<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:
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
        int n = nums.size();
        
        if(n == 0 || k < 0  || t < 0) return false;
        
        unordered_map<int,int> buckets;
        
        for(int i=0; i<n; ++i) {
            int bucket = nums[i] / ((long)t + 1);
            
			// For negative numbers, we need to decrement bucket by 1
			// to ensure floor division.
			// For example, -1/2 = 0 but -1 should be put in Bucket[-1].
			// Therefore, decrement by 1.
            if(nums[i] < 0) --bucket;
            
            if(buckets.find(bucket) != buckets.end()) return true;
            else {
                buckets[bucket] = nums[i];
                if(buckets.find(bucket-1) != buckets.end() && (long) nums[i] - buckets[bucket-1] <= t) return true;
                if(buckets.find(bucket+1) != buckets.end() && (long) buckets[bucket+1] - nums[i] <= t) return true;
                
                if(buckets.size() > k) {
                    int key_to_remove = nums[i-k] / ((long)t + 1);
                    
                    if(nums[i-k] < 0) --key_to_remove;
                    
                    buckets.erase(key_to_remove);
                }
            }
        }
        
        return false;
    }
};
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