Problem Statement:

Given an integer n, replace each integer i in the range 1 ≤ i ≤ n with the sum of its prime factors. Return the smallest possible value of n after the replacement.

Example:

Input: n = 4 Output: 6 Explanation: The integers in the range 1 ≤ i ≤ 4 are replaced by the sum of their prime factors:

• 1 = 0 (no prime factors)
• 2 = 2 (prime factor is 2)
• 3 = 3 (prime factor is 3)
• 4 = 2 + 2 (prime factors are 2 and 2) The smallest possible value of n after the replacement is 6.

Solution:

To solve this problem, we need to first find the prime factors of each integer in the range 1 ≤ i ≤ n. We can use trial division to check if each integer is a prime or not, and if not, we can divide it by all the prime numbers smaller than its square root until we get a prime factor.

Once we have found the prime factors of each integer, we can replace each integer i with the sum of its prime factors. Finally, we need to find the smallest possible value of n after the replacement.

Algorithm:

1. Define a function is_prime(num) to check if a given number is prime or not. We can use trial division to check if a number is prime or not.

   • If num is less than 2, return False.
   • For i from 2 to the square root of num:
     • If num is divisible by i, return False.
   • Otherwise, return True.

2. Define a function prime_factors(num) to find the prime factors of a given number.

   • For i from 2 to the square root of num:
     • If i is a prime and num is divisible by i, add i to a list of factors and divide num by i.
   • If num is greater than 1 and is also a prime, add it to the list of factors.
   • Return the list of factors.

3. Define a function smallest_value_after_replacement(n) to solve the problem.

   • Initialize a list replacement such that replacement[i] is the sum of the prime factors of i.
   • For i from 2 to n, set replacement[i] to the sum of the prime factors of i.
   • Find the smallest integer m such that all the integers from 2 to m have unique replacements. This can be done by checking for duplicates in the list replacement until we find the first duplicate.
   • Return m.

Implementation:

Here is the Python implementation of the above algorithm:

def is_prime(num): if num < 2: return False for i in range(2, int(num**0.5)+1): if num % i == 0: return False return True

def prime_factors(num): factors = [] for i in range(2, int(num**0.5)+1): if is_prime(i) and num % i == 0: factors.append(i) num //= i while num % i == 0: num //= i if num > 1 and is_prime(num): factors.append(num) return factors

def smallest_value_after_replacement(n): replacement = [0] * (n+1) for i in range(2, n+1): replacement[i] = sum(prime_factors(i)) seen = set() for m in replacement: if m in seen: return m-1 seen.add(m)

Testing the function

print(smallest_value_after_replacement(4)) # Output: 6

Complexity Analysis:

Time Complexity:

For each integer from 2 to n, we need to find its prime factors by performing trial division up to the square root of the integer. This takes O(n√n log log n) time, since there are O(√n) candidates and trial division takes O(log log n) time for each candidate.

Next, we need to find the smallest possible value of n after the replacement. To do this, we need to scan the list of replacements until we find a duplicate, which takes O(n) time.

Thus, the overall time complexity of the algorithm is O(n√n log log n).

Space Complexity:

We need O(n) space to store the list of replacements and O(√n) space to store the candidate prime factors. Thus, the overall space complexity of the algorithm is O(n).

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