grid47 Exploring patterns and algorithms
Oct 2, 2024
5 min read
Solution to LeetCode 357: Count Numbers with Unique Digits Problem
Given an integer n, return the count of all numbers with unique digits x, such that 0 <= x < 10^n.
Problem
Approach
Steps
Complexity
Input: The input consists of a single integer n, where 0 <= n <= 8.
Example: n = 3
Constraints:
• 0 <= n <= 8
Output: The output is the count of numbers with unique digits in the range 0 <= x < 10^n.
Example: Input: n = 3
Output: 739
Constraints:
• The output should be the total number of valid numbers with unique digits.
Goal: To count all the numbers from 0 to 10^n-1 that have unique digits.
Steps:
• Start with the total valid numbers for n=1 (which is 10).
• Iterate for each subsequent value of n and compute the number of unique digit numbers for that value.
• For each n, reduce the available digits and multiply by the previous number of unique digits to get the total count.
Goal: The input is constrained to be between 0 and 8 inclusive, and the solution should handle these limits efficiently.
Steps:
• 0 <= n <= 8
Assumptions:
• The number of digits n is small enough that the approach of reducing the available digits and multiplying is efficient.
• Input: Input: n = 3
Output: 739
• Explanation: For n = 3, the total numbers in the range of 0 to 999 excluding numbers with repeated digits (such as 11, 22, etc.) is 739.
Approach: We will iterate through the possible numbers of digits and calculate the count of numbers with unique digits using combinatorics.
Observations:
• The numbers with unique digits are formed by selecting distinct digits from the available pool of digits (0-9).
• We can count how many numbers have unique digits for each number of digits from 1 to n using a decrementing approach.
Steps:
• 1. If n = 0, return 1 since only the number 0 is valid.
• 2. Initialize a result to 10 for the base case when n = 1.
• 3. For each additional digit, calculate how many numbers can be formed by choosing unique digits and multiplying by the previous count.
Empty Inputs:
• When n = 0, the answer should be 1, as there is only one number: 0.
Large Inputs:
• When n = 8, the result needs to be computed efficiently using the combinatorial approach.
Special Values:
• If n = 1, the result should be 10 (the numbers 0-9).
Constraints:
• Ensure the solution handles n values from 0 to 8.
classSolution {
publicintcountNumbersWithUniqueDigits(int n) {
if(n ==0) return1;
int res =10;
int available =9;
int unqNums =9;
while(n-->1&& available >0) {
unqNums = unqNums * available;
res += unqNums;
available--;
}
return res;
}
1 : Class Declaration
classSolution {
This line begins the declaration of the `Solution` class, which contains the method `countNumbersWithUniqueDigits`.
2 : Method Declaration
publicintcountNumbersWithUniqueDigits(int n) {
This is the declaration of the method `countNumbersWithUniqueDigits`, which takes an integer `n` as input and returns an integer representing the number of unique digit numbers that can be formed.
3 : Conditional Check
if(n ==0) return1;
This line checks if `n` is zero, and if so, it returns 1, as there is only one number (0) with zero digits.
4 : Variable Initialization
int res =10;
Initializes the variable `res` to 10, which accounts for the numbers with one digit (0-9).
5 : Variable Initialization
int available =9;
Initializes the variable `available` to 9, which represents the number of available digits (1-9) for the next digit in multi-digit numbers.
6 : Variable Initialization
int unqNums =9;
Initializes the variable `unqNums` to 9, which represents the number of possible unique digits for the next digit.
7 : Variable Initialization
This is a placeholder for a blank line or spacing for readability.
8 : Loop Declaration
while(n-->1&& available >0) {
This is a `while` loop that continues as long as `n` is greater than 1 and there are available digits to use. It iterates through each digit position.
9 : Variable Assignment
unqNums = unqNums * available;
Multiplies the `unqNums` variable by the number of available digits for the next digit, updating the possible unique numbers.
10 : Variable Update
res += unqNums;
Adds the newly calculated `unqNums` value to `res`, which keeps track of the total number of unique numbers formed.
11 : Variable Update
available--;
Decreases the `available` variable by 1, as one less digit is available for use in the next digit position.
12 : Return Statement
return res;
Returns the total number of unique digit numbers (`res`) after completing the loop.
Best Case: O(1) when n = 0.
Average Case: O(n) for values of n up to 8, where the time complexity increases linearly with the value of n.
Worst Case: O(8) since n is capped at 8.
Description: The solution iterates up to n times, where n is at most 8.
Best Case: O(1) when n = 0.
Worst Case: O(1) as the space complexity is constant and only a few variables are used.
Description: Space complexity is constant as we only store a few variables for calculation.
