Sync LeetCode submission Runtime - 199 ms (14.24%), Memory - 34.4 MB (45.31%)

Author: jimit105

2626-count-the-number-of-good-subarrays/README.md

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+<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the number of <strong>good</strong> subarrays of</em> <code>nums</code>.</p>
+
+<p>A subarray <code>arr</code> is <strong>good</strong> if there are <strong>at least </strong><code>k</code> pairs of indices <code>(i, j)</code> such that <code>i &lt; j</code> and <code>arr[i] == arr[j]</code>.</p>
+
+<p>A <strong>subarray</strong> is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,1,1,1,1], k = 10
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> The only good subarray is the array nums itself.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,1,4,3,2,2,4], k = 2
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> There are 4 different good subarrays:
+- [3,1,4,3,2,2] that has 2 pairs.
+- [3,1,4,3,2,2,4] that has 3 pairs.
+- [1,4,3,2,2,4] that has 2 pairs.
+- [4,3,2,2,4] that has 2 pairs.
+</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], k &lt;= 10<sup>9</sup></code></li>
+</ul>

2626-count-the-number-of-good-subarrays/solution.py

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+# Approach 1: Two pointers
+
+# Time: O(n)
+# Space: O(n)
+
+from collections import Counter
+
+class Solution:
+    def countGood(self, nums: List[int], k: int) -> int:
+        n = len(nums)
+        same, right = 0, -1
+        count = Counter()
+        ans = 0
+
+        for left in range(n):
+            while same < k and right + 1 < n:
+                right += 1
+                same += count[nums[right]]
+                count[nums[right]] += 1
+            if same >= k:
+                ans += n - right
+            count[nums[left]] -= 1
+            same -= count[nums[left]]
+
+        return ans