<p>You are given an integer array <code>nums</code> and two integers <code>indexDiff</code> and <code>valueDiff</code>.</p>

<p>Find a pair of indices <code>(i, j)</code> such that:</p>

<ul>
	<li><code>i != j</code>,</li>
	<li><code>abs(i - j) &lt;= indexDiff</code>.</li>
	<li><code>abs(nums[i] - nums[j]) &lt;= valueDiff</code>, and</li>
</ul>

<p>Return <code>true</code><em> if such pair exists or </em><code>false</code><em> otherwise</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
<strong>Output:</strong> true
<strong>Explanation:</strong> We can choose (i, j) = (0, 3).
We satisfy the three conditions:
i != j --&gt; 0 != 3
abs(i - j) &lt;= indexDiff --&gt; abs(0 - 3) &lt;= 3
abs(nums[i] - nums[j]) &lt;= valueDiff --&gt; abs(1 - 1) &lt;= 0
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3
<strong>Output:</strong> false
<strong>Explanation:</strong> After trying all the possible pairs (i, j), we cannot satisfy the three conditions, so we return false.
</pre>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= indexDiff &lt;= nums.length</code></li>
	<li><code>0 &lt;= valueDiff &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    int clothest(vector<int>& v, int& n) {
        int uidx = upper_bound(v.begin(), v.end(), n) - v.begin();
        int lidx = lower_bound(v.begin(), v.end(), n) - v.begin();

        if (uidx == v.size()) uidx--;
        if (lidx == v.size()) lidx--;

        int ans = min(abs(n - v[uidx]), abs(n - v[lidx]));
        if (uidx > 0) ans = min(ans, abs(n - v[uidx - 1]));
        if (lidx > 0) ans = min(ans, abs(n - v[lidx - 1]));
        if (uidx < v.size() - 1) ans = min(ans, abs(n - v[uidx + 1]));
        if (lidx < v.size() - 1) ans = min(ans, abs(n - v[lidx + 1]));

        return ans;
    }

    void insertN(vector<int>& v, int& n) {
        v.insert(lower_bound(v.begin(), v.end(), n), n);
    }

    void deleteN(vector<int>& v, int& n) {
        v.erase(lower_bound(v.begin(), v.end(), n));
    }

    bool containsNearbyAlmostDuplicate(vector<int>& nums, int indexDiff, int valueDiff) {
        // priority_queue<int> maxHeap;
        // priority_queue < int, vector<int>, greater<int> > minHeap;
        // unordered_map<int, int> umap;

        vector<int> v = {};

        int l = 0;
        for (int r = 0; r < nums.size(); r++) {
            if (r > indexDiff) {
                deleteN(v, nums[l]);
                l++;
            }
    
            if (r > 0) {
                int x = clothest(v, nums[r]);
                if (x <= valueDiff) return true;
            }

            insertN(v, nums[r]);
        }

        // cout << maxHeap.top() << " " << minHeap.top();
        // sort(nums.begin(), nums.end());
        // int n = 10;
        // cout << clothest(nums, n);
        return false;
    }
};
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