🔄🔢Reverse Weighted Alphabet Sum Calculation | Simulating |Optimal | Greedy Approach 🚀
IntuitionThe problem involves computing a specific "reverse degree" for a given string, where each character's contribution is based on its position and its dis
### **Intuition**  
The problem involves computing a specific "reverse degree" for a given string, where each character's contribution is based on its position and its distance from the letter 'a'.  

### **Approach**  
1. We iterate through each character of the string.  
2. For each character `s[i]`, calculate its **reverse distance** from 'a':  
   \[
   26 - (s[i] - 'a')
   \]
   - This formula ensures that 'a' contributes 26, 'b' contributes 25, ..., 'z' contributes 1.  
3. Multiply this value by its **1-based position** in the string (`i + 1`).  
4. Accumulate the values in a `count` variable.  
5. Return `count` as the final result.

---

### **Complexity Analysis**  
- **Time Complexity:** \(O(n)\)  
  - We traverse the string once, performing constant-time operations at each step.  
- **Space Complexity:** \(O(1)\)  
  - Only a few integer variables are used; no extra space is required.

---

### **Example Walkthrough**  
#### **Example 1**  
**Input:** `"abc"`  
**Steps:**  
- 'a': \((26 - (0)) \times 1 = 26\)  
- 'b': \((26 - (1)) \times 2 = 50\)  
- 'c': \((26 - (2)) \times 3 = 72\)  
**Total:** \(26 + 50 + 72 = 148\)  
**Output:** `148`

---

### **Code Explanation**
```cpp
class Solution {
public:
    int reverseDegree(string s) {
        int count = 0;
        int n = s.size();
        for (int i = 0; i < n; i++) {
            count += (26 - (s[i] - 'a')) * (i + 1);
        }
        return count;
    }
};
```
**Why This Works:**  
- `s[i] - 'a'` gives the zero-based index of the character in the alphabet.  
- `26 - (s[i] - 'a')` gives the reverse distance.  
- Multiplying by `(i + 1)` incorporates the positional weight.  

# Upvote For the Effort🚀
['Math', 'String', 'Greedy', 'String Matching', 'Simulation', 'Counting', 'C++']