Jobs

Interviews

Discussions

Smallest Value After Replacing With Sum of Prime Factors

Medium

Amazon

1 view

Topics:

Greedy AlgorithmsArrays

You are given a positive integer n.

Continuously replace n with the sum of its prime factors.

Note that if a prime factor divides n multiple times, it should be included in the sum as many times as it divides n.

Return the smallest value n will take on.

Example 1:

Input: n = 15
Output: 5
Explanation: Initially, n = 15.
15 = 3 * 5, so replace n with 3 + 5 = 8.
8 = 2 * 2 * 2, so replace n with 2 + 2 + 2 = 6.
6 = 2 * 3, so replace n with 2 + 3 = 5.
5 is the smallest value n will take on.

Example 2:

Input: n = 3
Output: 3
Explanation: Initially, n = 3.
3 is the smallest value n will take on.

Constraints:

2 <= n <= 10^5

Smallest Value After Replacing With Sum of Prime Factors

Medium

Amazon

1 view

Topics:

Greedy AlgorithmsArrays

You are given a positive integer n.

Continuously replace n with the sum of its prime factors.

Note that if a prime factor divides n multiple times, it should be included in the sum as many times as it divides n.

Return the smallest value n will take on.

Example 1:

Input: n = 15
Output: 5
Explanation: Initially, n = 15.
15 = 3 * 5, so replace n with 3 + 5 = 8.
8 = 2 * 2 * 2, so replace n with 2 + 2 + 2 = 6.
6 = 2 * 3, so replace n with 2 + 3 = 5.
5 is the smallest value n will take on.

Example 2:

Input: n = 3
Output: 3
Explanation: Initially, n = 3.
3 is the smallest value n will take on.

Constraints:

2 <= n <= 10^5

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the input integer n?
If n is 1, what should the return value be?
Can the input integer n be negative or zero?
If the sum of prime factors is equal to the original number, do I need to perform the replacement again, and if so, how many times?
If there is an issue with integer overflow during the sum of the prime factors, how should I handle that?

Brute Force Solution

Approach

The brute force way to find the smallest value involves repeatedly finding the prime factors of a number and summing them. We keep replacing the original number with this sum until the number stops changing, at which point we have our answer.

Here's how the algorithm would work step-by-step:

Start with the given number.
Find all the prime numbers that divide evenly into the number.
Add these prime numbers together.
Replace the original number with this new sum.
Repeat the process of finding prime factors and summing them for this new number.
Keep doing this until the number no longer changes after replacing it with the sum of its prime factors.
The final number is the smallest value you're looking for.

Code Implementation

def smallest_value_after_replacing_with_sum_of_prime_factors(number):
    current_number = number

    while True:
        prime_factors_sum = 0
        temporary_number = current_number
        divisor = 2

        # Find prime factors and calculate their sum
        while divisor * divisor <= temporary_number:
            while temporary_number % divisor == 0:
                prime_factors_sum += divisor
                temporary_number //= divisor
            divisor += 1

        # If temporary_number is still greater than 1, it's a prime factor itself
        if temporary_number > 1:
            prime_factors_sum += temporary_number

        #Check if the number has converged
        if prime_factors_sum == current_number:
            break

        #Update for the next iteration
        current_number = prime_factors_sum

    return current_number

Big(O) Analysis

Time Complexity

O(n log n) – The outer loop continues until the number n converges to a stable value. In the worst case, this might require several iterations, but it's difficult to give an exact bound. The inner loop, which finds the prime factors, iterates up to the square root of n in the worst case to find a divisor. Factoring each number takes O(sqrt(n)) time. Because prime factorization is related to the number of primes less than n, which grows logarithmically, and the outer loop runs an unknown number of times, the overall time complexity can be approximated to O(n log n) since sqrt(n) is less than log n for sufficiently large n.

