<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <code>true</code> <em>if there are two <strong>distinct indices</strong> </em><code>i</code><em> and </em><code>j</code><em> in the array such that </em><code>nums[i] == nums[j]</code><em> and </em><code>abs(i - j) &lt;= k</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,1], k = 3
<strong>Output:</strong> true
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,0,1,1], k = 1
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,1,2,3], k = 2
<strong>Output:</strong> false
</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>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        map<int,int> m;
        queue<int> q;

        for (int i = 0; i < k + 1; i++) {
            if (i >= nums.size()) {
                return false;
            }
            if (m[nums[i]] > 0) {
                return true;
            }
            else {
                m[nums[i]]++;
                q.push(nums[i]);
            }
        }
        for (int j = k + 1; j < nums.size(); j++) {
            int old = q.front();
            m[old]--;
            q.pop();

            if(m[nums[j]] > 0) {
                return true;
            }
            else {
                m[nums[j]]++;
                q.push(nums[j]);
            }
        }
        return false;
    }
};
86061