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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 −
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
Updated on: 18-Mar-2021
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