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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 −
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 ]
Updated on: 07-Apr-2021
299 Views
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