How do you solve Numbers With Repeated Digits on LeetCode?

Numbers With Repeated Digits (LeetCode #1012) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(d²) time complexity and O(d) space complexity. Key insight: Counting unique digits is more efficient than checking each number individually.

What is the brute force solution for Numbers With Repeated Digits?

The brute force approach for Numbers With Repeated Digits is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each number from 1 to n to see if it has repeated digits. This is straightforward but inefficient for large n. The optimal Optimal Solution approach improves this to O(d²).

What algorithm or data structure is used to solve Numbers With Repeated Digits?

Numbers With Repeated Digits uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Numbers With Repeated Digits?

Explain your thought process clearly. Consider edge cases like small values of n. Be ready to optimize your solution if asked.

What are common mistakes when solving Numbers With Repeated Digits?

Forgetting to handle leading zeros in multi-digit numbers. Not considering the case where digits can repeat in the same position.

#1012

Numbers With Repeated Digits

Hard

MathDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(d²)
Space	O(1)	O(d)

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Intuition

Time O(d²)Space O(d)

Instead of counting numbers with repeated digits directly, we can count the numbers without repeated digits and subtract from n. This is more efficient as it avoids checking each number individually.

⚙️

Algorithm

4 steps

1Step 1: Count the total numbers with unique digits up to the number of digits in n.
2Step 2: For each digit position, calculate how many unique digit numbers can be formed.
3Step 3: Handle the case where the number has the same prefix as n.
4Step 4: Subtract the count of unique digit numbers from n to get the count of numbers with repeated digits.

solution.py17 lines

1def countDupDigits(n):
2    str_n = str(n)
3    length = len(str_n)
4    count = 0
5    for i in range(1, length):
6        count += 9 * (10 ** (i - 1)) * (9 ** (i - 1))
7    used = set()
8    for i in range(length):
9        for j in range(1 if i == 0 else 0, int(str_n[i])):
10            if j not in used:
11                count += (9 - i) * (9 ** (length - i - 1))
12        if str_n[i] in used:
13            break
14        used.add(str_n[i])
15    return n - count
16
17print(countDupDigits(20)

ℹ

Complexity note: The time complexity is O(d²) where d is the number of digits in n. This is because we iterate over the digits and for each digit, we may check against previously used digits.

1Counting unique digits is more efficient than checking each number individually.
2Using sets helps track used digits effectively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.