CODEWITHSUNDEEP
javascriptarraysloopsfor-loopiteration
Somename
Somename

Reputation: 3444

Javascript for loop check condition

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.

Upvotes: 0

Views: 3468

Answers (6)

Ihor
Ihor

Reputation: 491

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();

Upvotes: 1

Telmo Trooper
Telmo Trooper

Reputation: 5694

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))

Upvotes: 3

Edmund1645
Edmund1645

Reputation: 349

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()

Upvotes: 0

Vinesh Goyal
Vinesh Goyal

Reputation: 637

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);

Upvotes: 0

zfrisch
zfrisch

Reputation: 8670

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);

Upvotes: 0

Amardeep Bhowmick
Amardeep Bhowmick

Reputation: 16908

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]));

Upvotes: 0

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