Find First Missing Number In A Sequence Of Numbers

I am trying to figure out how to find the first missing number of a sequence of numbers like this (1,2,3,5,6,9,10,15) I want to put the first missing number, #4, into an variable f

Solution 1:

How about this

var mynumbers = new Array(1,2,3,6,9,10);
var missing;

for(var i=1;i<=32;i++)
{    
   if(mynumbers[i-1] != i){
        missing = i;
        alert(missing);
        break;
   }
}

Solution 2:

The O(n) solutions are easy , but this is a common interview question and often we look for O(log n) time solution. Here is the javascript code. It's basically a modified binary search.

function misingNumInSeq(source, min = 0, max = source.length - 1){
    if(min >= max){
        return min + 1;
    }
    let pivot = Math.floor((min + max)/2);
    // problem is in right side. Only look at right sub array
    if(source[pivot] === pivot + 1){
        return misingNumInSeq(source, pivot + 1, max);
    } else {
        return misingNumInSeq(source, min , pivot);
    }
} 

Output

misingNumInSeq([1,2,3,5,6,9,10,15])
4

Solution 3:

By if(mynumbers[i] - mynumbers[i-1] != 1), you mean to say the series will always be incrementing by 1?

var missing = (function (arr) {
    var i;
    for (i = 0; i < arr.length; ++i) {
        if (i + arr[0] !== arr[i]) return i + arr[0];
    }
    if (i < 32)            // if none missing inside array and not yet 32nd
        return i + arr[0]; // return next
}([1,2,3,6,9,10])); // 4
alert(missing);

Solution 4:

You're going to need the break no matter what. That's what it's there for; to stop the loop from continuing on to the end. And you should use the length of the array instead of hardcoding 32 as the end condition, because your numbers only go up to 32, but there are possibly holes in the list so there will not be 32 elements in the array.

Since you know that each element should be 1 more than the previous element, then the number in the hole is clearly mynumbers[i - 1] + 1.

var mynumbers = new Array(1,2,3,6,9,10);
for(var i = 1; i < mynumbers.length; i++) {
    if(mynumbers[i] - mynumbers[i-1] != 1) {
        alert("First missing number id: " + (mynumbers[i - 1] + 1));
        break;
    }
}

EDIT: This only holds true for the missing number not being 1. To catch that, you will need to check if (mynumbers[0] != 1)

Solution 5:

Edit:

function findFirstMissing(array) {
    for (var i = 0; i < array.length; i++) {
        if (i+1 !== array[i]) {
            return i+1;
        }
    }
}

---

function findFirstMissing(array) {
    for (var i = 0; i < array.length; i++) {
        if (array[i+1] - array[i] !== 1) {
            return array[i] + 1;
        }
    }
}

If you do it this way then storing it in a variable is easy:

var missing = findFirstMissing(array);