Easy Javascript Solution with Full Explanation
IntuitionApproachComplexity
Time complexity: O(n∗log(max))

The function iterates through the nums array, which has a length of n. This gives us an O(n) factor
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->

# Approach
<!-- Describe your approach to solving the problem. -->

# Complexity
- Time complexity: $$O(n * log(max))$$
1. The function iterates through the `nums` array, which has a length of `n`. This gives us an O(n) factor for the loop.
2. Inside the loop, the `sumOfDigits` function is called for each element. The `sumOfDigits` function converts the number to a string, splits it into an array of characters, and then uses the `reduce` method to calculate the sum of its digits. The time complexity of converting a number to a string and splitting it into an array is O(d), where `d` is the number of digits in the number. In the worst case, if the numbers are large, `d` can be proportional to log10(num), which is O(log(num)).
3. Therefore, the overall time complexity of the function is O(n * log(max)), where `max` is the largest number in the `nums` array.
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: $$O(log(max))$$
1. The space used by the `min` variable is O(1) since it only stores a single number.
2. The `sumOfDigits` function creates an array of characters from the string representation of the number, which takes O(d) space, where `d` is the number of digits in the number.
3. However, since this space is used temporarily for each call to `sumOfDigits`, it does not accumulate across calls. Thus, the overall space complexity is O(d) for the largest number processed, which can be considered O(log(max)) in terms of the largest number in the `nums` array.
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```javascript []
/**
 * @param {number[]} nums
 * @return {number}
 */
var minElement = function (nums) {
    let min = Infinity;

    function sumOfDigits(num) {

        // find sum of digits of num by coverting num to string and then array and use reduce method
        let digitSum = num.toString().split("").reduce((a, b) => +a + +b, 0);
        return digitSum;
    }
    for (let i = 0; i < nums.length; i++) {

        // invoke sumOfDigits() function to get sum
        let sum = sumOfDigits(nums[i]);

        // find minimum value between min and sum
        min = Math.min(min, sum);
    }

    return min;
};
```
['Array', 'Math', 'JavaScript']