Sync LeetCode submission Runtime - 23 ms (38.46%), Memory - 17.8 MB (94.93%)

Author: jimit105

2856-count-complete-subarrays-in-an-array/README.md

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+<p>You are given an array <code>nums</code> consisting of <strong>positive</strong> integers.</p>
+
+<p>We call a subarray of an array <strong>complete</strong> if the following condition is satisfied:</p>
+
+<ul>
+	<li>The number of <strong>distinct</strong> elements in the subarray is equal to the number of distinct elements in the whole array.</li>
+</ul>
+
+<p>Return <em>the number of <strong>complete</strong> subarrays</em>.</p>
+
+<p>A <strong>subarray</strong> is a contiguous non-empty part of an array.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,3,1,2,2]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> The complete subarrays are the following: [1,3,1,2], [1,3,1,2,2], [3,1,2] and [3,1,2,2].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [5,5,5,5]
+<strong>Output:</strong> 10
+<strong>Explanation:</strong> The array consists only of the integer 5, so any subarray is complete. The number of subarrays that we can choose is 10.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 2000</code></li>
+</ul>

2856-count-complete-subarrays-in-an-array/solution.py

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+# Approach: Sliding Window
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def countCompleteSubarrays(self, nums: List[int]) -> int:
+        res = 0
+        count = {}
+        n = len(nums)
+        right = 0
+        distinct = len(set(nums))
+
+        for left in range(n):
+            if left > 0:
+                remove = nums[left - 1]
+                count[remove] -= 1
+                if count[remove] == 0:
+                    count.pop(remove)
+            while right < n and len(count) < distinct:
+                add = nums[right]
+                count[add] = count.get(add, 0) + 1
+                right += 1
+            if len(count) == distinct:
+                res += n - right + 1
+
+        return res
+
+