Factorial Trailing Zeroes
To calculate the number of trailing zeroes in factorial of a number, we need to find how many times the number 10 is a factor in the factorial result. Since 10 is the product of 2 and 5, we can simply count the number of pairs of 2 and 5.
However, because factorials have more factors of 2 than 5, we just need to count the number of 5s in the factors of the factorial. More specifically, for each number n, we add floor(n/5) to our count. Then we continue with n = floor(n/5) until n becomes 0. This will count the number of factors of 5 in n, n/5, n/25, n/125, and so on.
Here is the code:


This solution works in O(log n) time complexity and O(1) space complexity.
Identifying Problem Isomorphism
“Factorial Trailing Zeroes” can be approximately mapped to “Preimage Size of Factorial Zeroes Function”.
In the “Factorial Trailing Zeroes”, you are given an integer n and you have to return the number of trailing zeroes in n!. The main concept behind this problem is to count the number of factors of 5 in the factorial, as each factor of 5 contributes a trailing zero.
In the “Preimage Size of Factorial Zeroes Function”, you are also dealing with trailing zeroes of a factorial. Here you’re given an integer k and you need to find how many nonnegative integers x have the property that f(x) = k, where f(x) is the number of trailing zeroes in x!.
The reason for this mapping is that both problems deal with the number of trailing zeroes in the factorial of a number. While “Factorial Trailing Zeroes” asks for the number of zeroes in a given factorial, “Preimage Size of Factorial Zeroes Function” uses this concept as a function to determine the count of numbers that have a certain number of trailing zeroes in their factorial.
“Factorial Trailing Zeroes” is simpler since it directly involves counting the factors of 5 in a given number’s factorial. “Preimage Size of Factorial Zeroes Function” requires a deeper understanding of the mathematical concept and might involve binary search to find the preimages efficiently.
10 Prerequisite LeetCode Problems
“172. Factorial Trailing Zeroes” deals with factorials, prime factors, and has a bit of mathematical insights related to numbers. Here are some problems to prepare:
7. Reverse Integer: This problem will familiarize you with dealing with the digits of a number.
9. Palindrome Number: Similar to problem 7, it deals with analyzing digits of a number.
50. Pow(x, n): This problem will help you understand the concept of power and its implications.
69. Sqrt(x): It is a relatively simpler problem which will help understand logarithmic time complexity.
168. Excel Sheet Column Title: It helps to understand how to convert a number to another number system.
204. Count Primes: This problem is good for understanding the concept of prime numbers, which is useful for the problem.
231. Power of Two: This problem teaches you to understand properties of numbers which are powers of 2.
326. Power of Three: Similar to problem 231, but this time it’s about the powers of 3.
342. Power of Four: This problem, again, is about understanding properties of numbers which are powers of 4.
628. Maximum Product of Three Numbers: This problem helps in understanding the multiplication property of numbers in an array.
While solving “172. Factorial Trailing Zeroes”, you should understand how the trailing zeroes in the factorial of a number are related to the number of pairs of 2 and 5 in its prime factors, as 2*5=10. The prime factor 2 appears more frequently, so the number of trailing zeroes is actually determined by the number of prime factor 5.
Clarification Questions
What are the clarification questions we can ask about this problem?
Problem Analysis and Key Insights
What are the key insights from analyzing the problem statement?
Problem Boundary
What is the scope of this problem?
How to establish the boundary of this problem?
Problem Classification
The problem falls under the category of Mathematics and Number Theory, specifically dealing with factorials and their properties.
What Components
 Input: An integer
n
where0 <= n <= 104
.  Output: An integer representing the number of trailing zeroes in
n!
.  Constraints: The problem asks for an optimized logarithmic time complexity solution in the followup.
 Operations: Factorial calculation, zerocounting.
This is an Analytical Problem, requiring mathematical insight into the structure of factorials to solve efficiently. It’s not just about computing the factorial and then counting zeroes, as that would be computationally expensive for large numbers. The problem subtly hints at the necessity of an optimized solution, especially with the followup question about logarithmic time complexity.
It has an Optimization Component. It asks for the ’number of trailing zeroes’, which means we are looking for a minimal representation of something (zeroes, in this case), making it an optimization query.
Distilling the Problem to Its Core Elements
Fundamental Concept
The fundamental concept here is Counting Factors within the factorial number. Specifically, the problem focuses on finding the number of times the factor 10 can divide n!
without a remainder, as each factor of 10 will add a trailing zero.
Simplest Description
If you multiply all numbers from 1 up to a given number n
(this is called factorial), how many zeroes will be at the end of that big number?
Core Problem
The core problem is to find how many times the number 10 is a factor in n!
, without actually calculating n!
.
Key Components
 Input Number: The number
n
for which we need to find the factorial.  Factor Counting: The mechanism to count factors of 10 in
n!
.  Output: The number of trailing zeroes in
n!
.
Minimal Set of Operations
 Initialization: Start with zero count for trailing zeros.
