<p>You are given an array of people, <code>people</code>, which are the attributes of some people in a queue (not necessarily in order). Each <code>people[i] = [h<sub>i</sub>, k<sub>i</sub>]</code> represents the <code>i<sup>th</sup></code> person of height <code>h<sub>i</sub></code> with <strong>exactly</strong> <code>k<sub>i</sub></code> other people in front who have a height greater than or equal to <code>h<sub>i</sub></code>.</p>

<p>Reconstruct and return <em>the queue that is represented by the input array </em><code>people</code>. The returned queue should be formatted as an array <code>queue</code>, where <code>queue[j] = [h<sub>j</sub>, k<sub>j</sub>]</code> is the attributes of the <code>j<sup>th</sup></code> person in the queue (<code>queue[0]</code> is the person at the front of the queue).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> people = [[7,0],[4,4],[7,1],[5,0],[6,1],[5,2]]
<strong>Output:</strong> [[5,0],[7,0],[5,2],[6,1],[4,4],[7,1]]
<strong>Explanation:</strong>
Person 0 has height 5 with no other people taller or the same height in front.
Person 1 has height 7 with no other people taller or the same height in front.
Person 2 has height 5 with two persons taller or the same height in front, which is person 0 and 1.
Person 3 has height 6 with one person taller or the same height in front, which is person 1.
Person 4 has height 4 with four people taller or the same height in front, which are people 0, 1, 2, and 3.
Person 5 has height 7 with one person taller or the same height in front, which is person 1.
Hence [[5,0],[7,0],[5,2],[6,1],[4,4],[7,1]] is the reconstructed queue.
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= people.length &lt;= 2000</code></li>
	<li><code>0 &lt;= h<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
	<li><code>0 &lt;= k<sub>i</sub> &lt; people.length</code></li>
	<li>It is guaranteed that the queue can be reconstructed.</li>
</ul>

class BIT {
private:
    int size;
    vector<int> bit;
    
public:
    BIT (int n) : size(n), bit(n, 0) {
        for (int i = 0; i < size; i++) {
            add(i, 1);
        }
    }

    void add(int idx, int val) {
        for (; idx < size; idx |= (idx + 1)) {
            bit[idx] += val;
        }
    }

    void unset(int idx) {
        add(idx, -1);
    }

    int pref_sum(int idx) {
        int sum = 0;
        for (; idx >= 0; idx = (idx & (idx + 1)) - 1) sum += bit[idx];
        return sum;
    }

    int getKthEmptyPosition(int k) {
        int low = 0, high = size - 1;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (pref_sum(mid) >= k) high = mid;
            else low = mid + 1;
        }
        return low;
    }
};


class Solution {
public:
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        sort(people.begin(), people.end());
        vector<vector<int>> result(people.size());

        BIT bit = BIT(people.size());
        int sameCount = 0;
        for (int i = 0; i < people.size(); i++) {
            if (i && people[i][0] == people[i - 1][0]) sameCount++;
            else sameCount = 0;
            
            int idx = bit.getKthEmptyPosition(people[i][1] - sameCount + 1);
            result[idx] = people[i];
            bit.unset(idx);
        }

        return result;
    }
};
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