JavaScript: Simple function to find positive missing integers in Array of numbers.

missing-integers.js

/* Write a function solution(A); that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A.
For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5.
Given A = [1, 2, 3], the function should return 4.
Given A = [−1, −3], the function should return 1.

Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [−1,000,000..1,000,000]. */

const A = [-1, -4];
function solution(A) {
    let max = A.reduce((acc, val) => acc > val ? acc : val);
    let missing = 1;
    for (let i = max + 1; i > 0; i--) {
        found = A.find(el => el == i);
        if(found === undefined) {
            missing = i;
        }
    }
    return missing; 
}
console.log(solution(A));

It is failing in performance....

i have and easy solution and the best run-time possible.
const A = [1, 3, 6, 4, 1, 2];
function solution (A){
var i =1;
while (A.indexOf(i) != -1){
i++
}
return i;
}

below come with 100%:

function solution(A){
let counter = 0;
A.sort(function (a, b) {
return a - b;
});

for (let i = 0; i < A.length; i++) {
const element = A[i];
if (element > 0) {
if (element != counter) {
counter++;
}
if (element == counter) {
continue;
}
return counter;
}
}
counter++;
return counter;
}