Sync LeetCode submission Runtime - 321 ms (82.46%), Memory - 46.4 MB (65.40%)

Author: jimit105

1364-tuple-with-same-product/README.md

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+<p>Given an array <code>nums</code> of <strong>distinct</strong> positive integers, return <em>the number of tuples </em><code>(a, b, c, d)</code><em> such that </em><code>a * b = c * d</code><em> where </em><code>a</code><em>, </em><code>b</code><em>, </em><code>c</code><em>, and </em><code>d</code><em> are elements of </em><code>nums</code><em>, and </em><code>a != b != c != d</code><em>.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,3,4,6]
+<strong>Output:</strong> 8
+<strong>Explanation:</strong> There are 8 valid tuples:
+(2,6,3,4) , (2,6,4,3) , (6,2,3,4) , (6,2,4,3)
+(3,4,2,6) , (4,3,2,6) , (3,4,6,2) , (4,3,6,2)
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,4,5,10]
+<strong>Output:</strong> 16
+<strong>Explanation:</strong> There are 16 valid tuples:
+(1,10,2,5) , (1,10,5,2) , (10,1,2,5) , (10,1,5,2)
+(2,5,1,10) , (2,5,10,1) , (5,2,1,10) , (5,2,10,1)
+(2,10,4,5) , (2,10,5,4) , (10,2,4,5) , (10,2,5,4)
+(4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)
+</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;= 10<sup>4</sup></code></li>
+	<li>All elements in <code>nums</code> are <strong>distinct</strong>.</li>
+</ul>

1364-tuple-with-same-product/solution.py

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+# Approach 3: Product Frequency Hash Map
+
+# Time: O(n^2)
+# Space: O(n^2)
+
+class Solution:
+    def tupleSameProduct(self, nums: List[int]) -> int:
+        n = len(nums)
+
+        pair_products_freq = {}
+        total_num_of_tuples = 0
+
+        for first_idx in range(n):
+            for second_idx in range(first_idx + 1, n):
+                product_val = nums[first_idx] * nums[second_idx]
+
+                pair_products_freq[product_val] = pair_products_freq.get(product_val, 0) + 1
+
+            
+        for product_freq in pair_products_freq.values():
+            pairs_of_equal_product = (product_freq - 1) * product_freq // 2
+
+            total_num_of_tuples += 8 * pairs_of_equal_product
+
+        return total_num_of_tuples
+