Sync LeetCode submission Runtime - 70 ms (83.51%), Memory - 29.5 MB (77.76%)

Author: jimit105

3213-count-subarrays-where-max-element-appears-at-least-k-times/README.md

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+<p>You are given an integer array <code>nums</code> and a <strong>positive</strong> integer <code>k</code>.</p>
+
+<p>Return <em>the number of subarrays where the <strong>maximum</strong> element of </em><code>nums</code><em> appears <strong>at least</strong> </em><code>k</code><em> times in that subarray.</em></p>
+
+<p>A <strong>subarray</strong> is a contiguous sequence of elements within an array.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,3,2,3,3], k = 2
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> The subarrays that contain the element 3 at least 2 times are: [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,4,2,1], k = 3
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> No subarray contains the element 4 at least 3 times.
+</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>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
+</ul>

3213-count-subarrays-where-max-element-appears-at-least-k-times/solution.py

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+# Approach 1: Sliding Window
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def countSubarrays(self, nums: List[int], k: int) -> int:
+        max_elem = max(nums)
+        ans = start = max_elem_in_window = 0
+
+        for end in range(len(nums)):
+            if nums[end] == max_elem:
+                max_elem_in_window += 1
+            while max_elem_in_window == k:
+                if nums[start] == max_elem:
+                    max_elem_in_window -= 1
+                start += 1
+            ans += start
+
+        return ans
+