 Loop through Divisors: Loop to find the power of 5 in the prime factorization of
n!
. Each pair of 2 and 5 in the factorization creates a 10, which adds a trailing zero.  Accumulation: Add the count of such pairs to a counter.
 Output: Return the counter as the number of trailing zeroes.
This avoids calculating n!
, instead focusing on counting relevant factors.
Visual Model of the Problem
To visualize this problem, think of it as a sequence of multiplication from 1 to n
. Now, imagine each number in this sequence breaking down into its prime factors.
 Factorial Sequence: Picture a line or a row of numbers from 1 to
n
. This representsn!
when all these numbers are multiplied together.
1, 2, 3, 4, 5, ..., n
 Prime Factorization: Replace each number with its prime factors. Pay special attention to the numbers that are multiples of 2 or 5, as these contribute to the trailing zeroes.
1, (2), 3, (2 * 2), (5), 6 (2 * 3), ..., (2 * 5), ...
 Pairing 2s and 5s: Visualize pairing a ‘2’ and a ‘5’ to make a ‘10’. Each ‘10’ will contribute a trailing zero to the result.
10, 10, 10, ...
 Counting 10s: Finally, picture a counter that increments every time you make a ‘10’.
By visualizing the problem this way, you see that you don’t need to calculate the actual factorial. Instead, you can focus on the key elements that contribute to trailing zeroes: the 2s and 5s that you can pair up to make 10s.
Problem Restatement
The problem asks for the number of trailing zeros in the factorial of a given integer n
. In other words, you calculate n!
(n factorial), which is the product of all integers from 1 to n
, and then count how many zeros are at the end of that number. The value of n
can be as low as 0 and as high as 10,000. The problem also hints at a more efficient, logarithmictime solution as a followup.
Abstract Representation of the Problem
The abstract representation of this problem centers around finding the count of a specific digit pattern—in this case, trailing zeros—at the end of a calculated factorial number ( N! ) for a given ( N ). The challenge lies in doing this within computational boundaries defined by the value range for ( N ), which is [0, 10^4].
Key elements:
 Input Integer ( N )
 Factorial Calculation ( N! )
 Pattern Counting (Trailing zeros)
 Constraints on ( N )
 Efficiency (hinted as logarithmic time complexity)
This problem can be generalized to finding patterns in the results of mathematical operations under certain constraints.
Terminology
Factorial (n!): The product of all positive integers less than or equal to ( n ). It is central to the problem as we’re analyzing the trailing zeros in ( n! ).
Trailing Zeros: These are the zeros that appear at the end of a number. The core problem is to count these in ( n! ).
Logarithmic Time Complexity: This is a hint about the efficiency of the desired solution. In computational complexity theory, an algorithm has logarithmic time complexity if the time to run it is proportional to the logarithm of the size of the input.
Constraints: These are the limitations imposed on the input, in this case, ( 0 \leq n \leq 10^4 ). Constraints are crucial for determining the boundaries within which the solution must operate.
Understanding these terms is essential for formulating both a bruteforce and an optimized solution for counting the trailing zeros in ( n! ).
Problem Simplification and Explanation
At its core, the problem is asking how many zeros are at the end of a really big number, specifically a factorial number ( n! ).
Key Concepts:
Factorial: Imagine a chain of multiplication where you start with the number ( n ) and keep multiplying it by one less each time until you reach 1. For example, ( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 ).
Trailing Zeros: These are the zeros at the end of a number. For example, in 120, there is one trailing zero.
Efficiency: The challenge is to do this as fast as possible, ideally in a way that takes less time as the numbers get bigger (logarithmic time complexity).
Interaction:
 Factorials result in a large number, and that number may or may not have trailing zeros.
 Trailing zeros come from multiplying 10s, and 10s are the result of multiplying 2s and 5s.
 To find out the number of trailing zeros in ( n! ), you essentially need to find out how many times ( n! ) is divisible by 10.
Analogy:
Imagine you’re making a “recipe” that results in a big cake, and the cake can have different layers of flavors. The flavors are your factorial numbers. The question is how many vanilla layers are at the bottom of this cake. The vanilla layers represent the trailing zeros, and you can only get a vanilla layer by mixing chocolate and strawberry flavors together (2s and 5s in our original number). You want to find the most efficient way to figure out the number of vanilla layers without having to make the entire cake.
By breaking down the ingredients (factors of the number), you can estimate how many vanilla layers you’ll end up with without actually making the cake (computing the large factorial number).
Constraints
Understanding the characteristics and constraints of a problem can often lead to a more efficient solution. In this problem, there are a few aspects that can be exploited for an efficient solution.
Key Characteristics to Exploit:
Limited Range: ( n ) is constrained between 0 and 10^4, which gives us a finite set to work with.
Multiplication of 2 and 5: Trailing zeros are formed by the multiplication of 2 and 5. Since there are usually more 2’s than 5’s in the factors of ( n! ), the number of trailing zeros is essentially determined by the number of times 5 appears as a factor in ( n! ).
