Find Building Where Alice and Bob Can Meet
<p>You are given a <strong>0-indexed</strong> array <code>heights</code> of positive integers, where <code>heights[i]</code> represents the height of the <code>i<sup>th</sup></code> building.</p>

<p>If a person is in building <code>i</code>, they can move to any other building <code>j</code> if and only if <code>i &lt; j</code> and <code>heights[i] &lt; heights[j]</code>.</p>

<p>You are also given another array <code>queries</code> where <code>queries[i] = [a<sub>i</sub>, b<sub>i</sub>]</code>. On the <code>i<sup>th</sup></code> query, Alice is in building <code>a<sub>i</sub></code> while Bob is in building <code>b<sub>i</sub></code>.</p>

<p>Return <em>an array</em> <code>ans</code> <em>where</em> <code>ans[i]</code> <em>is <strong>the index of the leftmost building</strong> where Alice and Bob can meet on the</em> <code>i<sup>th</sup></code> <em>query</em>. <em>If Alice and Bob cannot move to a common building on query</em> <code>i</code>, <em>set</em> <code>ans[i]</code> <em>to</em> <code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> heights = [6,4,8,5,2,7], queries = [[0,1],[0,3],[2,4],[3,4],[2,2]]
<strong>Output:</strong> [2,5,-1,5,2]
<strong>Explanation:</strong> In the first query, Alice and Bob can move to building 2 since heights[0] &lt; heights[2] and heights[1] &lt; heights[2]. 
In the second query, Alice and Bob can move to building 5 since heights[0] &lt; heights[5] and heights[3] &lt; heights[5]. 
In the third query, Alice cannot meet Bob since Alice cannot move to any other building.
In the fourth query, Alice and Bob can move to building 5 since heights[3] &lt; heights[5] and heights[4] &lt; heights[5].
In the fifth query, Alice and Bob are already in the same building.  
For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.
</pre>

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

<pre>
<strong>Input:</strong> heights = [5,3,8,2,6,1,4,6], queries = [[0,7],[3,5],[5,2],[3,0],[1,6]]
<strong>Output:</strong> [7,6,-1,4,6]
<strong>Explanation:</strong> In the first query, Alice can directly move to Bob&#39;s building since heights[0] &lt; heights[7].
In the second query, Alice and Bob can move to building 6 since heights[3] &lt; heights[6] and heights[5] &lt; heights[6].
In the third query, Alice cannot meet Bob since Bob cannot move to any other building.
In the fourth query, Alice and Bob can move to building 4 since heights[3] &lt; heights[4] and heights[0] &lt; heights[4].
In the fifth query, Alice can directly move to Bob&#39;s building since heights[1] &lt; heights[6].
For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.

</pre>

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

<ul>
	<li><code>1 &lt;= heights.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= heights[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= queries.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>queries[i] = [a<sub>i</sub>, b<sub>i</sub>]</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= heights.length - 1</code></li>
</ul>
Stack; Binary Indexed Tree; Segment Tree; Array; Binary Search; Monotonic Stack; Heap (Priority Queue)
class BinaryIndexedTree {
    private final int inf = 1 << 30;
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
        Arrays.fill(c, inf);
    }

    public void update(int x, int v) {
        while (x <= n) {
            c[x] = Math.min(c[x], v);
            x += x & -x;
        }
    }

    public int query(int x) {
        int mi = inf;
        while (x > 0) {
            mi = Math.min(mi, c[x]);
            x -= x & -x;
        }
        return mi == inf ? -1 : mi;
    }
}

class Solution {
    public int[] leftmostBuildingQueries(int[] heights, int[][] queries) {
        int n = heights.length;
        int m = queries.length;
        for (int i = 0; i < m; ++i) {
            if (queries[i][0] > queries[i][1]) {
                queries[i] = new int[] {queries[i][1], queries[i][0]};
            }
        }
        Integer[] idx = new Integer[m];
        for (int i = 0; i < m; ++i) {
            idx[i] = i;
        }
        Arrays.sort(idx, (i, j) -> queries[j][1] - queries[i][1]);
        int[] s = heights.clone();
        Arrays.sort(s);
        int[] ans = new int[m];
        int j = n - 1;
        BinaryIndexedTree tree = new BinaryIndexedTree(n);
        for (int i : idx) {
            int l = queries[i][0], r = queries[i][1];
            while (j > r) {
                int k = n - Arrays.binarySearch(s, heights[j]) + 1;
                tree.update(k, j);
                --j;
            }
            if (l == r || heights[l] < heights[r]) {
                ans[i] = r;
            } else {
                int k = n - Arrays.binarySearch(s, heights[l]);
                ans[i] = tree.query(k);
            }
        }
        return ans;
    }
}