Longest Subsequence With Limited Sum - Greedy - Sorting & Two Pointers [JS]

Description 

Solution: Greedy - Sorting & Two Pointers

It is always optimal to take the smallest numbers possible, since we want the more quantity.
Sort nums in asc order.
Sort queries in asc order and use two pointers (i = index in nums, j = index in queries) to process the queries efficiently.
Move i up until it exceeds queries[j].

n = length of nums, m = length of queries
Time Complexity: O(n log(n) + m log(m))
Space Complexity: O(m)

var answerQueries = function(nums, queries) {
  nums.sort((a, b) => a - b);
  queries = queries.map((query, idx) => [query, idx]).sort((a, b) => a[0] - b[0]);
  let n = nums.length, m = queries.length;
  let res = Array(m).fill(0);
  let sum = 0;
  for (let j = 0, i = 0; j < m; j++) {
    let [maxSum, index] = queries[j];
    while (i < n && sum + nums[i] <= maxSum) {
      sum += nums[i++];
    }
    res[index] = i;
  }
  return res;
};