<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>

class Solution {
public:
    std::vector<int> smallerNumbersThanCurrent(std::vector<int>& nums) {
        std::vector<std::pair<int, int>> indexedNums;
        std::vector<int> ans(nums.size());

        // Fill the vector with pairs (number, original_index)
        for (int i = 0; i < nums.size(); ++i) {
            indexedNums.push_back({nums[i], i});
        }

        // Sort based on the number values
        std::sort(indexedNums.begin(), indexedNums.end());

        // Traverse the sorted vector to compute the number of smaller elements
        int count = 0; // Number of elements smaller than the current element
        for (int i = 0; i < indexedNums.size(); ++i) {
            if (i > 0 && indexedNums[i].first != indexedNums[i - 1].first) {
                count = i; // Number of elements smaller than current
            }
            ans[indexedNums[i].second] = count; // Assign the count to the result vector at the original index
        }

        return ans;
    }
};
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