Day 11 of JavaScriptmas - Avoid Obstacles Solution

Day 11 is finding the minimum missing number from the sequence of number

For instance, an array with [5,3,6,7,9].
So, the minimum missing number will be 4.

This is JavaScript solution

function avoidObstacles(nums) {

    // sorted input
    let sortedNums = nums.sort();

    // create sequence number from sortedNums value (min to max)
    let sequenceNum = [];
    for(let i = sortedNums[0]; i <= sortedNums[sortedNums.length - 1]; i++) {
        sequenceNum.push(i);
    }

    // find intersection that returns unmatch number between 2 arrays
    let intersection = sequenceNum.filter(el => sortedNums.indexOf(el) == -1);

    // find the minimum value of intersection
    let min = Math.min(...intersection);

    return min;
}