<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:
    def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
        # buffer = arr[:m]
        buffer = 0
        for c in arr[:m]:
            buffer *= 100
            buffer += (c % 100)
        counts = {buffer: {0:1}}
        for i, c in enumerate(arr[m:]):
            
            if m == 1:
                buffer = 0
            else:
                buffer = buffer % (100 ** (m-1))
                buffer *= 100
            buffer += c % 100
            # print(i+m, buffer, counts)
            if buffer not in counts:
                counts[buffer] = {i+m: 1}
                if k <= 1:
                    return True
            elif i+m - m in counts[buffer]:
                counts[buffer][i+m] = counts[buffer][i] + 1
                if counts[buffer][i+m] >= k:
                    return True
            else:
                counts[buffer][i+m] = 1
            
        return False
        
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