<p>You are given a <strong>0-indexed</strong> array <code>nums</code> consisting of <strong>positive</strong> integers. You can choose two indices <code>i</code> and <code>j</code>, such that <code>i != j</code>, and the sum of digits of the number <code>nums[i]</code> is equal to that of <code>nums[j]</code>.</p>

<p>Return <em>the <strong>maximum</strong> value of </em><code>nums[i] + nums[j]</code><em> that you can obtain over all possible indices </em><code>i</code><em> and </em><code>j</code><em> that satisfy the conditions.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [18,43,36,13,7]
<strong>Output:</strong> 54
<strong>Explanation:</strong> The pairs (i, j) that satisfy the conditions are:
- (0, 2), both numbers have a sum of digits equal to 9, and their sum is 18 + 36 = 54.
- (1, 4), both numbers have a sum of digits equal to 7, and their sum is 43 + 7 = 50.
So the maximum sum that we can obtain is 54.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,12,19,14]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There are no two numbers that satisfy the conditions, so we return -1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def maximumSum(self, nums: List[int]) -> int:
        # function that computes the sum of the digits in a number. e.g. 32 --> 5
        def sum_from_digits(num):
            digit_sum = 0
            for digit in str(num):
                digit_sum += int(digit)
            return digit_sum

        hmap = defaultdict(int)
        ans = -1

        for num in nums:
            # compute the digit sum from the current number
            curr = sum_from_digits(num)
            # check if the current sum exists in the map already (indicating a match has been found)
            if curr in hmap:
                # if so, set the value of answer to the 2 ints summed or ans
                ans = max(ans, num + hmap[curr])
            
            # Here we want to update the value in the hmap for this digit sum to be the largest between whats in there already and whats currently
            # seen.
            hmap[curr] = max(num, hmap[curr])
        
        return ans

        # This version is O(n * d) time and O(n) space






        # This is not bad but theres a better way to do it, like i was thinking before
        # hmap = defaultdict(list)

        # for num in nums:
        #     curr = 0
        #     for digit in str(num):
        #         curr += int(digit)
        #     hmap[curr].append(num)
        
        # res = -1
        # for k, v in hmap.items():
        #     if len(v) > 1:
        #         v.sort(reverse = True)
        #         res = max(res, v[0] + v[1])

        # return res

        # # Time: O (n log n)
        # # Space: O(n)
943