CODEWITHSUNDEEP
javascript
Narek
Narek

Reputation: 263

Get quantity of missed elements from an array

I'm trying to solve the task, get quantity of missed elements from an array. For example, if given an array [1,3,6], the quantity of missed elements is 3 (2,4,5). But somewhere code goes wrong and system doesn't accept the code. I tried some methods, but unfortunately they are useless.

function arr(x){
 let y = [];
 for (let i = x[0]; i <= x[x.length-1]; i++){
    y.push(i);
 }
 return y.length-x.length;
}
let m = arr([1,2,3,4,5,6]);
console.log(m);

Or...

function arr(x){
    let y = [];
    for (let i = 0; i < x.length; i++){
        for (let j = 0; j < i; j++){
            if (x[i] == x[j]){
                x.splice(i,1);
                i--;
            }
        }
    }
    console.log(x);
    for (let i = x[0]; i <= x[x.length-1]; i++){
        y.push(i);
    }
    console.log(y);
    return y.length-x.length;
}
let l = arr([1,3,2,4,9]);
console.log(l);

I also tried to sort array, but there are no changes

Upvotes: 2

Views: 160

Answers (4)

Icepickle
Icepickle

Reputation: 12796

To be honest, you don't really need a for loop. I think you can calculate the nr by checking the maximum number, the minimum number and the length of the array.

Would this also work for you?

const source = [1,3,6];

/**
 * @method nrOfMissedItems
 * @param {Array<Number>} an array containing only numbers
 * @returns -Infinity when the parameter arr is null or undefined, otherwise number of non-mentioned numbers, ie [5,5] returns 0, [1,1,1,3] returns 1
 * When the array contains non-numbers it will return NaN
 */
function nrOfMissedItems( arr ) {
  const noDuplicates = [...new Set(arr)];
  const highestNumber = Math.max( ...noDuplicates );
  const lowestNumber = Math.min( ...noDuplicates );
  return highestNumber - lowestNumber - noDuplicates.length + 1;
}

console.log( nrOfMissedItems( source ) );         // 3
console.log( nrOfMissedItems( [1] ) );            // 0
console.log( nrOfMissedItems( [0,1,4] ) );        // 2
console.log( nrOfMissedItems( [5,3,1] ) );        // 2
console.log( nrOfMissedItems( [1,1,1,1,5] ) );    // 3
console.log( nrOfMissedItems( null ) );           // -Infinity
console.log( nrOfMissedItems( undefined ) );      // -Infinity
console.log( nrOfMissedItems() );                 // -Infinity 
console.log( nrOfMissedItems( ['a','b', 1] ) );   // NaN
console.log( nrOfMissedItems( ['a', null, 1] ) ); // NaN
console.log( nrOfMissedItems( [undefined, 1] ) ); // NaN

Upvotes: 4

Maheer Ali
Maheer Ali

Reputation: 36584

You can do that in following steps:

  • Remove the duplicate elements from array.
  • Get the range of array by subtracting first value from last.
  • Subtract the range from the length of array.

Note: I am considering array is sorted.

function missed(arr){
  arr = [...new Set(arr)];
  return arr[arr.length - 1] - arr[0] - arr.length + 1
}

console.log(missed([1,3,6]))

If you need to use this on a unsorted array. Then use Math.max() and Math.min().

function missed(arr){
  arr = [...new Set(arr)];
  let max = Math.max(...arr);
  let min = Math.min(...arr);
  return max - min - arr.length + 1
}

console.log(missed([1,3,6]))
console.log(missed([6,1,3]))

Upvotes: 2

Vishnu
Vishnu

Reputation: 897

You can try like :

function arr(x){
 let y = [];
  let start = Math.min(...x);
  let end = Math.max(...x);
 for (let i = start; i <= end; i++){
    y.push(i);
 }
 return y.length-x.length;
}

let m = arr([1,3,6]);

console.log(m);

Upvotes: 0

Jorge Fuentes Gonz&#225;lez
Jorge Fuentes Gonz&#225;lez

Reputation: 11807

I don't know if optimization is a must, bust the easiest non-ES6 (for all browsers support) way I can think of is to sort the array and iterate through it logging the numbers that you expect but are missing. Something like this:

var arr = [4,1,6,8];
arr.sort();
var miss = [];
var expect = arr[0]; // The starting number is the first element in the array I guess. Change to 1 if you want it to be 1

// Loop through the sorted array
for (var i = 0; i < arr.length; i++) {
    while(arr[i] > expect++) { // keep looping while the current number is not the one expected, and increment the expected by one
        miss.push(expect-1); // If missed, record the missed number (which is the actual expected minus 1, as we already incremented it)
    }
}

console.log(miss);

Upvotes: 0

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