## Arrays - Find missing numbers in a Sequence

find missing number in unsorted array

find missing number in sorted array

find missing number in a sequence in java

find 2 missing number in array

find missing number in array with duplicates

find missing element in a sorted array of consecutive numbers

how do you find the duplicate number on a given integer array?

I'm trying to find an easy way to loop (iterate) over an array to find all the missing numbers in a sequence, the array will look a bit like the one below.

`var numArray = [0189459, 0189460, 0189461, 0189463, 0189465];`

For the array above I would need `0189462`

and `0189464`

logged out.

**UPDATE** : this is the exact solution I used from Soufiane's answer.

var numArray = [0189459, 0189460, 0189461, 0189463, 0189465]; var mia= []; for(var i = 1; i < numArray.length; i++) { if(numArray[i] - numArray[i-1] != 1) { var x = numArray[i] - numArray[i-1]; var j = 1; while (j<x) { mia.push(numArray[i-1]+j); j++; } } } alert(mia) // returns [0189462, 0189464]

**UPDATE**

Here's a neater version using .reduce

var numArray = [0189459, 0189460, 0189461, 0189463, 0189466]; var mia = numArray.reduce(function(acc, cur, ind, arr) { var diff = cur - arr[ind-1]; if (diff > 1) { var i = 1; while (i < diff) { acc.push(arr[ind-1]+i); i++; } } return acc; }, []); console.log(mia);

If you know that the numbers are sorted and increasing:

for(var i = 1; i < numArray.length; i++) { if(numArray[i] - numArray[i-1] != 1) { //Not consecutive sequence, here you can break or do whatever you want } }

**How to Find Multiple Missing Integers in Given Array of Numbers,** How do you find the missing number in an array of 1 to 100? Find missing number in a sequence in C#. Programming Server Side Programming Csharp. Set a list. List<int> myList = new List<int> () {1, 2, 3, 5, 8, 9}; Now, get the first and last elements −. int a = myList.OrderBy (x => x).First (); int b = myList.OrderBy (x => x).Last ();

ES6-Style

var arr = [0189459, 0189460, 0189461, 0189463, 0189465]; var [min,max] = [Math.min(...arr), Math.max(...arr)]; var out = Array.from(Array(max-min),(v,i)=>i+min).filter(i=>!arr.includes(i));

Result: [189462, 189464]

**Quickest way to find missing number in an array of numbers,** Hence four numbers from 1 to N are missing in the array. Find the 4 missing numbers in sorted order. Examples: Input : arr[] = {2, 5, 6, 3, 9} Output Explanation : 1. In this example, maximum number of the sequence is given.If the maximum number is not available, we can easily find 2. Call the findMissingNumber method to find out the missing number in the sequence. 3. To find the missing number, we will find out the sum of all numbers in the

Watch your leading zeroes, they will be dropped when the array is interpreted-

var A= [0189459, 0189460, 0189461, 0189463, 0189465]

(A returns [189459,189460,189461,189463,189465])

function absent(arr){ var mia= [], min= Math.min.apply('',arr), max= Math.max.apply('',arr); while(min<max){ if(arr.indexOf(++min)== -1) mia.push(min); } return mia; }

**var A= [0189459, 0189460, 0189461, 0189463, 0189465];
alert(absent(A))**

/* returned value: (Array) 189462,189464 */

**Find four missing numbers in an array containing elements from 1 to ,** You can do this in O(n). Iterate through the array and compute the sum of all numbers. Now, sum of natural numbers from 1 to N, can be expressed as Nx(N+1)/2 Find missing number m = N-S Example : suppose array given is {1,2,3,4,5,6,8,9,10} and range is 10. So N will be sum of 1 to 10 = 10 (10+1)/2 = 55 S will be sum of all the array elements which is = 48

To find a missing number in a sequence, First of all, We need to sort an array. Then we can identify what number is missing. I am providing here full code with some test scenarios. this code will identify only missing positive number, if you pass negative values even then it gives positive number.

