Smallest Value After Repeated Prime Factor Sum Replacement
Intuition\nThe problem involves continuously replacing a number n with the sum of its prime factors until n becomes a prime number. The intuition behind this ap
## Intuition\nThe problem involves continuously replacing a number `n` with the sum of its prime factors until `n` becomes a prime number. The intuition behind this approach is that by summing the prime factors, we can iteratively reduce the number until it converges to a prime.\n\n## Approach\n1. **Prime Factor Sum Calculation**: \n   - Create a helper function `sum_of_prime_factors` to calculate the sum of prime factors of `n`. This involves dividing `n` by its smallest divisors starting from 2 and increasing, ensuring all prime factors are summed.\n   \n2. **Repeated Replacement**:\n   - Continuously replace `n` with the sum of its prime factors. If the new sum equals the current number `n`, it indicates `n` has become a prime.\n   - Use caching to avoid redundant calculations for the same numbers, improving efficiency.\n\n## Complexity\n- **Time complexity**: O (log \uD835\uDC5B \u22C5 \uD835\uDC5B)\n- **Space complexity**: O (n)\n\n## Code\n```python\nclass Solution:\n    def smallestValue(self, n: int) -> int:\n        def sum_of_prime_factors(n: int) -> int:\n            if n < 2:\n                return n\n\n            total = 0\n            while n % 2 == 0:\n                total += 2\n                n //= 2\n\n            factor = 3\n            while factor * factor <= n:\n                while n % factor == 0:\n                    total += factor\n                    n //= factor\n                factor += 2\n            \n            if n > 1:\n                total += n\n\n            return total\n\n        cache = {}\n\n        while True:\n            if n in cache:\n                n = cache[n]\n            else:\n                new_n = sum_of_prime_factors(n)\n                cache[n] = new_n\n\n            if new_n == n:\n                break\n            n = new_n\n\n        return n\n```
['Python3']