Count the Number of K-Big Indices
<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and a positive integer <code>k</code>.</p>

<p>We call an index <code>i</code> <strong>k-big</strong> if the following conditions are satisfied:</p>

<ul>
	<li>There exist at least <code>k</code> different indices <code>idx1</code> such that <code>idx1 &lt; i</code> and <code>nums[idx1] &lt; nums[i]</code>.</li>
	<li>There exist at least <code>k</code> different indices <code>idx2</code> such that <code>idx2 &gt; i</code> and <code>nums[idx2] &lt; nums[i]</code>.</li>
</ul>

<p>Return <em>the number of k-big indices</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [2,3,6,5,2,3], k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are only two 2-big indices in nums:
- i = 2 --&gt; There are two valid idx1: 0 and 1. There are three valid idx2: 2, 3, and 4.
- i = 3 --&gt; There are two valid idx1: 0 and 1. There are two valid idx2: 3 and 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,1], k = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no 3-big indices in nums.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i], k &lt;= nums.length</code></li>
</ul>
Binary Indexed Tree; Segment Tree; Array; Binary Search; Divide and Conquer; Ordered Set; Merge Sort
class BinaryIndexedTree {
    private int n;
    private int[] c;

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

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

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= x & -x;
        }
        return s;
    }
}

class Solution {
    public int kBigIndices(int[] nums, int k) {
        int n = nums.length;
        BinaryIndexedTree tree1 = new BinaryIndexedTree(n);
        BinaryIndexedTree tree2 = new BinaryIndexedTree(n);
        for (int v : nums) {
            tree2.update(v, 1);
        }
        int ans = 0;
        for (int v : nums) {
            tree2.update(v, -1);
            if (tree1.query(v - 1) >= k && tree2.query(v - 1) >= k) {
                ++ans;
            }
            tree1.update(v, 1);
        }
        return ans;
    }
}