How do you solve Count Distinct Integers After Removing Zeros on LeetCode?

Count Distinct Integers After Removing Zeros (LeetCode #3747) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: Removing zeros simplifies the problem to counting valid digit combinations.

What is the time complexity of Count Distinct Integers After Removing Zeros?

The optimal solution for Count Distinct Integers After Removing Zeros runs in O(log n) time and uses O(1) space. This approach runs in O(log n) because we only process the digits of n, making it efficient for large values.

What is the brute force solution for Count Distinct Integers After Removing Zeros?

The brute force approach for Count Distinct Integers After Removing Zeros is Brute Force, with O(n²) time complexity and O(n) space complexity. We can generate all integers from 1 to n, remove zeros, and store them in a set to count distinct values. This is straightforward but inefficient for large n. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Count Distinct Integers After Removing Zeros?

Count Distinct Integers After Removing Zeros uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Distinct Integers After Removing Zeros?

Explain your thought process clearly. Consider edge cases, like small values of n. Discuss time and space complexity during your solution.

What are common mistakes when solving Count Distinct Integers After Removing Zeros?

Not considering leading zeros in combinations. Overcounting distinct integers by not using a set.

#3747

Count Distinct Integers After Removing Zeros

Medium

MathDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(n)	O(1)

💡

Intuition

Time O(log n)Space O(1)

Instead of iterating through all numbers, we can count the distinct integers formed by digits 1-9, avoiding zeros altogether. This is efficient and leverages combinatorial counting.

⚙️

Algorithm

3 steps

1Step 1: Count the number of digits in n.
2Step 2: For each digit position, calculate combinations of digits (1-9) that can be formed without zeros.
3Step 3: Sum the counts of valid numbers formed for all digit lengths up to the length of n.

solution.py10 lines

1def countDistinctIntegers(n):
2    count = 0
3    str_n = str(n)
4    length = len(str_n)
5    for i in range(1, length):
6        count += 9 * (10 ** (i - 1))
7    for i in range(1, int(str_n[0]) + 1):
8        if i > 1:
9            count += 10 ** (length - 1) - 10 ** (length - 2)
10    return count

ℹ

Complexity note: This approach runs in O(log n) because we only process the digits of n, making it efficient for large values.

1Removing zeros simplifies the problem to counting valid digit combinations.
2Understanding digit positions helps in optimizing the counting process.

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