Finding sequential digit numbers within a range in JavaScript

Sequential Digits Number

A number has sequential digits if and only if each digit in the number is one more than the previous digit.

Problem

We are required to write a JavaScript function that takes in an array, arr, of exactly two elements specifying a range.

Our function should return a sorted array of all the integers in the range arr (limits inclusive) that have sequential digits.

For example, if the input to the function is −

const arr = [1000, 13000];

Then the output should be −

const output = [1234, 2345, 3456, 4567, 5678, 6789, 12345];

Example

The code for this will be −

 Live Demo

const arr = [1000, 13000]; const sequentialDigits = ([low, high] = [1, 1]) => {    const findCount = (num) => {       let count = 0;       while(num > 0){          count += 1          num = Math.floor(num / 10)       };       return count;    };    const helper = (count, start) => {       let res = start;       while(count > 1 && start < 9){          res = res * 10 + start + 1;          start += 1;          count -= 1;       };       if(count > 1){          return 0;       };       return res;    };    const count1 = findCount(low);    const count2 = findCount(high);    const res = [];    for(let i = count1; i <= count2; i++){       for(let start = 1; start <= 8; start++){          const num = helper(i, start);          if(num >= low && num <= high){             res.push(num);          };       };    };    return res; }; console.log(sequentialDigits(arr));

Output

And the output in the console will be −

[    1234, 2345,    3456, 4567,    5678, 6789,    12345 ]