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小墨 提交于 2022-03-29 12:55 +08:00 . update
<p>Given the array <code>nums</code>, for each <code>nums[i]</code> find out how many numbers in the array are smaller than it. That is, for each <code>nums[i]</code> you have to count the number of valid <code>j&#39;s</code>&nbsp;such that&nbsp;<code>j != i</code> <strong>and</strong> <code>nums[j] &lt; nums[i]</code>.</p>
<p>Return the answer in an array.</p>
<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<pre>
<strong>Input:</strong> nums = [8,1,2,2,3]
<strong>Output:</strong> [4,0,1,1,3]
<strong>Explanation:</strong>
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3).
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1).
For nums[3]=2 there exist one smaller number than it (1).
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
</pre>
<p><strong>Example 2:</strong></p>
<pre>
<strong>Input:</strong> nums = [6,5,4,8]
<strong>Output:</strong> [2,1,0,3]
</pre>
<p><strong>Example 3:</strong></p>
<pre>
<strong>Input:</strong> nums = [7,7,7,7]
<strong>Output:</strong> [0,0,0,0]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>2 &lt;= nums.length &lt;= 500</code></li>
<li><code>0 &lt;= nums[i] &lt;= 100</code></li>
</ul>
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