Counting n digit Numbers with all unique digits in JavaScript

Problem

We are required to write a JavaScript function that takes in a number, let’s say num, as the only argument. The function should count all such numbers that have num digits and all of their digits are unique.

For example, if the input to the function is −

const num = 1;

Then the output should be −

const output = 10;

Output Explanation:

The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 all have 1 digit and all are unique.

Example

The code for this will be −

 Live Demo

const num = 1;
const uniqueDigits = (num = 1) => {
   const dp = [1, 10];
   const sum = [1, 11];
   for (let i = 2; i <= num; i++) {
      dp[i] = sum[i - 1] + (10 - i) * (dp[i - 1]);
      sum[i] = sum[i - 1] + dp[i];
   };
   return dp[num];
};
console.log(uniqueDigits(num));
console.log(uniqueDigits(2));
console.log(uniqueDigits(3));

Code Explanation:

We have here used Dynamic Programming to keep track of desired numbers.

Output

And the output in the console will be −

10
91
739

AmitDiwan

Published on 18-Mar-2021 09:15:23

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