I need to generate array from letters A,H,O,J and numbers 0-9. I want to generate letters into new array in order AHOJ. The individual characters can appear elsewhere in new array but always in AHOJ order and they dont need to follow each other. Remaining indexes of new array will be filled with digits. Examples - A123H85O2J3, AHOJ9854273, 012AH851OJ3 and so on. The order of AHOJ must not be randomized within new array. I have this code, but is probably completly wrong. It only works as expected when condition equal to 7 is met.
let letters = ['A','H','O','J'];
let newArray =new Array(11);
let lastIndex = 10;
for (let i = 0; i < letters.length; i++) {
newIndex = Math.floor(Math.random() * lastIndex);
console.log(newIndex);
if (newIndex == 7) {
for (let j = 0; j < letters.length; j++) {
newArray.splice(j + 7,1,letters[j]);
}
for ( let k = 0; k < 7; k++) {
newArray.splice(k,1,Math.floor(Math.random() * 10).toString());
}
break;
}
else if (newIndex > 7) {
lastIndex = Math.floor(Math.random() * 6);
newArray.splice(lastIndex,1,letters[i]);
}
else {
newArray.splice(newIndex,1,letters[i]);
}
console.log(letters[i]);
}
console.log(newArray);
It can be simplified to one loop :
let letters = ['A','H','O','J'];
for (let l = 5; l < 12; l++) // for random indexes from 4 to 11
{
let position = Math.floor(Math.random() * l);
let digit = Math.floor(Math.random() * 10);
letters.splice(position, 0, digit);
}
console.log( letters.join(''), letters.length );
for (let i = 0; i < 100; i += 1) {
console.log(generate('AHOJ', 11))
}
function generate(word, length) {
// create array [0, 1, 2, ..., length - 1]
const indexes = [...Array(length).keys()]
// get unique random indexes for each character of the word
const reserved = [...word].map(() => {
// random index of indexes array
const index = Math.floor(Math.random() * indexes.length)
// remove the number at index from indexes array
const res = indexes.splice(index, 1)
// return removed number
return res[0]
})
const result = []
for (let i = 0, w = 0; i < length; i += 1) {
// if index is reserved, push a letter from word
if (reserved.includes(i)) {
result.push(word[w])
w += 1
// push random digit otherwise
} else {
result.push(Math.floor(Math.random() * 9))
}
}
return result.join('')
}
I implemented the Largest Triple Products algorithm, but I use sort which makes my time complexity O(nlogn). Is there a way to implement it without a temporary sorted array?
The problem:
You're given a list of n integers arr[0..(n-1)]. You must compute a list output[0..(n-1)] such that, for each index i (between 0 and n-1, inclusive), output[i] is equal to the product of the three largest elements out of arr[0..i] (or equal to -1 if i < 2, as arr[0..i] then includes fewer than three elements).
Note that the three largest elements used to form any product may have the same values as one another, but they must be at different indices in arr.
Example:
var arr_2 = [2, 4, 7, 1, 5, 3];
var expected_2 = [-1, -1, 56, 56, 140, 140];
My solution:
function findMaxProduct(arr) {
// Write your code here
if(!arr || arr.length === 0) return [];
let helper = arr.slice();
helper.sort((a,b)=>a-b); // THIS IS THE SORT
let ans = [];
let prod = 1;
for(let i=0; i<arr.length; i++) {
if(i < 2) {
prod *= arr[i];
ans.push(-1);
}
else {
if(i === 3) {
prod *= arr[i];
ans.push(prod);
} else if(arr[i] < helper[0]) {
ans.push(prod);
} else {
const min = helper.shift();
prod /= min;
prod *= arr[i];
ans.push(prod);
}
}
}
return ans;
}
Thanks
You don't need to sort it. You just maintain an array of the largest three elements at each index.
For the first three elements it is simple you just assign the product of them to the third element in the result.
For the next elements, you add the current element to the three-largest-element-array and sort it and take the elements from 1 to 3 ( the largest three ) and assign the product of those at that index in result array. Then update the three-element-array with largest three.
Complexity :
This sort and slice of three-element-array should be O(1) because each time atmost 4 elements are there in the array.
