[LeetCode Sync] Runtime - 883 ms (18.91%), Memory - 41.4 MB (77.93%)

Author: NinePiece2

Solutions/3885-count-special-triplets/README.md

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+<p>You are given an integer array <code>nums</code>.</p>
+
+<p>A <strong>special triplet</strong> is defined as a triplet of indices <code>(i, j, k)</code> such that:</p>
+
+<ul>
+	<li><code>0 &lt;= i &lt; j &lt; k &lt; n</code>, where <code>n = nums.length</code></li>
+	<li><code>nums[i] == nums[j] * 2</code></li>
+	<li><code>nums[k] == nums[j] * 2</code></li>
+</ul>
+
+<p>Return the total number of <strong>special triplets</strong> in the array.</p>
+
+<p>Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [6,3,6]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The only special triplet is <code>(i, j, k) = (0, 1, 2)</code>, where:</p>
+
+<ul>
+	<li><code>nums[0] = 6</code>, <code>nums[1] = 3</code>, <code>nums[2] = 6</code></li>
+	<li><code>nums[0] = nums[1] * 2 = 3 * 2 = 6</code></li>
+	<li><code>nums[2] = nums[1] * 2 = 3 * 2 = 6</code></li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [0,1,0,0]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The only special triplet is <code>(i, j, k) = (0, 2, 3)</code>, where:</p>
+
+<ul>
+	<li><code>nums[0] = 0</code>, <code>nums[2] = 0</code>, <code>nums[3] = 0</code></li>
+	<li><code>nums[0] = nums[2] * 2 = 0 * 2 = 0</code></li>
+	<li><code>nums[3] = nums[2] * 2 = 0 * 2 = 0</code></li>
+</ul>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [8,4,2,8,4]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">2</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>There are exactly two special triplets:</p>
+
+<ul>
+	<li><code>(i, j, k) = (0, 1, 3)</code>
+
+	<ul>
+		<li><code>nums[0] = 8</code>, <code>nums[1] = 4</code>, <code>nums[3] = 8</code></li>
+		<li><code>nums[0] = nums[1] * 2 = 4 * 2 = 8</code></li>
+		<li><code>nums[3] = nums[1] * 2 = 4 * 2 = 8</code></li>
+	</ul>
+	</li>
+	<li><code>(i, j, k) = (1, 2, 4)</code>
+	<ul>
+		<li><code>nums[1] = 4</code>, <code>nums[2] = 2</code>, <code>nums[4] = 4</code></li>
+		<li><code>nums[1] = nums[2] * 2 = 2 * 2 = 4</code></li>
+		<li><code>nums[4] = nums[2] * 2 = 2 * 2 = 4</code></li>
+	</ul>
+	</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= n == nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+</ul>

Solutions/3885-count-special-triplets/solution.py

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+class Solution:
+    def specialTriplets(self, nums: List[int]) -> int:
+        left = Counter()
+        right = Counter(nums)
+        result = 0
+        mod = 10**9 + 7
+
+        for num in nums:
+            right[num] -= 1
+            result = (result + left[num * 2] * right[num * 2] % mod) % mod
+            left[num] += 1
+        
+        return result