lists intact after sorting.