Overall complexity is O(n).
You can do it as follows :
function findMaxProduct(arr) {
if(!arr) return [];
if (arr.length < 3) return arr.slice().fill(-1)
let t = arr.slice(0,3)
let ans = arr.slice().fill(-1,0,2) //fill first two with -1
ans[2] = t[0]*t[1]*t[2];
for(let i=3; i<arr.length; i++) {
t.push(arr[i]);
t = t.sort().slice(1,4);
ans[i] = t[0]*t[1]*t[2];
}
return ans;
}
I am keeping the array ordered (manually). Then just get the first 3 elements.
function findMaxProduct(arr) {
let results = [];
let heap = [];
for (let i = 0; i < arr.length; i++) {
// Insert the new element in the correct position
for (let j = 0; j < heap.length; j++) {
if (arr[i] >= heap[j]) {
heap.splice(j, 0, arr[i]);
break;
}
}
// No position found, insert at the end
if (heap.length != i + 1) {
heap.push(arr[i]);
}
if (i < 2) {
results.push(-1);
} else {
results.push(heap[0] * heap[1] * heap[2]);
}
}
return results;
}
You can make an array that holds three currently largest integers, and update that array as you passing through original array. That's how you will always have three currently largest numbers and you will be able to solve this with O(n) time complexity.
I think there's a faster and more efficient way to go about this. This is a similar thought process as #Q2Learn, using Python; just faster:
def findMaxProduct(arr):
#create a copy of arr
solution = arr.copy()
# make first 2 elements -1
for i in range(0,2):
solution[i] = -1
#for each item in copy starting from index 2, multiply item from 2 indices b'4 (notice how each index of arr being multiplied is reduced by 2, 1 and then 0, to accommodate each move)
for i in range(2, len(arr)):
solution[i] = arr[i-2] * arr[i-1] * arr[i]
return solution
check = findMaxProduct(arr)
print(check)
Single Scan Algorithm O(n)
We don't need to necessarily sort the given array to find the maximum product. Instead, we can only find the three largest values (x, y, z) in the given stage of iteration:
JavaScript:
function findMaxProduct(arr) {
let reults = []
let x = 0
let y = 0
let z = 0
for(let i=0; i<arr.length; i++) {
n = arr[i]
if (n > x) {
z = y
y = x
x = n
}
if (n < x && n > y) {
z = y
y = n
}
if (n < y && n > z) {
z = n
}
ans = x*y*z
if (ans === 0) {
results.push(-1)
} else {
results.push(ans)
}
return ans;
}
Python:
def findMaxProduct(arr):
results = []
if not arr:
return []
x = 0
y = 0
z = 0
for i, n in enumerate(arr):
if n > x:
z = y
y = x
x = n
if n < x and n > y:
z = y
y = n
if n < y and n > z:
z = n
ans = x*y*z
if ans == 0:
results.append(-1)
else:
results.append(ans)
print(results)
public int[] LargestTripleProducts(int[] input)
{
var ansArr = new int[input.Length];
var firstLargetst = input[0];
var secondLargetst = input[1];
ansArr[0] = ansArr[1] = -1;
for (int i = 2; i < input.Length; i++)
{
ansArr[i] = firstLargetst * secondLargetst * input[i];
if (firstLargetst < input[i] && firstLargetst < secondLargetst)
{
firstLargetst= input[i];
continue;
}
if (secondLargetst < input[i] && secondLargetst < firstLargetst)
{
secondLargetst= input[i];
}
}
return ansArr;
}
Python solution based on #SomeDude answer above. See explanation there.
def findMaxProduct(arr):
if not arr:
return None
if len(arr) < 3:
for i in range(len(arr)):
arr[i] = -1
return arr
three_largest_elem = arr[0:3]
answer = arr.copy()
for i in range(0, 2):
answer[i] = -1
answer[2] = three_largest_elem[0] * three_largest_elem[1] * three_largest_elem[2]
for i in range(3, len(arr)):
three_largest_elem.append(arr[i])
three_largest_elem = sorted(three_largest_elem)
three_largest_elem = three_largest_elem[1:4]
answer[i] = three_largest_elem[0] * three_largest_elem[1] * three_largest_elem[2]
return answer #Time: O(1) n <= 4, to Overall O(n) | Space: O(1)
Python has it's in-built package heapq, look at it for it.
