LeetCode: Advantage Shuffle Solution

1/**
2 * @param {number[]} A
3 * @param {number[]} B
4 * @return {number[]}
5 */
6var advantageCount = function (A, B) {
7  A = A.sort((a, b) => a - b)
8  B = B.map((el, i) => [el, i]).sort(([a], [b]) => b - a) // keep the index for later sort to the start position
9
10  for (const b of B) {
11    const a = A.slice(-1)[0]
12    if (a > b[0]) {
13      b.push(a)
14      A.pop()
15    }
16  }
17
18  return B.map(([b, i, a]) => [b, i, a != null ? a : A.pop()])
19    .sort(([, iA], [, iB]) => iA - iB)
20    .map(([, , el]) => el)
21}