function findMissingNumber(inputAr) { // Sort array sortArray(inputAr); // finding missing number here var result = 0; if (inputAr[0] > 1 || inputAr[inputAr.length - 1] < 1) { result = 1; } else { for (var i = 0; i < inputAr.length; i++) { if ((inputAr[i + 1] - inputAr[i]) > 1) { result = inputAr[i] + 1; } } } if (!result) { result = inputAr[inputAr.length - 1] + 1; } return result; } function sortArray(inputAr) { var temp; for (var i = 0; i < inputAr.length; i++) { for (var j = i + 1; j < inputAr.length; j++) { if (inputAr[j] < inputAr[i]) { temp = inputAr[j]; inputAr[j] = inputAr[i]; inputAr[i] = temp; } } } } console.log(findMissingNumber([1, 3, 6, 4, 1, 2])); console.log(findMissingNumber([1, 2, 3])); console.log(findMissingNumber([85])); console.log(findMissingNumber([86, 85])); console.log(findMissingNumber([0, 1000]));

**Find the missing number in an array,** Given an array of n-1 distinct integers in the range of 1 to n, find the missing number in the array in linear time. For example, consider the array {1, 2, 3, 4, 5, 7, 8, Like this: Type this array formula into a blank cell, and then press Ctrl + Shift + Enter keys in your keyboard. The missing numbers are listed in cells. Note: this formula will check the given sequence from 1 to 20 if there are missing numbers. If so, returns the missing numbers.

const findMissing = (arr) => { const min = Math.min(...arr); const max = Math.max(...arr); // add missing numbers in the array let newArr = Array.from(Array(max-min), (v, i) => { return i + min }); // compare the full array with the old missing array let filter = newArr.filter(i => { return !arr.includes(i) }) return filter; };

**How to Find Multiple Missing Integers in Given Array ,** Write a Java program to print the missing number from the sequence. For example, if the given array is {1, 1, 2, 3, 5, 5, 7, 9, 9, Approach: The length of the array is n-1. So the sum of all n elements, i.e sum of numbers from 1 to n can be calculated using the formula n* (n+1)/2. Now find the sum of all the elements in the array and subtract it from the sum of first n natural numbers, it will be the value of the missing element.

**Find Missing Number From Series/Array in Java,** how to find multiple numbers missing in a sequence using java 8 streams? Reply. Subelal. December 26, 2019. why he is taken N=12 One number from set {1, 2, …n} is missing and one number occurs twice in the array. Find these two numbers. Examples: Input: arr[] = {3, 1, 3} Output: Missing = 2, Repeating = 3 Explanation: In the array, 2 is missing and 3 occurs twice Input: arr[] = {4, 3, 6, 2, 1, 1} Output: Missing = 5, Repeating = 1

**How to find the missing number in a given Array from number 1 to n ,** If a single number is missing in an integer array that contains a sequence of numbers values, you can find it basing of the sum of numbers or, basing on the xor of the numbers. The sum of n sequential numbers will be [n*(n+1)]/2. Using this get the sum of the numbers the n numbers. Add all the elements in the array. The cell reference in the ROW (A1) part of the formula is relative, so as you copy the formula down column C, ROW (A1) becomes ROW (A2) which =2 and returns the second smallest missing number, ROW (A3) which is 3, returns the third smallest missing number and so on.

**How to find the missing number in a sorted array,** Given a sorted array of consecutive numbers from 1 1 1 to N N N, find the first missing number, i.e., the hole. The naive approach is to linearly traverse the array Find and highlight missing numbers from sequence: Kutools for Excel ’s Find Missing Sequence Number feature can helper you quickly and easily find the missing sequence, and insert the missing numbers or blank rows into the existing data sequence, or fill background color when encountering the missing sequence.

##### Comments

- You can iterate over the array and compare each two elements.
- Thanks Soufiane, is that going to catch each occurence in the array as the array will contain approx 100 numbers and how would i log the actualy number that isn't there, again i might be being thick.
- @Mark: Every number between
`numArray[i]`

and`numArray[i-1]`

, if the difference is greater than`1`

, is not in the array. - Yes, you are going to go through every item in the array, and to catch the missing number, when the difference is X (X != 1), your missing numbers are : numArray[i-1] + j (j>0 and j<X)
- The answer works, so my +1. Would be good to have how it works explained though ;-)
- This doesn't account for sequential missing values, amongst a few other issues. Try with this example: const numArray = [3,5,1,8,9]