Credit: Martin
> Helper function for any type of calculations
import math
> Heap algorithm
import heapq
> Create empty list to append output values
output = []
def findMaxProduct(arr):
out = []
h = []
for e in arr:
heapq.heappush(h, e)
if len(h) < 3:
out.append(-1)
else:
if len(h) > 3:
heapq.heappop(h)
out.append(h[0] * h[1] * h[2])
return out
Hope this helps!
I have an integer X and a list of all the factors of X. I want to output the ratio between X and a factor, and for each time set X to that ratio. I want this algorithm to continue till the ratio is bigger than or equal to 1.
For example: X = 36.
All factors (but 1 and 36): 2, 3, 4, 6, 9, 12, 18,
Algorithm: 36 / 2 = 18 --> 18 / 3 = 6 --> 6 / 4 = 1.5 --> 1.5 / 6 < 1 --> Stop
Output: [18, 6, 1.5]. Question:
How do I get this output?
What i have written:
var arr = [];
for (var i = 2; i < X; i++) {
if (X % i == 0) {
arr.push(i);
}
}
var temp = [];
var index = 0;
while (X / arr[index] >= 1) {
index += 1;
X = X / arr[index];
temp.push(X / arr[index]);
}
var arr = [];
var X = 36
for (var i = 2; i < X; i++) {
if (X % i == 0) {
arr.push(i);
}
}
var temp = arr.reduce((acc, item, index) => {
if(!(X/item < 1)){
acc.push(X/item)
X = X/item;
}
return acc;
}, [])
console.log(temp)
You could calculate the first value before you go into the while loop an check the asssigned x. If the value passes the condition, you could push that value to the result array and increment the index and calculate a new value.
function fn(x) {
var array = [], // declare all variable on top
i,
result = [];
for (var i = 2; i < x; i++) {
if (x % i == 0) {
array.push(i);
}
}
i = 0; // prepare index for looping
x /= array[i]; // take a first possible value
while (x >= 1) { // check value
result.push(x); // push value
i++; // increment index
x /= array[i]; // take the next value for checking
}
return result;
}
console.log(fn(36))
A bit shorter with Array#every
function fn(x) {
var array = [], // declare all variable on top
i,
result = [];
for (var i = 2; i < x; i++) {
if (x % i == 0) {
array.push(i);
}
}
array.every(v => (x /= v) >= 1 && result.push(x));
return result;
}
console.log(fn(36))
Your question is a little bit confuse. Is this what you want?
Basically, I update the factor after each iteration (if >= 1) and store into an array the output.
// Generate a range of integers (starting at 2, and we should not including the input)
// get only values divisible by the input (filter)
const generateRange = (input) => {
return Array.from({length: (input - 2)}, (v, k) => k + 2).filter(i => {
return input % i === 0;
})
}
const divideFactors = (input) => {
const list = generateRange(input) // filtered range
let output = []
let factor = input
for (const item of list) {
let ratio = factor / item
if (ratio < 1) { //if ratio < 1, break the loop
break;
}
output.push(ratio) // update output array with new ratio
factor = ratio; // update factor variable with current ratio
}
return output
}
console.log(divideFactors(36))
More info:
Array.from: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/from
For of: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/for...of
My goal is to make a randomly generated 2D Array in Javascript, that has an X amount of the same one character value while the rest of the values are equal to another character.
In this example, there are 10 rows and 10 columns for the 2D Array. 20 out of the possible 100 values of the Array should be equal to 'Y' (for yes) and the 80 others should be 'N' (for no). I want the 'Y's to be randomly placed all over the Array, and I absolute need exactly 20 of them to be 'Y's and the rest 'N's.
I had a less efficient way before, and I thought to try this approach, where after I define the Array, I make the first X amount of values a 'Y' and then the rest all 'N's. Then I shuffle the array, (using the shuffle from the underscore library) so that the 'Y's are all spread out randomly everywhere.