Factors of 5: Some numbers contribute more than one 5. For example, 25 contributes two (5 x 5), 125 contributes three (5 x 5 x 5), and so on. These should be counted multiple times.
Divisibility by 5: A pattern emerges where you can find how many numbers up to ( n ) are divisible by 5, then how many are divisible by 25, 125, and so on, to calculate the total number of 5s in the factors of ( n! ).
Logarithmic Possibility: The number of 5s can be found without iterating through every number up to ( n ), which allows for a more efficient, logarithmic time complexity solution.
By understanding these key characteristics, you can aim to solve the problem in a way that takes advantage of these specific patterns or numerical ranges.
From analyzing the constraints, the key insights are:
Limited Input Range: With ( n ) capped at 10^4, the problem doesn’t require handling extremely large numbers. However, factorial calculations grow rapidly, making direct calculation impractical. This constraint suggests that an optimized approach is necessary but within reach.
Trailing Zero Formation: Understanding that trailing zeros are formed by the multiplication of 2 and 5 narrows down our focus to counting the number of times 5 appears as a factor in ( n! ).
Multiple Contributions: Some numbers like 25, 125, etc., contribute more than one 5 as a factor. This suggests we can’t just simply count numbers divisible by 5; we have to count their multiplicative contributions as well.
Pattern in 5’s: Since 5’s are our focus and numbers that are powers of 5 contribute more than one 5, a pattern exists that we can exploit to count these efficiently, possibly even in logarithmic time.
Efficient Solution Feasible: The constraints make it clear that a more efficient, even logarithmic, solution is feasible. We don’t need to calculate the factorial or iterate through each of its components to find the trailing zeros.
These insights offer clues to approach the problem efficiently, reducing the computational load by targeting specific characteristics of the problem.
Case Analysis
Additional examples can reveal nuances in the problem, especially in terms of constraints and edge cases. Here are some:
Small Values of ( n )
Case: Zero Factorial (Edge Case)
 Input: ( n = 0 )
 Output: 0
 Reasoning: 0! is defined as 1, which has no trailing zeros.
Case: Single Digit Factorials
 Input: ( n = 3 )
 Output: 0
 Reasoning: 3! = 6, which has no trailing zeros.
Case Where ( n ) is a Power of 5
 Case: ( n ) is 25
 Input: ( n = 25 )
 Output: 6
 Reasoning: 25 contributes two 5’s (5 x 5), other numbers like 20, 15, 10, and 5 contribute one 5 each, total 6.
High Value of ( n )
 Case: Upper Bound (Edge Case)
 Input: ( n = 10^4 )
 Output: Will depend on the exact algorithm, but this is to test performance.
Case Where ( n ) is a Multiple of 10
 Case: ( n ) is 30
 Input: ( n = 30 )
 Output: 7
 Reasoning: 25 contributes two 5’s, 20, 15, 10, and 5 contribute one each, and 30 contributes one more, total 7.
Case Where ( n ) is a Prime Number
 Case: ( n ) is 7
 Input: ( n = 7 )
 Output: 1
 Reasoning: Only 5 contributes one 5 to the factorial.
Odd Cases
 Case: ( n ) is 9
 Input: ( n = 9 )
 Output: 1
 Reasoning: Again, only 5 contributes one 5 to the factorial.
Edge Cases Summary
 ( n = 0 ): The factorial is 1, so zero trailing zeros.
 Upper Bound ( n = 10^4 ): Tests algorithm performance.
By considering a range of values, including edge cases and multiples of 5, we can make sure the solution is robust and optimized for various scenarios.
Visualizing these cases can help in understanding the problem better. Here are some ways to visualize:
Graphical Visualization
 Number Line:
 Place numbers from 0 to ( n ) on a number line.
 Mark the multiples of 5 with special indicators. These are the key numbers that contribute to the trailing zeros.
 Bar Chart:
 For each test case, display a bar chart where the height represents the number of trailing zeros in ( n! ).
 Pie Chart:
 For test cases with larger ( n ), show a pie chart representing the contribution of different multiples of 5 to the total trailing zeros.
Table Visualization
Create a table with columns for:
 ( n )
 ( n! ) (if computationally feasible)
 Number of trailing zeros
 Special multiples of 5 (those that are squared or cubed, etc., as they contribute more to the count)
For example:
( n )  Special Multiples of 5  Trailing Zeros 

0    0 
3    0 
25  25 (5x5)  6 
30  25 (5x5)  7 
Visual Code or Pseudocode Blocks
 You can use code blocks to visually depict the stepbystep progress for a particular ( n ), showing how each multiple of 5 is found and contributes to the trailing zeros.
These visual aids can clarify the patterns and behaviors, especially focusing on multiples of 5 and their powers, which are crucial for solving this problem.
Analyzing different cases provides several key insights:
Low Numbers: When ( n ) is less than 5, there are no trailing zeros. This is a quick check condition.
Multiples of 5: A simple multiple of 5 adds one trailing zero. For example, if ( n ) is 10, you get one trailing zero, contributed by the factor 5 in 10!.
