I was asked to find the missing number from 1..N array.

For instance, for array: let numArr = [2,4,6,8,3,5,1,9,10]; the missing number is 7

let numArr=[2,4,6,8,3,5,1,9,10];
numArr.sort(function(a,b){  //sort numArr
  return a-b;
});

let newNumArr=[];
for(let i=1;i<=10;i++){
  newNumArr.push(i);
}

for(let i=0;i<newNumArr.length;i++){  //pare with new arr
  if(newNumArr[i] !== numArr[i]){
    console.log('The missing num is:'+newNumArr[i]);  //The missing num is:7
    break;
  }
}

I was asked to find the missing number from 1..N array.

For instance, for array: let numArr = [2,4,6,8,3,5,1,9,10]; the missing number is 7

let numArr=[2,4,6,8,3,5,1,9,10];
numArr.sort(function(a,b){  //sort numArr
  return a-b;
});

let newNumArr=[];
for(let i=1;i<=10;i++){
  newNumArr.push(i);
}

for(let i=0;i<newNumArr.length;i++){  //pare with new arr
  if(newNumArr[i] !== numArr[i]){
    console.log('The missing num is:'+newNumArr[i]);  //The missing num is:7
    break;
  }
}

• javascript
• ecmascript-6

• 2 Question best suited for codereview. Moreover from what question you are asking, it is not clear that 7 is the missing number unless and until you say so. Try rewording it to fit the code you have written – Krishna Prashatt Commented May 10, 2018 at 9:09

4 Answers 4

You can use MAP and FILTER to find out the missing number in seperate array

const numArr = [2, 4, 6, 8, 3, 5, 1, 9, 10];
const missingNumberArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10].map(number => {
    if (!numArr.includes(number)) {
        return number;
    }
}).filter(y => y !== undefined);

You can use the simple logic of sum of consecutive n numbers is n*(n+1)/2. Subtracting the sum of array numbers from above will give the missing number

let numArr=[2,4,6,8,3,5,1,9,10];
var sum = numArr.reduce((a,c) => a+c, 0);

// As the array contains n-1 numbers, here n will be numArr.length + 1
console.log(((numArr.length + 1) * (numArr.length + 2))/2 - sum);

It would be easier to use .find:

function findMissing(input) {
  input.sort((a, b) => a - b);
  const first = input[0];
  return input.find((num, i) => first + i !== num) - 1;
}
console.log(findMissing([2, 4, 6, 8, 3, 5, 1, 9, 10]));
console.log(findMissing([3, 4, 5, 6, 8, 9, 2]));

(note that this also works for finding missing values from arrays that don't start at 1)

You can use XOR features.

• XOR all the array elements, let the result of XOR be arr_xor.
• XOR all numbers from 1 to n, let XOR be interval_xor .
• XOR of arr_xor and interval_xor gives the missing number.

let numArr=[2,4,6,8,3,5,1,9,10];
function getMissingNo(arr){
    arr = arr.sort()  
    n = arr.length
    arr_xor = arr[0]
    interval_xor = 1
    for(i = 0; i < n; i++)
        arr_xor ^= arr[i]
    for( i = 0; i<n + 2; i++)
        interval_xor ^= i
    return arr_xor ^ interval_xor
}
    
console.log(getMissingNo(numArr));