Is this an efficient way of getting what I need done? Are there any better solutions? I tried making a JSFiddle with my example, but the site appears to be down at the moment.
(I was unable to test my code yet to see if the shuffle worked correctly on my 2D array)
var rows = 10;
var cols = 10;
var elements = 20;
//Define Empty Array
var test = new Array(rows);
for (var k = 0; k < rows; k++)
{
test[k] = Array(cols);
}
var i = 1;
for (var x = 0; x < rows; x++)
{
for (var y = 0; y < cols; y++)
{
if (i <= elements)
{
test[x][y] = "Y";
}
else
{
test[x][y] = "N";
}
}
}
//Shuffle all those values so they're no longer in order
var shuffledTest = _.shuffle(test);
//Print in rows
for (var x = 0; x < rows; x++)
{
console.log(shuffledTest[x]);
}
A very simple solution is to first create an array, fill it with a number of "N"s, insert the "Y"s at random indexes, and then finally splitting it into the 2-dimensional array that you want:
var tmpArr = [], // Temporary 1-dimensional array to hold all values
arr = [], // The final 2-dimensional array
rows = 10,
cols = 10,
elements = 20; // Number of "Y"s
// 1. Fill temporary array with "N"s
for (var i = 0; i < rows * cols - elements; i += 1) {
tmpArr.push("N");
}
// 2. Insert "Y"s at random indexes in the temporary array
for (var i = 0; i < elements; i += 1) {
var index = Math.round(Math.random() * (tmpArr.length + 1));
tmpArr.splice(index, 0, "Y");
}
// 3. Split temporary array into 10 seperate arrays
// and insert them into the final array
for (var i = 0; i < rows; i += 1) {
var row = tmpArr.slice(i * cols, (i + 1) * cols);
arr.push(row);
}
JSBin to illustrate: http://jsbin.com/luyacora/1/edit
You can try this solution, it uses underscores range to create a pair of arrays to use as iterators, though their values don't matter.
Play around with the randomizer function to get an even distribution of 'y's
JSBIN: http://jsbin.com/yaletape/1/
var rows = _.range(0, 10, 0);
var columns = _.range(0, 10, 0);
function randomizer(mult){
return Math.floor((Math.random()*mult)+1);
}
var y_count = 0;
var matrix = _.map(rows, function(){
return _.map(columns, function(v, i){
var value;
var y_allowed = randomizer(3);
var current_y_count = 0;
if(y_count < 20 && current_y_count < y_allowed){
var rand = randomizer(5);
if(rand > 4){
value = 'y';
current_y_count++;
y_count++;
}
}
if(!value){
value = 'n';
}
return value;
});
});
//The above could be simplified to
var matrix = _.range(0,10,0).map(function(){
return _.range(0,10,0).map(function(){
//put the logic code from above here
});
});
Maybe shuflle a 2D array is not the best way. As #Zeb mentioned, here is some code that fill random positions with the 'Y' value. After that, the other positions are filled with 'N'.
http://plnkr.co/edit/avyKfgsgOSdAkRa1WOsk
var arr = [];
var cols = 10;
var rows = 10;
var positions = rows*cols; // 100
var YQty = 10; // only 10 'Y' are needed
// 'Y' values.
for(i = 0; i < YQty; i++)
{
do
{
x = parseInt(Math.random() * cols);
y = parseInt(Math.random() * rows);
filled = false;
if (typeof(arr[x]) == "undefined")
{
arr[x] = [];
}
if (typeof(arr[x][y]) == "undefined")
{
arr[x][y] = 'Y';
filled = true;
}
}
while (!filled);
}
// 'N' values.
for (x = 0; x < cols; x++)
{
if (typeof(arr[x]) == "undefined")
{
arr[x] = [];
}
for (y = 0; y < rows; y++)
{
if (arr[x][y] != 'Y')
{
arr[x][y] = 'N';
}
}
}
Shuffling the multidimensional array is not the best approach. Seeing as any sort is worse than linear time complexity. The easiest solution would be to create your multidimensional array and then set each index value to the char you want the 'rest' of the values to be. Then for 1 -> the number of other char value choose a random index and set that to the char.
