How Many Numbers Are Smaller Than the Current Number
<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 class="example">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 class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [6,5,4,8]
<strong>Output:</strong> [2,1,0,3]
</pre>

<p><strong class="example">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>
Array; Hash Table; Counting Sort; Sorting
function smallerNumbersThanCurrent(nums: number[]): number[] {
    const search = (nums: number[], x: number) => {
        let l = 0,
            r = nums.length;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    const arr = nums.slice().sort((a, b) => a - b);
    for (let i = 0; i < nums.length; ++i) {
        nums[i] = search(arr, nums[i]);
    }
    return nums;
}
