Max Sum of a Pair With Equal Sum of Digits
<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 the <strong>maximum</strong> value of<em> </em><code>nums[i] + nums[j]</code><em> </em>that you can obtain over all possible indices <code>i</code> and <code>j</code> that satisfy the conditions. If no such pair of indices exists, return -1.</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>
Array; Hash Table; Sorting; Heap (Priority Queue)
class Solution {
    public int maximumSum(int[] nums) {
        int[] d = new int[100];
        int ans = -1;
        for (int v : nums) {
            int x = 0;
            for (int y = v; y > 0; y /= 10) {
                x += y % 10;
            }
            if (d[x] > 0) {
                ans = Math.max(ans, d[x] + v);
            }
            d[x] = Math.max(d[x], v);
        }
        return ans;
    }
}