Note: If the randomly picked spot has already been changed you need to choose a new one to make sure you have the right amount at the end.
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
}
}
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]
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
*/
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]));
This can now be done easily as a one-liner with the find method:
const arr = [1,2,3,5,6,7,8,9];
return arr.find((x,i) => arr[i+1]-x > 1) + 1
//4
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;
};
const findMissing = (numarr) => {
for(let i = 1; i <= numarr.length; i++) {
if(i - numarr[i-1] !== 0) {
console.log('found it', i)
break;
} else if(i === numarr.length) console.log('found it', numarr.length + 1)
}
};
console.log(findMissing([1,2,3,4,5,6,7,8,9,10,11,12,13,14]))
It would be fairly straightforward to sort the array:
numArray.sort();
Then, depending upon what was easiest for you:
You could just traverse the array, catching sequential patterns and checking them as you go.
You could split the array into multiple arrays of sequential numbers and then check each of those separate arrays.
You could reduce the sorted array to an array of pairs where each pair is a start and end sequence and then compare those sequence start/ends to your other data.
function missingNum(nums){
const numberArray = nums.sort((num1, num2)=>{
return num1 - num2;
});
for (let i=0; i < numberArray.length; i++){
if(i !== numberArray[i]){
return i;
}
}
}
console.log(missingNum([0,3,5,8,4,6,1,9,7]))
Please check below code.....
function solution(A) {
var max = Math.max.apply(Math, A);
if(A.indexOf(1)<0) return 1;
var t = (max*(max+1)/2) - A.reduce(function(a,b){return a+b});
return t>0?t:max+1;
}
Try as shown below
// Find the missing number
let numArray = [0189459, 0189460, 0189461, 0189463, 0189468];
let numLen = numArray.length;
let actLen = Number(numArray[numLen-1])-Number(numArray[0]);
let allNumber = [];
for(let i=0; i<=actLen; i++){
allNumber.push(Number(numArray[0])+i);
}
[...allNumber].forEach(ele=>{
if(!numArray.includes(ele)){
console.log('Missing Number -> '+ele);
}
})
I use a recursive function for this.
function findMissing(arr, start, stop) {
var current = start,
next = stop,
collector = new Array();
function parseMissing(a, key) {
if(key+1 == a.length) return;
current = a[key];
next = a[key + 1];
if(next - current !== 1) {
collector.push(current + 1);
// insert current+1 at key+1
a = a.slice( 0, key+1 ).concat( current+1 ).concat( a.slice( key +1 ) );
return parseMissing(a, key+1);
}
return parseMissing(a, key+1);
}
parseMissing(arr, 0);
return collector;
}
Not the best idea if you are looking through a huge set of numbers. FAIR WARNING: recursive functions are resource intensive (pointers and stuff) and this might give you unexpected results if you are working with huge numbers. You can see the jsfiddle. This also assumes you have the array sorted.
Basically, you pass the "findMissing()" function the array you want to use, the starting number and stopping number and let it go from there.
So:
var missingArr = findMissing(sequenceArr, 1, 10);
let missing = [];
let numArray = [3,5,1,8,9,36];
const sortedNumArray = numArray.sort((a, b) => a - b);
sortedNumArray.reduce((acc, current) => {
let next = acc + 1;
if (next !== current) {
for(next; next < current; next++) {
missing.push(next);
}
}
return current;
});
Assuming that there are no duplicates
let numberArray = [];
for (let i = 1; i <= 100; i++) {
numberArray.push(i);
}
let deletedArray = numberArray.splice(30, 1);
let sortedArray = numberArray.sort((a, b) => a - b);
let array = sortedArray;
function findMissingNumber(arr, sizeOfArray) {
total = (sizeOfArray * (sizeOfArray + 1)) / 2;
console.log(total);
for (i = 0; i < arr.length; i++) {
total -= arr[i];
}
return total;
}
console.log(findMissingNumber(array, 100));
Here is the most efficient and simple way to find the missing numbers in the array. There is only one loop and complexity is O(n).
