In order to satisfy the O(N) time-complexity, construct a Set() in O(N) time and space complexity, then use a while loop which is considered constant time relative to N, since the specified range of numbers in A is independent of N:

function solution(A) {
  const set = new Set(A);
  let i = 1;

  while (set.has(i)) {
    i++;
  }

  return i;
}

In order to satisfy the O(N) time-complexity, construct a Set() in O(N) time and space complexity, then use a while loop which is considered constant time relative to N O(N) as well (thank you, wchargin), since the maximum possible number of iterations is equal to N and average performance of a Set#has() operation is O(1). Because O(N + N) = O(N), regarding time complexity, this solution is overall O(N) performance in both time and space:

function solution(A) {
  const set = new Set(A);
  let i = 1;

  while (set.has(i)) {
    i++;
  }

  return i;
}

^{While this is a relatively simplistic and deceivingly elegant implementation, insertusernamehere's solution is admittedly an order of magnitude faster, when using an array as a perfect hash table for non-negative integers instead.}

In order to satisfy the O(N) time-complexity, construct a Set() in O(N) time and then use a while loop which is considered constant time relative to N, since the specified range of numbers in A is independent of N:

function solution(A) {
  const set = new Set(A);
  let i = 1;

  while (set.has(i)) {
    i++;
  }

  return i;
}

In order to satisfy the O(N) time-complexity, construct a Set() in O(N) time and space complexity, then use a while loop which is considered constant time relative to N, since the specified range of numbers in A is independent of N:

function solution(A) {
  const set = new Set(A);
  let i = 1;

  while (set.has(i)) {
    i++;
  }

  return i;
}