<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 indexDiff, int valueDiff) {
        //sort first indexdiff numbers and store them
        vector<int> sorted;
        for(int i = 0; i < indexDiff; i++) {
            sorted.push_back(nums[i]);
        }
        sort(sorted.begin(), sorted.end());
        //CHECK FOR A PAIR
        printf("sorted: ");
        for(int i = 0; i < sorted.size()-1; i++) {
            if(sorted[i+1] - sorted[i] <= valueDiff) {
                return true;
            }
            printf("%d ", sorted[i]);
        }
        printf("%d\n", sorted[sorted.size()-1]);

        //SLIDING WINDOW
        for(int i = indexDiff; i < nums.size(); i++) {
            int toRemove = nums[i-indexDiff];
            int toAdd = nums[i];
            //remove toRemove via binary search 
            int l = 0, r = indexDiff-1;
            while(l <= r) {
                int mid = l + (r-l)/2;
                if(sorted[mid] <= toAdd) {
                    l = mid+1;
                } else {
                    r = mid-1;
                }
            }
            sorted.insert(sorted.begin()+l, toAdd);
            if(l>0 && sorted[l] - sorted[l-1] <= valueDiff) {
                return true;
            }
            if(l<indexDiff && sorted[l+1] - sorted[l] <= valueDiff) {
                return true;
            }

            //remove toRemove via binary search
            l = 0, r = indexDiff;
                while(l <= r) {
                int mid = l + (r-l)/2;
                if(sorted[mid] == toRemove) {
                    sorted.erase(sorted.begin()+mid);
                    break;
                } else if(sorted[mid] < toRemove) {
                    l = mid+1;
                } else {
                    r = mid-1;
                }
            }
        }
        printf("returning false\n");
        return false;
    }
};

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