Complete Solution with detailed Explanation beats 100%....
IntuitionAfter Reading the problem we understood that we have to return the minimum number after performing the operations.Approach
Step 1: Declare the integer
# Intuition
After Reading the problem we understood that we have to return the minimum number after performing the operations.

# Approach
1. **Step 1:** Declare the integer which we would return hence I declared and defined `min` to any random value.
2. **Step 2:** Iterate along the given `nums` array while finding the sum.
3. **Step 3:** Compared it with our `min`.
4. **Step 4:** Return the minimum value.

# Complexity
- Time complexity:
    1. Initialization: `int min=nums[0];`this operation takes $$O(1)$$ time.
    2. For Loop: `for(int i=0;i<nums.length;i++)`
        - The loop runs `n` times, where *n* is the length of nums array.
        - Inside the loop, the statement `min>nums[i]%10+(nums[i]/10)%10+(nums[i]/100)%10+(nums[i]/1000)%10+(nums[i]/10000)` involves constant-time operations since each division and modulus operation is $$O(1)$$

Therefore the total Time complexity is $$O(1) + O(n) = O(n)$$
- Space complexity:
    1. Variables: The variable min takes $$O(1)$$ space.

    2. Input Array: The input array nums is provided as input and   doesn't add to the space complexity.

The algorithm doesn't use any additional data structures that grow with the size of the input array, so the space complexity is: $$O(1)$$

# Code
```java []
class Solution {
    public int minElement(int[] nums) {
        int min=nums[0];
        for(int i=0;i<nums.length;i++){
            if (min>nums[i]%10+(nums[i]/10)%10+(nums[i]/100)%10+(nums[i]/1000)%10+(nums[i]/10000)){
                min = nums[i]%10+(nums[i]/10)%10+(nums[i]/100)%10+(nums[i]/1000)%10+(nums[i]/10000);
            }
        }
        return min;
    }
}
```
['Array', 'Math', 'Java']