LeetCode 357: Count Numbers with Unique Digits Solution
Master LeetCode problem 357 (Count Numbers with Unique Digits), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
357. Count Numbers with Unique Digits
Problem Explanation
Explanation
To solve this problem, we can use a dynamic programming approach. We need to count the numbers with unique digits up to 10^n. We can iterate from 0 to n and calculate the count of unique numbers with the given number of digits. We can use the concept of permutations to calculate the count.
- For n = 0, the answer is always 1 (which is 0 itself).
- For n = 1, the answer is 10 (0 to 9).
- For n = 2, we have 9 choices for the first digit (1-9) and 9 choices for the second digit (0-9 excluding the first digit). So the count is 9 * 9 = 81.
- For n = 3, we have 9 choices for the first digit, 9 choices for the second digit, and 8 choices for the third digit (0-9 excluding the first two digits). So the count is 9 * 9 * 8 = 648.
- This pattern continues for higher values of n.
Therefore, we can sum up the counts for each value of n to get the total count of numbers with unique digits up to 10^n.
Time complexity: O(n) Space complexity: O(1)
Solution Code
class Solution {
public int countNumbersWithUniqueDigits(int n) {
if (n == 0) return 1;
int count = 10;
int uniqueDigits = 9;
int availableDigits = 9;
for (int i = 2; i <= n && availableDigits > 0; i++) {
uniqueDigits = uniqueDigits * availableDigits;
count += uniqueDigits;
availableDigits--;
}
return count;
}
}Try It Yourself
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