Iterative Prime Factor Summation for Number Reduction
Intuition\nThe problem seems to involve repeatedly summing the prime factors of a number until the sum remains unchanged. This suggests a process of reducing th
# Intuition\nThe problem seems to involve repeatedly summing the prime factors of a number until the sum remains unchanged. This suggests a process of reducing the number by its prime factors iteratively.\n\n# Approach\n- Prime Factorization: Create a function PrimeFactors that calculates the sum of all prime factors of a given number n.\n* Iterative Reduction: Use the PrimeFactors function in another function smallestValue to iteratively reduce the number until the sum of its prime factors equals the number itself.\n# Complexity\n\n- Time complexity: The time complexity for finding prime factors of a number n is approximately O(sqrt(n)\u200B). Since we are iterating this process until the number stabilizes, the overall time complexity can be considered as O(k\u22C5sqrt(n)\u200B), where k is the number of iterations.\n- Space complexity: The space complexity is O(1)\n as we are using a constant amount of extra space.\n# Code\n```cpp []\nclass Solution {\npublic:\n    int PrimeFactors(int n) {\n        int sum = 0;\n        while (n % 2 == 0) {\n            sum += 2;\n            n /= 2;\n        }\n        for (int i = 3; i <= sqrt(n); i += 2) {\n            while (n % i == 0) {\n                sum += i;\n                n /= i;\n            }\n        }\n        if (n > 2) {\n            sum += n;\n        }\n        return sum;\n    }\n\n    int smallestValue(int n) {\n        int sum;\n        while (true) {\n            sum = PrimeFactors(n);\n            if (sum == n) {\n                return sum;\n            } else {\n                n = sum;\n            }\n        }\n        return n; \n    }\n};\n\n```
['Math', 'Number Theory', 'C++']