<p>Given an array of positive integers <code>arr</code>, find a pattern of length <code>m</code> that is repeated <code>k</code> or more times.</p>

<p>A <strong>pattern</strong> is a subarray (consecutive sub-sequence) that consists of one or more values, repeated multiple times <strong>consecutively </strong>without overlapping. A pattern is defined by its length and the number of repetitions.</p>

<p>Return <code>true</code> <em>if there exists a pattern of length</em> <code>m</code> <em>that is repeated</em> <code>k</code> <em>or more times, otherwise return</em> <code>false</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [1,2,4,4,4,4], m = 1, k = 3
<strong>Output:</strong> true
<strong>Explanation: </strong>The pattern <strong>(4)</strong> of length 1 is repeated 4 consecutive times. Notice that pattern can be repeated k or more times but not less.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,2,1,2,1,1,1,3], m = 2, k = 2
<strong>Output:</strong> true
<strong>Explanation: </strong>The pattern <strong>(1,2)</strong> of length 2 is repeated 2 consecutive times. Another valid pattern <strong>(2,1) is</strong> also repeated 2 times.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,2,1,2,1,3], m = 2, k = 3
<strong>Output:</strong> false
<strong>Explanation: </strong>The pattern (1,2) is of length 2 but is repeated only 2 times. There is no pattern of length 2 that is repeated 3 or more times.
</pre>

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

<ul>
	<li><code>2 &lt;= arr.length &lt;= 100</code></li>
	<li><code>1 &lt;= arr[i] &lt;= 100</code></li>
	<li><code>1 &lt;= m &lt;= 100</code></li>
	<li><code>2 &lt;= k &lt;= 100</code></li>
</ul>

class Solution {
public:
    bool containsPattern(vector<int>& arr, int m, int k) {
        int size = arr.size();
        vector<int> ans;
        map<vector<int>, int> containPattern;

        for(int i=0; i<size; i++){
            ans.push_back(arr[i]);
            if(i+1 >= m){
                int count = 1;
                for(int j=i+1; j<size-m+1; j+=m){
                    vector<int> temp(arr.begin()+j, arr.begin()+j+m);
                    if(ans == temp) count++;
                    else break;
                }
                containPattern[ans] = max(count, containPattern[ans]);
                ans.erase(ans.begin());
            }
        }
        for(auto i: containPattern){
            if(i.second >= k) return true;
        }
        return false;
    }
};
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