Higher Powers of 5: Numbers like 25, 125, etc., which are higher powers of 5, contribute more than one trailing zero. For instance, 25 contributes two (5x5), and 125 contributes three (5x5x5).
Range Variability: As ( n ) increases, the number of trailing zeros increases, but not linearly. The increment is based on the number and type of multiples of 5 present.
Unaffected Factors: Factors other than multiples of 5 don’t contribute to trailing zeros. Thus, we don’t need to calculate ( n! ) fully, which is computationally expensive.
Upper Bound: The problem’s constraint of ( n \leq 10^4 ) suggests that the number of trailing zeros will also have an upper bound. This can be useful for setting array sizes if needed.
Edge Cases: For ( n = 0 ), the number of trailing zeros is 0. It’s a special case and should be handled separately.
Understanding these insights helps in formulating a more efficient algorithm that avoids unnecessary calculations and focuses on the problem’s core aspects.
Identification of Applicable Theoretical Concepts
The problem is counting trailing zeros in factorial numbers, which is essentially a number theory problem. Here are some mathematical concepts that can help:
Prime Factorization: Any integer greater than 1 can be uniquely represented by its prime factors. For example, 10! can be broken down into its prime factors, and then you can count the number of times 2 and 5 pair up to make 10, each contributing to a trailing zero.
Counting Divisors: The problem can be simplified to counting the number of times a number is divisible by 5, as each division contributes to a trailing zero.
Geometric Progression: When considering multiples of higher powers of 5 (like 25, 125, etc.), it’s like a geometric series (5, 25, 125, \ldots), where each term can contribute one or more trailing zeros.
Logarithmic Time Complexity: The constraint (n \leq 10^4) implies a need for efficiency. Logarithmic time algorithms can be highly efficient in this case.
Dynamic Programming: For very large numbers, you can use dynamic programming to store previously calculated values of smaller factorials to reduce computation time. However, for this specific problem, it may not be needed due to the manageable constraint.
Integer Overflow: When dealing with factorials, numbers can become huge. But in this problem, you don’t actually need to calculate (n!), which avoids integer overflow issues.
Mathematical Proofs: Knowing the loop invariant, as well as preconditions and postconditions, can ensure the algorithm’s correctness. This comes from formal methods in computer science.
These mathematical and algorithmic concepts can help devise an efficient solution for the problem. They provide avenues to focus on specific calculations rather than bruteforce methods.
Simple Explanation
Imagine you have a big number that ends with some zeros at the end. Think of it like a long train with many cars, and the last few cars are empty. Your job is to find out how many empty cars are at the end of this train.
Now, to make that big number, you start with a number—let’s say 5—and multiply it by every number smaller than it (4, 3, 2, 1). That gives you the long train of cars.
But you don’t have to make the whole train to find out how many empty cars are at the end. You can be smart about it by looking at how often you multiply by a 5, because every time you do, you add an empty car to the end of the train.
So, the problem is finding a clever way to count those empty cars without having to build the whole train.
Problem Breakdown and Solution Methodology
The overall approach for solving this problem involves counting how many times 5 is a factor in each of the numbers that contribute to n!. Since each 5 introduces a zero at the end of the result, this count will give us the number of trailing zeros.
Breakdown into Steps
Initialize Counter: Start with a counter set to zero. This counter will keep track of the number of times 5 is a factor.
Iterate and Divide: For each number between
n
and 1, check how many times 5 divides it evenly and update the counter accordingly.Special Cases: For numbers that are powers of 5 (25, 125, etc.), each will contribute more than one 5 as a factor. Account for this.
Final Count: The counter at the end will give the number of trailing zeros in n!.
Metaphor
Imagine you’re an apple picker. Each tree you visit has a certain number of apples (factors of 5) on it. Some special trees (like 25, 125) have extra apples because they are powers of 5. Your job is to pick all these apples and count them; the total number of apples will be the number of trailing zeros in n!.
Parameters Affecting Solution
 Larger values of
n
will require more checks, but the methodology won’t change.  Smaller values (especially under 5) are edge cases, as there won’t be any trailing zeros.
Example Case
Let’s take n=10.
 10 has one 5 as a factor.
 5 has one 5 as a factor.
 Numbers 9 to 1 don’t have 5 as a factor.
 10! has two trailing zeros.
Thus, the counter will be 2 at the end.
In this manner, we can systematically count the trailing zeros in n! by focusing on the factors of 5 involved in the multiplication.
Inference of ProblemSolving Approach from the Problem Statement
Key Terms and Concepts
Factorial: The product of all positive integers less than or equal to
n
. Understanding the factorial helps us realize that we’ll deal with a series of multiplications of integers from 1 ton
.Trailing Zero: A zero at the end of a number. We’re counting these, so understanding what generates a trailing zero (a factor of 10, which is 5 * 2) is crucial.
Factor: A number that divides another without leaving a remainder. We’re particularly interested in the number 5 as a factor because it, combined with a 2, generates a trailing zero.
