easiest approch you can get (beginner friendly)
Intuition\n\nThe problem involves finding a transformation of a number based on the sum of its prime factors. The idea is to continuously replace the number wit
Intuition\n\nThe problem involves finding a transformation of a number based on the sum of its prime factors. The idea is to continuously replace the number with the sum of its prime factors until the value no longer changes. This essentially reduces the number by breaking it down into smaller prime components and summing them.\n\nApproach\n    Prime Checking: A helper function is_prime determines if a given number is prime. It checks divisibility up to sqrt(n) for optimization.\n    Prime Factorization: The function primefactor calculates the sum of all prime factors of n. It iterates over potential factors, checking if they are prime and if they divide n.\n    Transformation: The function smallestValue repeatedly calls primefactor on n and updates n to the result until the value stabilizes (i.e., the sum of prime factors equals the original number).\n\nComplexity\nTime Complexity:\n        The is_prime function has a time complexity of O(n)(n\u200B).\nThe primefactor function iterates over the factors up to nn\n\u200B, and in the worst case, it may reduce the number logarithmically each time, leading to a complexity of O(nlog\u2061n)O(n\u200Blogn) per call.\nThe smallestValue function may call primefactor logarithmically many times based on the prime factors sum reductions, resulting in an overall complexity of O(nlog\u20612n)O(nlog2n).\n\nSpace Complexity:\n\nThe space complexity is O(1)O(1) as no extra space proportional to the input size is required.\n\n# Code\n```\nclass Solution {\npublic:\nbool is_prime(int n)\n{\n    if(n==1)\n    {\n        return false;\n    }\n    else\n    {\n        for(int i=2;i<=sqrt(n);i++)\n        {\n            if(n%i==0)\n            {\n                return false;\n            }\n        }\n    }\n    return true;\n}\n\n\n\nint primefactor(int n)\n{\n    if(is_prime(n))\n    {\n        return n;\n    }\n    int sml=0;\n    int i=1;\n    while(n>1)\n    {\n     if(is_prime(i) && n%i==0)\n     {\n         n=n/i;\n         sml=sml+i;\n         if(n%i==0)\n         {\n             i=i;\n         }\n         else\n         {\n             i++;\n         }\n     }\n     else\n     {\n         i++;\n     }\n    }\n    return sml;\n}\n    int smallestValue(int n) {\n    while (true) {\n        int sum = primefactor(n);\n        if (sum == n) break;  // No further reduction possible\n        n = sum;\n    }\n    return n;\n}\n};\n```
['Math', 'Number Theory', 'C++']