Find Missing Numbers from Unsorted Array
The simplest solution to this problem
miss = (arr) => {
let missArr=[];
let l = Math.max(...arr);
let startsWithZero = arr.indexOf(0) > -1 ? 0 : 1;
for(i = startsWithZero; i < l; i++) {
if(arr.indexOf(i) < 0) {
missArr.push(i);
}
}
return missArr;
}
miss([3,4,1,2,6,8,12]);
You could use a sparse array with 1-values at indexes that correspond to values in the input array. Then you could create yet another array with all numbers (with same length as the sparse array), and retain only those values that correspond to an index with a 1-value in the sparse array.
This will run in O(n) time:
function findMissingNumbers(arr) {
// Create sparse array with a 1 at each index equal to a value in the input.
var sparse = arr.reduce((sparse, i) => (sparse[i]=1,sparse), []);
// Create array 0..highest number, and retain only those values for which
// the sparse array has nothing at that index (and eliminate the 0 value).
return [...sparse.keys()].filter(i => i && !sparse[i]);
}
var someArr = [2, 5, 3, 1, 4, 7, 10, 15]
var result = findMissingNumbers(someArr);
console.log(result);NB: this requires EcmaScript2015 support.
Something like this will do what you want.
var X = [2, 5, 3, 1, 4, 7, 10, 15]; // Array of numbers
var N = Array.from(Array(Math.max.apply(Math, X)).keys()); //Generate number array using the largest int from X
Array.prototype.diff = function(a) {
return this.filter(function(i) {return a.indexOf(i) < 0;}); //Return the difference
};
console.log(N.diff(X));