Division and Remainder: We use these operations to identify factors of 5 in the numbers that contribute to the factorial.
Counter: A simple way to keep track of how many trailing zeros we’ll have.
Powers of 5: These are special cases that contribute more than one 5 as a factor. Recognizing and accounting for them is crucial for an accurate count.
Strategy or Method
Iteration: We iterate over each integer up to
n
to identify its factors.Modulo Operation: Helps in checking if 5 is a factor of the number.
Division: We divide the number by 5 to count how many times 5 is a factor.
Diagrams and Tables
You could use a table to track your progress for smaller values of n
. Columns could be the integer value, the number of 5’s as factors, and a running total for trailing zeros.
Integer Value  Factors of 5  Running Total 

1  0  0 
2  0  0 
3  0  0 
4  0  0 
5  1  1 
…  …  … 
By completing such a table, you’ll visualize how each integer contributes to the total count of trailing zeros. For larger values of n
, this method becomes impractical, but it’s helpful for understanding the problem and verifying your solution for smaller inputs.
How did you infer from the problem statement that this problem can be solved using ?
Simple Explanation of the Proof
I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?
Stepwise Refinement
Stepwise Refinement of Approach
Highlevel Approach:
 Count the factors of 5 in
n!
to find the number of trailing zeros.
Granular Steps:
 Initialize a counter variable as zero.
 Loop from 1 to
n
to consider each integer value: Check if the current integer is divisible by 5.
 If yes, find how many times it is divisible by 5.
 Add the count to the counter variable.
 Return the counter variable as the answer.
Granular, Actionable Steps:
Initialization:
 Set
counter = 0
.
 Set
Main Loop:
 Start a loop
i
from 1 ton
. Inside the loop, set
current = i
.  Check if
current % 5 == 0
.  If yes, then enter another loop:
 While
current % 5 == 0
: Add 1 to
counter
.  Update
current = current / 5
.
 Add 1 to
 While
 Inside the loop, set
 Start a loop
Final Step:
 Once the loop ends, the value in
counter
will be the number of trailing zeros. Returncounter
.
 Once the loop ends, the value in
Independent Parts
 The divisibility of each integer between 1 and
n
by 5 can be checked independently. Parallelization could speed up this step for very largen
, though it’s not necessary for most cases given the constraints.
Repeatable Patterns
 The recurring task is checking each integer for factors of 5 and adding the count to a counter. This can be encapsulated in a separate function for modularity.
By breaking down the problem in this manner, the solution becomes straightforward and easier to understand. This process also makes it easier to write and debug the code.
Solution Approach and Analysis
Detailed Approach to Solving the Problem
Step 1: Understand the Problem
We need to find the number of trailing zeros in the factorial of a given number n
.
Step 2: Recognize the Math Behind the Problem
A trailing zero is produced by the product of a pair of 2 and 5. In the factorial sequence, we will have more 2s than 5s. Therefore, the number of trailing zeros depends on the number of 5s in the sequence.
Metaphor
Imagine you’re making ‘Zero Cookies,’ each requiring one unit of “2” and one unit of “5.” You have an abundance of “2s,” but “5s” are rare. The number of cookies you can make (trailing zeros) depends on the number of “5s” you have.
Step 3: Initialize Counter Variable
Initialize a variable called counter
to zero. This will store the count of trailing zeros.
Step 4: Loop Through 1 to n
Iterate through each number from 1 to n
. For each number, check how many times it can be divided evenly by 5. Each division represents one trailing zero.
 Loop from 1 to n: For each integer
i
in this range, Initialize a variable
current = i
.  Inner Loop: While
current
is divisible by 5, Increment
counter
by 1.  Divide
current
by 5.
 Increment
 Initialize a variable
Step 5: Return Counter
Return the counter
value.
How Changes Affect the Solution
 If
n
is higher, the computation will require more iterations.  The maximum constraint
n <= 104
is efficient enough for this approach, but for larger values, we may need to consider optimization.
Example Cases
Example 1: n = 5
 5! = 5 x 4 x 3 x 2 x 1 = 120
 The number of trailing zeros = 1
 Our counter would also result in 1.
current = 5
, it’s divisible by 5 one time.
Example 2: n = 10
 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,628,800
 The number of trailing zeros = 2 (from 5 and 10)
 Our counter would also result in 2.
current = 5
(one 5),current = 10
(one more 5).
By following this stepbystep process, we systematically count the number of trailing zeros for any given n!
.
Identify Invariant
The loop invariant in this problem is that, at the start of each iteration of the outer loop, the variable counter
contains the total number of times that 5 can divide evenly into the factorials of all the numbers from 1 to i1
.
This invariant is useful because it helps us accumulate the count of trailing zeros for the factorial at each step, ensuring that we don’t miss any along the way.
Establishing the Loop Invariant
Before entering the loop, we initialize counter
to 0, establishing our loop invariant. At this point, we’ve considered zero numbers, and thus have zero trailing zeros, which is what counter
indicates.
