Simple C++ Solution 🤯🤯🤯
🧠 IntuitionWe start by making a function that checks whether a number is divisible by a divisor. If it is, we add the divisor to the sum (as it's a factor) and
# 🧠 Intuition  
<!-- Describe your first thoughts on how to solve this problem. -->  
We start by making a function that checks whether a number is divisible by a divisor. If it is, we add the divisor to the sum (as it's a factor) and divide the number by the same divisor. 🔢  
We continue this process until the number becomes `1`. Inside the main function, we repeatedly call this factor sum function. If our input integer is equal to the sum of its factors, we stop 🛑; otherwise, we replace the integer with the computed sum and continue.  

# 🛠️ Approach  
<!-- Describe your approach to solving the problem. -->  
1️⃣ Define a helper function `factorSum(n)` that finds the sum of the prime factors of `n`.  
2️⃣ In `smallestValue(n)`, repeatedly call `factorSum(n)` until `n` stabilizes (i.e., does not change between iterations).  
3️⃣ Return the final stable value. ✅  

# 🔍 Step-by-Step Execution  

### **Example: n = 20**  
Let's break down how the loops work step by step:

#### **Factorization Process (factorSum function)**
| 🏁 Step | 🎯 `n` (Remaining Number) | ➗ `divisor` | 🔢 `ans` (Sum of Factors) |
|------|----------------|-----------|----------------|
| 1️⃣  | 20            | 2         | 2              |
| 2️⃣  | 10            | 2         | 4              |
| 3️⃣  | 5             | 5         | 9              |
| ✅  | 1             | -         | **9** (Final) |

🔹 **Output of `factorSum(20)`**: **9**  
Since `n ≠ sum`, we replace `n = 9` and repeat 🔄.

#### **Next Iteration (factorSum(9))**
| 🏁 Step | 🎯 `n` (Remaining Number) | ➗ `divisor` | 🔢 `ans` (Sum of Factors) |
|------|----------------|-----------|----------------|
| 1️⃣  | 9             | 3         | 3              |
| 2️⃣  | 3             | 3         | 6              |
| ✅  | 1             | -         | **6** (Final) |

🔹 **Output of `factorSum(9)`**: **6**  
Since `n ≠ sum`, we replace `n = 6` and repeat 🔄.

#### **Next Iteration (factorSum(6))**
| 🏁 Step | 🎯 `n` (Remaining Number) | ➗ `divisor` | 🔢 `ans` (Sum of Factors) |
|------|----------------|-----------|----------------|
| 1️⃣  | 6             | 2         | 2              |
| 2️⃣  | 3             | 3         | 5              |
| ✅  | 1             | -         | **5** (Final) |

🔹 **Output of `factorSum(6)`**: **5**  
Since `n ≠ sum`, we replace `n = 5` and repeat 🔄.

#### **Next Iteration (factorSum(5))**
| 🏁 Step | 🎯 `n` (Remaining Number) | ➗ `divisor` | 🔢 `ans` (Sum of Factors) |
|------|----------------|-----------|----------------|
| 1️⃣  | 5             | 5         | 5              |
| ✅  | 1             | -         | **5** (Final) |

🔹 **Output of `factorSum(5)`**: **5**  
Now, `n == sum`, so we stop 🛑 and return **5** as the answer 🎯.

# ⏳ Complexity Analysis  
- **Time Complexity**:  
  - The worst-case scenario involves factorizing `n`, which takes at most **`O(log n)`**, and iterating this process **`O(log n)`** times.  
  - Thus, the overall time complexity is **`O((log n)^2)`**. ⚡  

- **Space Complexity**:  
  - We use only a few integer variables, leading to a constant space usage of **`O(1)`**. 💾  

# 🏆 Code  
```cpp
class Solution {
public:
    // Function to compute the sum of prime factors of n
    int factorSum(int n) {
        int ans = 0;
        int divisor = 2;
        while (n > 1) {
            if (n % divisor == 0) {
                ans += divisor;
                n /= divisor;
            } else {
                divisor++;
            }
        }
        return ans;
    }

    int smallestValue(int n) {
        while (true) {
            int sum = factorSum(n);
            if (n == sum) break;
            n = sum;
        }
        return n;
    }
};


['C++']