count-values-with-equally-spaced-occurrences-i

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Algorithms

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

An integer <code>x</code> is called special if:

<ul>
	<li><code>x</code> appears exactly three times in <code>nums</code>.</li>
	<li>All three occurrences of <code>x</code> are equally spaced in <code>nums</code>. In other words, if all occurrences of <code>x</code> are at indices <code>i1 &lt; i2 &lt; i3</code>, then <code>i2 - i1 = i3 - i2</code>.</li>
</ul>

Return the number of distinct special integers in <code>nums</code>.

&nbsp;
Example 1:

<div class="example-block">
Input: nums = [1,8,1,5,1,5,8,5]

Output: 2

Explanation:

<ul>
	<li>1 is special because it occurs exactly three times at equally spaced indices 0, 2, and 4.</li>
	<li>5 is special because it occurs exactly three times at equally spaced indices 3, 5, and 7.</li>
	<li>8 is not special because it occurs only twice.</li>
</ul>

Therefore, the answer is 2.
</div>

Example 2:

<div class="example-block">
Input: nums = [8,8,8,8]

Output:&nbsp;0

Explanation:

8 is not special because it does not occur exactly three times. Therefore, the answer is 0.
</div>

Example 3:

<div class="example-block">
Input: nums = [8,6,6,8,8]

Output: 0

Explanation:

8 occurs at indices 0, 3, and 4, which are not equally spaced. 6 occurs only twice. Therefore, no integer is special.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>3 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>