Maintaining the Loop Invariant
During each loop iteration, we take the next number i
, check how many times it can be evenly divided by 5, and increment counter
accordingly. This keeps the invariant true as we move to the next iteration.
Exit Condition
The loop exits when we’ve checked all numbers from 1 to n
.
Loop Invariant + Exit Condition
The combination of the loop invariant and the exit condition ensures that counter
holds the correct total number of trailing zeros for n!
when the loop finishes. We return counter
as the final output.
Identify Loop Invariant
In the problem of finding the number of trailing zeros in (n!), the loop invariant could be: “At the start of each iteration, the variable count
holds the current number of fives that divide (i!) where (i) is the loop variable.”
This is significant because trailing zeros are created by the product of twos and fives, and factorials naturally contain more twos than fives. Hence, counting the number of fives will give us the number of trailing zeros.
Establishing the Loop Invariant
Before the loop starts, you would initialize count
to 0, effectively establishing the invariant.
Maintaining the Loop Invariant
Inside the loop, for each (i), you would factor out all the fives from (i) and add them to count
. This maintains the loop invariant because count
accurately keeps track of the number of fives that (i!) is divisible by as you progress through the loop.
Exit Condition
The loop would exit when (i > n), as you’ve examined all integers from 1 to (n).
Concluding the Problem
Combining the loop invariant and the exit condition ensures that count
will hold the total number of trailing zeros in (n!) when the loop exits, which is the problem’s desired output.
Thought Process
Thought Process
Identify Key Insight: We start by noticing that trailing zeros are formed by multiplying 5 and 2. Factorials have plenty of 2s but fewer 5s. So, counting the number of 5s that can divide (n!) gives us the number of trailing zeros.
Single Count vs Multiple Count: A number like 25 has two 5s in its factorization (5 * 5). Therefore, every multiple of 25 actually contributes two 5s. We need to account for such numbers multiple times.
Divisibility by 5: To find how many numbers up to (n) are divisible by 5, we can use integer division (n // 5).
Divisibility by Higher Powers of 5: For numbers that are divisible by higher powers of 5 (25, 125, …), they should be counted additional times. We can do this by further dividing (n) by 5, until (n) becomes less than 5.
Steps to Solve
 Initialize a counter variable to zero.
 While (n) is greater than or equal to 5:
 Divide (n) by 5 (integer division).
 Add the result to the counter.
 Return the counter as it now holds the number of trailing zeros.
Python Code
Here’s how you can code this in Python3:


Insights and Cues from the Problem Statement
 The problem statement’s constraint (0 \leq n \leq 10^4) indicates that the solution needs to be efficient. Our approach is (O(\log n)), which is sufficient.
 The followup question about logarithmic time complexity suggests that an (O(n)) solution might not be efficient enough, pointing us towards mathematical insights to improve the complexity.
By focusing on the number of 5s in the factorization of (n!), we can solve this problem efficiently, capturing the core concept behind trailing zeros.
Establishing Preconditions and Postconditions
Parameters:
 Inputs: The input to the method is an integer ( n ).
 Types: The input is of type
int
.  Representation: This integer ( n ) represents the number for which we want to find the count of trailing zeros in ( n! ).
Preconditions:
 State: There’s no requirement for a specific state of the program.
 Constraints: The input ( n ) must satisfy ( 0 \leq n \leq 10^4 ).
 Program State: The program doesn’t have to be in any particular state before this method is called.
Method Functionality:
 Task: The method is expected to count the number of trailing zeros in ( n! ).
 Interaction: The method takes ( n ) and performs integer division by 5 repeatedly to count the number of factors of 5 in ( n! ).
Postconditions:
 State: The program state remains unchanged; the method has no side effects.
 Return Value: The method returns an integer that represents the count of trailing zeros in ( n! ).
 Side Effects: None.
Error Handling:
 Preconditions Not Met: If ( n ) is outside the specified range, the behavior is undefined.
 Exception or Return: The code does not explicitly handle this scenario, but Python would likely throw an error if ( n ) is not an integer.
By understanding these various aspects, we get a thorough idea of how the method functions and interacts with the rest of the program. It also helps to identify potential areas where things could go wrong and plan for those scenarios.
Problem Decomposition
Problem Understanding:
 In My Words: The problem asks us to find the number of trailing zeros in the factorial of a given number ( n ).
 Key Components: Given ( n ), calculate ( n! ) and count the number of trailing zeros.
 Requirements: The input ( n ) is an integer between 0 and 10^4.
Initial Breakdown:
 Major Parts:
 Count factors of 5 in ( n! )
 Return the count
Subproblem Refinement:
Count Factors of 5:
 Initialize a variable ( count ) as 0.
 Iterate: ( i = 5 ) to ( n ) with ( i *= 5 )
 ( count += \frac{n}{i} )
Return Count:
 Return the ( count ) variable.
Task Identification:
 Iterate and Count: The iteration over multiples of 5 and counting can be seen as a repeated task but specific to this problem.