Space Complexity

O(sqrt(N)) – The primary space usage comes from finding the prime factors. The largest possible prime factor of a number N cannot exceed the square root of N. Therefore, the space needed to store the prime factors is at most proportional to the square root of N. No significant additional data structures are used. Thus, the auxiliary space complexity is O(sqrt(N)), where N is the input number.

Optimal Solution

Approach

The problem requires repeatedly replacing a number with the sum of its prime factors until it stops changing. The trick is to recognize that we can optimize by calculating prime factors efficiently and stopping early if the number becomes small enough.

Here's how the algorithm would work step-by-step:

First, create a way to find the smallest prime factor of any number. You can do this by checking for divisibility starting from 2 and going upwards.
Next, repeatedly find the smallest prime factor of the number. Add this prime factor to a running sum.
Divide the original number by the prime factor you found. Keep doing this until the original number becomes 1.
After you've found all the prime factors and their sum, replace the original number with this sum.
Now, repeat the process of finding the sum of prime factors for this new number.
Continue repeating the process until the number doesn't change anymore after replacing it with the sum of its prime factors. This means you've reached the smallest possible value.
You can optimize by stopping if the number becomes less than 10 because the prime factorization sum will only be 2, 3, 5 or 7, or it may stay the same. This reduces unecessary steps.

Code Implementation

def smallest_value_after_replacing_with_sum_of_prime_factors(number):
    while True:
        original_number = number
        sum_of_prime_factors = 0
        temporary_number = number

        while temporary_number > 1:
            smallest_prime_factor = find_smallest_prime_factor(temporary_number)

            sum_of_prime_factors += smallest_prime_factor
            temporary_number //= smallest_prime_factor

        number = sum_of_prime_factors

        # If the number doesn't change, we've reached the smallest value.
        if number == original_number:
            return number

def find_smallest_prime_factor(number):
    # Optimizing by only checking up to the square root
    for factor in range(2, int(number**0.5) + 1):
        if number % factor == 0:
            return factor

    # If no smaller factor is found, the number itself is prime.
    return number

Big(O) Analysis

Time Complexity

O(n log log n) – The outer loop repeatedly replaces the number with the sum of its prime factors until it converges. The number of iterations of this loop is difficult to precisely determine but in the worst case, it can be considered bounded by some constant. Inside the outer loop, the dominant operation is finding the prime factors, which involves repeatedly finding the smallest prime factor. The smallest prime factor calculation can be done up to the square root of n, and using an optimized approach that iterates through possible prime factors, the complexity is approximately O(sqrt(n)). The Sieve of Eratosthenes to precompute primes up to sqrt(n) takes O(sqrt(n) log log sqrt(n)), which is O(sqrt(n) log log n). Factoring a number using precomputed primes is faster, but the precomputation step dominates and it must be repeated, so overall, the total runtime is dominated by the prime factorization which is roughly bounded by O(n log log n).

Space Complexity

O(1) – The algorithm calculates prime factors and their sum iteratively, using a constant number of variables to store the smallest prime factor, the running sum of prime factors, and the current number being factored. No auxiliary data structures, such as lists or hash maps, are used to store intermediate results or prime factors. Therefore, the auxiliary space used remains constant regardless of the input number N, leading to a space complexity of O(1).