/**
*
* #param {*} item Takes only the sorted array
*/
function getAllMissingNumbers(item) {
let first = 0;
let second = 1;
let currentValue = item[0];
const container = [];
while (first < second && item[second]) {
if ((item[first] + 1) !== item[second]) { // Not in sequence so adds the missing numbers in an array
if ((currentValue + 1) === item[second]) { // Moves the first & second pointer
first = second;
second++;
currentValue = item[first];
} else { // Adds the missing number between two number
container.push(++currentValue);
}
} else { // Numbers are in sequence so just moves the first & second pointer
first = second;
second++;
currentValue = item[first];
}
}
return container;
}
console.log(getAllMissingNumbers([0189459, 0189460, 0189461, 0189463, 0189465].sort( (a, b) => a - b )));
console.log(getAllMissingNumbers([-5,2,3,9]));
Adding one more similar method
Find the min and max of the numbers in the array
Loop with the max and min numbers to get the full list
compare the full list of numbers with the input array to get the difference
const array = [0189459, 0189460, 0189461, 0189463, 0189465]
const max = Math.max(...array)
const min = Math.min(...array)
let wholeNumber = []
for(var i = min ;i<=max ;i++ ){
wholeNumber.push(i)
}
const missing = wholeNumber.filter((v)=>!array.includes(v))
console.log('wholeNumber',wholeNumber)
console.log('missingNumber',missing)
Here's a variant of #Mark Walters's function which adds the ability to specify a lower boundary for your sequence, for example if you know that your sequence should always begin at 0189455, or some other number like 1.
It should also be possible to adjust this code to check for an upper boundary, but at the moment it can only look for lower boundaries.
//Our first example array.
var numArray = [0189459, 0189460, 0189461, 0189463, 0189465];
//For this array the lowerBoundary will be 0189455
var numArrayLowerBoundary = 0189455;
//Our second example array.
var simpleArray = [3, 5, 6, 7, 8, 10, 11, 13];
//For this Array the lower boundary will be 1
var simpleArrayLowerBoundary = 1;
//Build a html string so we can show our results nicely in a div
var html = "numArray = [0189459, 0189460, 0189461, 0189463, 0189465]<br>"
html += "Its lowerBoundary is \"0189455\"<br>"
html += "The following numbers are missing from the numArray:<br>"
html += findMissingNumbers(numArray, numArrayLowerBoundary);
html += "<br><br>"
html += "simpleArray = [3, 5, 6, 7, 8, 10, 11, 13]<br>"
html += "Its lowerBoundary is \"1\".<br>"
html += "The following numbers are missing from the simpleArray:<br>"
html += findMissingNumbers(simpleArray, simpleArrayLowerBoundary);
//Display the results in a div
document.getElementById("log").innerHTML=html;
//This is the function used to find missing numbers!
//Copy/paste this if you just want the function and don't need the demo code.
function findMissingNumbers(arrSequence, lowerBoundary) {
var mia = [];
for (var i = 0; i < arrSequence.length; i++) {
if (i === 0) {
//If the first thing in the array isn't exactly
//equal to the lowerBoundary...
if (arrSequence[i] !== lowerBoundary) {
//Count up from lowerBoundary, incrementing 1
//each time, until we reach the
//value one less than the first thing in the array.
var x = arrSequence[i];
var j = lowerBoundary;
while (j < x) {
mia.push(j); //Add each "missing" number to the array
j++;
}
} //end if
} else {
//If the difference between two array indexes is not
//exactly 1 there are one or more numbers missing from this sequence.
if (arrSequence[i] - arrSequence[i - 1] !== 1) {
//List the missing numbers by adding 1 to the value
//of the previous array index x times.
//x is the size of the "gap" i.e. the number of missing numbers
//in this sequence.
var x = arrSequence[i] - arrSequence[i - 1];
var j = 1;
while (j < x) {
mia.push(arrSequence[i - 1] + j); //Add each "missing" num to the array
j++;
}
} //end if
} //end else
} //end for
//Returns any missing numbers, assuming that lowerBoundary is the
//intended first number in the sequence.
return mia;
}
<div id="log"></div> <!-- Just used to display the demo code -->