Task Abstraction:
 Iterate and Count: This task is abstract enough as it states what needs to be done and is also reusable.
Method Naming:
 Iterate and Count:
countFactorsOfFive
 Return Count:
returnCount
Subproblem Interactions:
 Order:
countFactorsOfFive
must be performed beforereturnCount
.  Dependencies:
returnCount
depends on the result fromcountFactorsOfFive
.
By breaking down the problem like this, you get a roadmap that you can follow to code the solution. This ensures that you don’t miss out on any crucial steps and also helps in making your code modular and clean.
From Brute Force to Optimal Solution
Brute Force Solution:
Code:
Here’s a Python code snippet for the brute force solution.


Inefficiencies:
 Calculating Factorial: This takes (O(n)) time.
 Counting Zeros: In the worst case, this takes another (O(n)) time.
 Space: Storing the factorial takes (O(n)) space.
Total Time Complexity: (O(n)) Total Space Complexity: (O(n))
Optimized Solution:
Step 1: Count Factors of 5
Trailing zeros in the factorial are formed by multiples of 10. Every 10 is a result of a pair 5 and 2. Factorials have more factors of 2 than 5. So we count factors of 5.
Code:


Step 2: Final Optimization
This single method performs the task efficiently. We don’t need further steps.
Code:


Impact on Time and Space Complexity:
 Time: The time complexity is reduced to (O(\log_5 n)), much better than (O(n)).
 Space: The space complexity is (O(1)), a significant improvement over (O(n)).
In summary, the optimized solution is more efficient in both time and space. It achieves the goal by focusing on the factors that contribute to trailing zeros, thus sidestepping the need to calculate the entire factorial.
Code Explanation and Design Decisions
Initial Parameters
n
: This is the number for which we are trying to find the count of trailing zeros in its factorial. The parameter signifies the upper limit for the factorial calculation.
Primary Loop
 The loop
while n // i > 0:
iterates over the divisors ofn
that are powers of 5. Each iteration represents a set of numbers in the factorial series that contribute a factor of 5.
Conditions and Branches
n // i > 0
: This condition checks if there are still numbers in the series contributing factors of 5. Ifn // i
is zero, it means we’ve covered all such numbers.
Updates or Modifications to Parameters
count += n // i
: This update aggregates the factors of 5 found at each stage of the loop.i *= 5
: This update prepares for the next iteration by multiplying the divisor by 5, so we check the next power of 5.
Invariant
count
remains an accurate count of the number of factors of 5 contributed by numbers up ton
in their factorial. This satisfies the objective to find the number of trailing zeros inn!
.
Final Output
 The final output is the variable
count
, which holds the total number of trailing zeros in the factorial ofn
. This directly answers the problem statement and satisfies its requirements by offering a timeefficient way to get the count of trailing zeros.
Coding Constructs
Highlevel Problemsolving Strategies
 The code employs mathematical insight to avoid direct computation of the factorial. It uses a loop to sum the count of numbers contributing factors of 5, optimizing the search for trailing zeros in the factorial.
Explaining to a Nonprogrammer
 This code calculates how many zeros are at the end of a really big number, which is a factorial of another number. Instead of doing the big calculation, it takes a shortcut to find the answer more quickly.
Logical Elements
 Iteration: A loop to sum counts.
 Condition: Checking when to stop the loop.
 Accumulator: A variable to hold the sum of counts.
 Division: Breaking down numbers to their divisible parts.
Algorithmic Approach in Plain English
 The code starts by setting a count to zero.
 It then looks at all numbers up to the target number (
n
), focusing only on those that are divisible by 5, 25, 125, and so on.  For each such number, it adds how many times that number can be divided by 5 to the count.
 The final count tells us how many zeros will be at the end of the big number (the factorial).
Key Steps or Operations
 Initialize
count
to zero: To start counting from a known point.  Loop through divisors of 5, 25, 125, etc.: To catch all numbers that contribute factors of 5.
 Sum counts: To keep a running total of factors of 5.
 Update divisor: To move to the next set of numbers in the factorial series that contribute more factors of 5.
Algorithmic Patterns
 Loop with geometric progression: The loop iterates through divisors in a geometric sequence (5, 25, 125, …).
 Accumulator Pattern: The
count
variable accumulates the number of factors of 5.  Early Exit: The loop breaks when there are no more numbers that contribute factors of 5, making it efficient.
Language Agnostic Coding Drills
Distinct Coding Concepts
Variable Initialization: This is the foundation of any code. You set up a space in memory to keep track of your data.
Basic Looping: Understanding how to iterate over a sequence of numbers is fundamental to many algorithms.
Conditional Statements: Adding logic to decide the flow of a program.
Mathematical Operations: Basic multiplication, division, and understanding of mathematical properties.
Accumulator Pattern: Keeping a running total or accumulating values is a common need in many algorithms.
Break Statement: Knowing when to exit a loop early is critical for efficiency.
Geometric Progression in Loops: A more advanced iteration pattern where each step in the loop is a constant factor different from the previous step.
