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max-sum-of-a-pair-with-equal-sum-of-digits.html
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<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> </p>
<p><strong>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>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> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
<li><code>1 <= nums[i] <= 10<sup>9</sup></code></li>
</ul>