Edge Cases

Case	How to Handle
Input is 1	Return 1 as 1 has no prime factors other than itself, and the sum of its prime factors is still 1.
Input is 0 or negative	Return the original number since the problem does not specify how to handle zero or negative numbers, or define the prime factors of negative numbers, so we leave them unchanged
Input is a very large number (close to integer limit)	Ensure that the prime factorization algorithm does not cause integer overflow; use long data type if necessary.
Input is already a prime number	Return the original number since the sum of prime factors of a prime number is the number itself.
Input contains a large prime factor	The prime factorization process could take longer for large primes so optimize the prime finding to search till sqrt(n).
Input is a power of 2 (e.g., 8, 16, 32)	The sum of prime factors will be repeatedly 2 until the number becomes 2 itself.
Input produces identical value after the replacement	The loop should terminate once the value no longer changes after prime factorization and summation.
Integer overflow when calculating the sum of prime factors	Use a data type that can hold larger values (e.g., long) to calculate the sum of prime factors to prevent overflow.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the input integer n?
If n is 1, what should the return value be?
Can the input integer n be negative or zero?
If the sum of prime factors is equal to the original number, do I need to perform the replacement again, and if so, how many times?
If there is an issue with integer overflow during the sum of the prime factors, how should I handle that?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the given number.
Find all the prime numbers that divide evenly into the number.
Add these prime numbers together.
Replace the original number with this new sum.
Repeat the process of finding prime factors and summing them for this new number.
Keep doing this until the number no longer changes after replacing it with the sum of its prime factors.
The final number is the smallest value you're looking for.

Code Implementation

def smallest_value_after_replacing_with_sum_of_prime_factors(number):
    current_number = number

    while True:
        prime_factors_sum = 0
        temporary_number = current_number
        divisor = 2

        # Find prime factors and calculate their sum
        while divisor * divisor <= temporary_number:
            while temporary_number % divisor == 0:
                prime_factors_sum += divisor
                temporary_number //= divisor
            divisor += 1

        # If temporary_number is still greater than 1, it's a prime factor itself
        if temporary_number > 1:
            prime_factors_sum += temporary_number

        #Check if the number has converged
        if prime_factors_sum == current_number:
            break

        #Update for the next iteration
        current_number = prime_factors_sum

    return current_number

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, create a way to find the smallest prime factor of any number. You can do this by checking for divisibility starting from 2 and going upwards.
Next, repeatedly find the smallest prime factor of the number. Add this prime factor to a running sum.
Divide the original number by the prime factor you found. Keep doing this until the original number becomes 1.
After you've found all the prime factors and their sum, replace the original number with this sum.
Now, repeat the process of finding the sum of prime factors for this new number.
Continue repeating the process until the number doesn't change anymore after replacing it with the sum of its prime factors. This means you've reached the smallest possible value.
You can optimize by stopping if the number becomes less than 10 because the prime factorization sum will only be 2, 3, 5 or 7, or it may stay the same. This reduces unecessary steps.

Code Implementation

def smallest_value_after_replacing_with_sum_of_prime_factors(number):
    while True:
        original_number = number
        sum_of_prime_factors = 0
        temporary_number = number

        while temporary_number > 1:
            smallest_prime_factor = find_smallest_prime_factor(temporary_number)

            sum_of_prime_factors += smallest_prime_factor
            temporary_number //= smallest_prime_factor

        number = sum_of_prime_factors

        # If the number doesn't change, we've reached the smallest value.
        if number == original_number:
            return number

def find_smallest_prime_factor(number):
    # Optimizing by only checking up to the square root
    for factor in range(2, int(number**0.5) + 1):
        if number % factor == 0:
            return factor

    # If no smaller factor is found, the number itself is prime.
    return number

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input is 1	Return 1 as 1 has no prime factors other than itself, and the sum of its prime factors is still 1.
Input is 0 or negative	Return the original number since the problem does not specify how to handle zero or negative numbers, or define the prime factors of negative numbers, so we leave them unchanged
Input is a very large number (close to integer limit)	Ensure that the prime factorization algorithm does not cause integer overflow; use long data type if necessary.
Input is already a prime number	Return the original number since the sum of prime factors of a prime number is the number itself.
Input contains a large prime factor	The prime factorization process could take longer for large primes so optimize the prime finding to search till sqrt(n).
Input is a power of 2 (e.g., 8, 16, 32)	The sum of prime factors will be repeatedly 2 until the number becomes 2 itself.
Input produces identical value after the replacement	The loop should terminate once the value no longer changes after prime factorization and summation.
Integer overflow when calculating the sum of prime factors	Use a data type that can hold larger values (e.g., long) to calculate the sum of prime factors to prevent overflow.