sum-of-elements-with-frequency-divisible-by-k

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Algorithms

You are given an integer array <code>nums</code> and an integer <code>k</code>.

Return an integer denoting the sum of all elements in <code>nums</code> whose frequency is divisible by <code>k</code>, or 0 if there are no such elements.

Note: An element is included in the sum exactly as many times as it appears in the array if its total frequency is divisible by <code>k</code>.

&nbsp;
Example 1:

<div class="example-block">
Input: nums = [1,2,2,3,3,3,3,4], k = 2

Output: 16

Explanation:

<ul>
	<li>The number 1 appears once (odd frequency).</li>
	<li>The number 2 appears twice (even frequency).</li>
	<li>The number 3 appears four times (even frequency).</li>
	<li>The number 4 appears once (odd frequency).</li>
</ul>

So, the total sum is <code>2 + 2 + 3 + 3 + 3 + 3 = 16</code>.
</div>

Example 2:

<div class="example-block">
Input: nums = [1,2,3,4,5], k = 2

Output: 0

Explanation:

There are no elements that appear an even number of times, so the total sum is 0.
</div>

Example 3:

<div class="example-block">
Input: nums = [4,4,4,1,2,3], k = 3

Output: 12

Explanation:

<ul>
	<li>The number 1 appears once.</li>
	<li>The number 2 appears once.</li>
	<li>The number 3 appears once.</li>
	<li>The number 4 appears three times.</li>
</ul>

So, the total sum is <code>4 + 4 + 4 = 12</code>.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
	<li><code>1 &lt;= k &lt;= 100</code></li>
</ul>