Finding sequential digit numbers within a range in JavaScript

JavascriptWeb DevelopmentFront End Technology

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
]
raja
Published on 07-Apr-2021 07:34:10
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