Minimum Element After Replacement with digit sum
Great! Let's break down your code step by step to understand the intuition, approach, and time complexity.✅ Problem Intuition:Your code is solving a modified mi
Great! Let's break down your code step by step to understand the **intuition, approach, and time complexity**.

---

### ✅ **Problem Intuition:**

Your code is solving a **modified minimum problem**:

> From a given array `nums`, replace each element with the **sum of its digits**, and then return the **minimum** of these new values.

🧠 For example,  
If `nums = [123, 40, 9]`,  
then after processing:
- 123 → 1+2+3 = 6  
- 40  → 4+0 = 4  
- 9   → 9  
So final array = `[6, 4, 9]`, and **min = 4**

---

### 🔍 **Approach Explained:**

1. **Transform step**:
   - Iterate through each element in the array.
   - Replace each element with the **sum of its digits** using the `add()` method.
   
2. **Minimum finding step**:
   - Start with `min = nums[0]`.
   - Iterate again through the array.
   - If any element is smaller than `min`, update `min`.

3. **Return the minimum value.**

---

### 💡 Breakdown of `add()` method:
```java
public int add(int n) {
    int sum = 0;
    while(n > 0) {
        int rem = n % 10;  // extract last digit
        sum += rem;        // add it to sum
        n /= 10;           // remove last digit
    }
    return sum;
}
```

This gives you the **digit sum** of a number.

---

### 🧮 **Time Complexity Analysis:**

Let:
- `n` = number of elements in the array
- `d` = average number of digits in each number

#### 1. Transform step:
- Each number is processed digit by digit → `O(d)`
- For `n` numbers: total = `O(n * d)`

#### 2. Finding minimum:
- One pass over `n` elements → `O(n)`

✅ So total **Time Complexity = O(n * d)**

If numbers are small (e.g. within 32-bit int), `d` is bounded, so you can consider this ≈ **O(n)**.

---

### 📦 **Space Complexity:**

- Only using a few variables: `sum`, `min`, etc.
- ✅ **Space Complexity = O(1)** (constant extra space)

---

### ✅ Summary:

| Part              | Description                             |
|-------------------|-----------------------------------------|
| **Purpose**       | Find min of digit-summed array          |
| **Steps**         | Digit-sum each element → find minimum   |
| **Time Complexity** | `O(n * d)` → ≈ `O(n)` for small numbers |
| **Space Complexity** | `O(1)` (in-place operations)            |

Let me know if you want to optimize this or modify it for a different case!

# Code
```java []
class Solution {
    public int minElement(int[] nums) {
        for(int i=0;i<nums.length;i++){
            nums[i]=add(nums[i]);
        }
        int min=nums[0];
        for(int i=0;i<nums.length;i++){
            if(nums[i]<min) min=nums[i];
        }return min;
    }public int add(int n){
        int sum=0;
        while(n>0){
            int rem=n%10;
            sum+=rem;
            n/=10;
        }return sum;
    }
}
```
['Java']