How do you solve Factorial Trailing Zeroes on LeetCode?

Factorial Trailing Zeroes (LeetCode #172) 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: The number of trailing zeroes is determined by the number of pairs of factors 2 and 5 in the factorial's prime factorization.

What is the brute force solution for Factorial Trailing Zeroes?

The brute force approach for Factorial Trailing Zeroes is Brute Force, with O(n²) time complexity and O(1) space complexity. To find the number of trailing zeroes in n!, we can compute the factorial and then count the trailing zeroes. This approach is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Factorial Trailing Zeroes?

Factorial Trailing Zeroes uses the following concepts: Math. The recommended patterns to study are: Mathematical Counting, Greedy Algorithms.

What are the interview tips for Factorial Trailing Zeroes?

Always think about the mathematical properties of the problem rather than brute-forcing through computation. Clarify the problem statement and constraints before jumping into coding. Practice recognizing patterns in problems, such as counting factors or using logarithmic reductions.

What are common mistakes when solving Factorial Trailing Zeroes?

Students often try to compute the factorial directly, leading to overflow and inefficiency. Confusing the counting of trailing zeroes with counting zeroes in the entire number.

#172

Factorial Trailing Zeroes

Medium

MathMathematical CountingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(log n)Space O(1)

The number of trailing zeroes in n! is determined by the number of times 10 is a factor in the numbers from 1 to n. Since 10 is made from 2 and 5, and there are usually more factors of 2 than 5, we only need to count the factors of 5.

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Algorithm

3 steps

1Step 1: Initialize a count variable to 0.
2Step 2: While n is greater than or equal to 5, divide n by 5 and add the result to count.
3Step 3: Return the count.

solution.py7 lines

1# Full working Python code
2def trailing_zeroes(n):
3    count = 0
4    while n >= 5:
5        n //= 5
6        count += n
7    return count

ℹ

Complexity note: The time complexity is O(log n) because we keep dividing n by 5 until it becomes less than 5, which reduces the problem size logarithmically.

1The number of trailing zeroes is determined by the number of pairs of factors 2 and 5 in the factorial's prime factorization.
2Since factors of 2 are more frequent than factors of 5, we only need to count the factors of 5.

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