CODEWITHSUNDEEP
javascriptarrays
Matthew Spahr
Matthew Spahr

Reputation: 105

Given a range, find missing number in array

Given a range, how is it possible in javascript to find the missing number in an unordered array? For example, if I know the range of an array is [48,79] and the array is:

[56, 76, 48, 69, 60, 68, 57, 58, 52,
  72, 61, 64, 65, 66, 73, 75, 77,
  49, 63, 50, 70, 51, 74, 54, 59,
  78, 79, 71, 55, 67]

The missing number/output would be 62,53.

Upvotes: 0

Views: 461

Answers (2)

Shaun E. Tobias
Shaun E. Tobias

Reputation: 527

EDIT: by the time I posted this, Vinay Kaklotar had posted a much better solution for the OPs updated question.

I would iterate until the missing value's index isn't found:

var arr = [1,2,3,4,5,7,8,9,10];
var i = 1;

while(arr.indexOf(i) !== -1) {
  i++;
}

console.log(i);

The above was an answer to the original question before OP edited it. Below is a solution to the updated question.

I'm iterating through the array and comparing each item to a new sequence (n). Every time a missing number is found, it's added to the missing array.

var arr = [1, 2, 4, 5, 6, 9];
var missing = [];
var n = 1;

for (var i = 0; i < arr.length; i++) {
  if (arr[i] !== n) {
    missing.push(n);
    i--;
  }
  n++;
}

console.log(missing);

Upvotes: 1

Vinay Kaklotar
Vinay Kaklotar

Reputation: 424

You should try this

function findNumbers(arr) {
    var sparse = arr.reduce((sparse, i) => (sparse[i]=1,sparse), []);
    return [...sparse.keys()].filter(i => i && !sparse[i]);
}

var myArr = [56, 76, 48, 69, 60, 68, 57, 58, 52,
  72, 61, 64, 65, 66, 73, 75, 77,
  49, 63, 50, 70, 51, 74, 54, 59,
  78, 79, 71, 55, 67]
var res= findNumbers(myArr );
console.log(res);

Upvotes: 2

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