Difficulty Level and Descriptions
Variable Initialization: Easiest. Just defining a variable to store data.
Mathematical Operations: Still easy but requires a basic understanding of math.
Basic Looping: Moderate. Requires understanding how to traverse a sequence.
Conditional Statements: Moderate. Need to know how to direct flow based on conditions.
Accumulator Pattern: Moderate. Builds on variables and looping to maintain state.
Break Statement: Moderatetodifficult. Requires an understanding of when it’s beneficial to stop looping.
Geometric Progression in Loops: Difficult. Requires a deep understanding of the problem domain and loop mechanics.
Problemsolving Approach
Understand the Problem: Before any code, understand what you’re trying to solve. For example, the need to find trailing zeros in the factorial of a large number.
Initialize Variables: Start by initializing a counter (Variable Initialization).
Math Insight: Understand the mathematical properties that allow you to avoid brute force computation of the factorial (Mathematical Operations).
Loop Setup: Set up a loop to iterate through numbers that are powers of 5 (Basic Looping).
Add Conditions: Within the loop, add a condition to break the loop if you reach a point where no more factors of 5 will be found (Conditional Statements).
Accumulate Counts: For each iteration, increment the counter by the number of times the current power of 5 divides into the number (Accumulator Pattern).
Loop Exit: Use a break statement to stop the loop once you’ve counted enough powers of 5 (Break Statement).
Advanced Looping: Modify the loop to iterate through geometrically progressing divisors for efficiency (Geometric Progression in Loops).
Test and Debug: Ensure that each piece works as expected and that they work together to solve the problem.
Review: Check if the solution meets the problem requirements and constraints.
Each drill contributes to the overall algorithm, helping it become more efficient, accurate, and understandable. They fit together like pieces of a puzzle to form the complete solution.
Targeted Drills in Python
General Coding Drills in Python
Variable Initialization
1
x = 0 # Initializing an integer variable
Mathematical Operations
1
result = 5 * 2 # Multiplying numbers
Basic Looping
1 2
for i in range(5): print(i) # Prints 0 to 4
Conditional Statements
1 2
if x > 10: print("x is greater than 10")
Accumulator Pattern
1 2 3
sum = 0 for i in range(5): sum += i # Summing up 0 to 4
Break Statement
1 2 3
for i in range(10): if i == 5: break # Stops loop at i = 5
Geometric Progression in Loops
1 2 3 4
x = 1 while x < 100: print(x) x *= 2 # Multiplies x by 2 in each iteration
ProblemSpecific Coding Drills
 Divide Until RemainderThis is essential for our problem because we often need to count how many times a number can be divided by another number (in our case, 5) without a remainder.
1 2 3 4 5
n = 100 count = 0 while n % 5 == 0: count += 1 n = n // 5
Integration of Drills to Solve the Problem
Start by understanding the problem and initialize an accumulator variable to zero.
 Use Variable Initialization drill.
Math insight to skip unnecessary computation.
 Use Mathematical Operations drill.
Set up a loop to iterate through potential divisors.
 Use Basic Looping and Geometric Progression in Loops drills.
Inside the loop, add a conditional check to decide whether to continue the loop.
 Use Conditional Statements and Break Statement drills.
While inside the loop, accumulate counts based on the number of times the current divisor can divide the original number.
 Use Accumulator Pattern and Divide Until Remainder drills.
By carefully integrating these drills, you can create an efficient and understandable solution to your problem. Each drill sets up the stage for the next, and they fit together like building blocks.
Q&A
Similar Problems
Certainly. Given that the original problem involved elements like loops, conditionals, and accumulators, here are 10 LeetCode problems that require similar problemsolving strategies:
Two Sum (LeetCode #1)
 Requires iterating through an array and uses a hash map as an accumulator to store previously visited elements.
Maximum Subarray (LeetCode #53)
 Involves iterating through an array and maintaining an accumulator for the maximum sum found so far.
Palindrome Number (LeetCode #9)
 Uses a loop to reverse the digits of a number and conditionals to check for palindromes.
Reverse Integer (LeetCode #7)
 Similar to Palindrome Number but focuses on reversing the digits.
Single Number (LeetCode #136)
 Requires looping through an array and using bitwise operations as an accumulator pattern to find the unique element.
Climbing Stairs (LeetCode #70)
 Involves using loops and a dynamic programming approach similar to the accumulator pattern to solve the problem.
Power of Two (LeetCode #231)
 Requires an understanding of binary representation and uses bitwise operations, with a focus on conditionals.
Squares of a Sorted Array (LeetCode #977)
 Needs loops to iterate through the array and mathematical operations to square each element.
Valid Parentheses (LeetCode #20)
 Requires stack operations within a loop and conditionals to check for matching brackets.
First Bad Version (LeetCode #278)
 Involves a binary search loop and conditionals to find the first bad software version.
These problems were chosen because they all involve loops, conditionals, and often some sort of accumulating state—concepts that were important in solving our original problem.