Javascript for loop check condition

Question

I want to create a function that will generate a random number between 0 to 9 which is not in the array.

My code so far:

 var myArr = [0,2,3,4];
 console.log("Arr: " + myArr);
 function newNum(){
 console.log("test");
 for (var i = 0; i < 10; i++) {
 var n = myArr.includes(i)
 // I want to return n if it's not present in the array
 }
 return n;
 }
newNum()

I want to return only 1 number. How do I do this?

Thanks.

• Somename: @karthick but how do I check if the number is present in the array? I want to generate a number that is not in the array.
• karthick: Yeah sorry didnt read the question properly
• Dominik Matis: Well includes returns boolean, so you need to do something like if(!myArr.includes(i)) return i that's all you need in for loop, if it gets outside, there is no possible num between 0-9

Answer 1

What about this?

const invalidValues = [0,2,3,4];
const getRandomInt = (min, max) => {
 min = Math.ceil(min);
 max = Math.floor(max);
 return Math.floor(Math.random() * (max - min + 1)) + min;
}
const getValidRandomInt = (min, max) => {
 while(true) {
 let temp = getRandomInt(min,max)
 if(!invalidValues.includes(temp)) {
 return temp;
 }
 }
}
console.log(getValidRandomInt(0,10))

• karthick: a very costly way to generate random number if the distribution is only from 0 -9 ?

Answer 2

var myArr = [0,2,3,4];
 function newNum(){
 for (var i = 0; i < 10; i++) {
 if (!myArr.includes(i)) {
 return i;
 }
 }
 // return -1 if all numbers present in array..
 return -1;
 }
newNum();

• DanGo: This appears to only be a partial solution as it does not show how to generate a random number between 0 and 9

Answer 3

Generate the number within the range by using Math.random() then loop and check whether the number generated is in the array or not, if not in the array return the number:

function getRandomArbitrary(min, max, arr) {
 arr = new Set(arr);
 while(true){
 let val = Math.floor(Math.random() * (max - min) + min);
 if(!arr.has(val)){ return val;}
 }
}
console.log(getRandomArbitrary(0, 10, [4,3,2]));

Answer 4

Answer:

Use Math.random() * (max - min) + min to get a number within a range.

You can wrap it with Math.floor to round down to an integer, or alternatively use a bitwise OR ( | ) for small numbers.

function newNum(n_arr) {
 let r = () => Math.random()*9 | 0,
 n = r();
 while(n_arr.includes(n)) {
 n = r();
 }
 return n;
}

---

Example:

var myArr = [0,2,3,4];
 function newNum(n_arr){
 let r = () => Math.random()*9 | 0,
 n = r();
 while(n_arr.includes(n)) {
 n = r();
 }
 return n;
 }
let result = newNum(myArr);
console.log(result);

Answer 5

var myArr= [0,2,5];
function randomNumber(myArr, n){
 n ? n : 1;
 var num = Math.random() * n;
 if(myArr.indexOf( num ) !==-1 ) {
 return randomNumber( myArr, n );
 }
 return num;
}
randomNumber(myArr, 10);

Answer 6

If you want to return the first number missing in the array, from your code above, you could just check if every value of i exists in the array and the moment that value doesn't exist, return it.

 var myArr = [0,2,3,4]; // original array
 function newNum(){
 for (var i = 0; i < 10; i++) { // loop through i from 0-9
 if (myArr.indexOf(i) === -1){ // check for the first missing number
 return i; //return it
 }
 }
 }
newNum()