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Author: jimit105

2150-kth-smallest-product-of-two-sorted-arrays/README.md

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+Given two <strong>sorted 0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code> as well as an integer <code>k</code>, return <em>the </em><code>k<sup>th</sup></code><em> (<strong>1-based</strong>) smallest product of </em><code>nums1[i] * nums2[j]</code><em> where </em><code>0 &lt;= i &lt; nums1.length</code><em> and </em><code>0 &lt;= j &lt; nums2.length</code>.
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [2,5], nums2 = [3,4], k = 2
+<strong>Output:</strong> 8
+<strong>Explanation:</strong> The 2 smallest products are:
+- nums1[0] * nums2[0] = 2 * 3 = 6
+- nums1[0] * nums2[1] = 2 * 4 = 8
+The 2<sup>nd</sup> smallest product is 8.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [-4,-2,0,3], nums2 = [2,4], k = 6
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> The 6 smallest products are:
+- nums1[0] * nums2[1] = (-4) * 4 = -16
+- nums1[0] * nums2[0] = (-4) * 2 = -8
+- nums1[1] * nums2[1] = (-2) * 4 = -8
+- nums1[1] * nums2[0] = (-2) * 2 = -4
+- nums1[2] * nums2[0] = 0 * 2 = 0
+- nums1[2] * nums2[1] = 0 * 4 = 0
+The 6<sup>th</sup> smallest product is 0.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [-2,-1,0,1,2], nums2 = [-3,-1,2,4,5], k = 3
+<strong>Output:</strong> -6
+<strong>Explanation:</strong> The 3 smallest products are:
+- nums1[0] * nums2[4] = (-2) * 5 = -10
+- nums1[0] * nums2[3] = (-2) * 4 = -8
+- nums1[4] * nums2[0] = 2 * (-3) = -6
+The 3<sup>rd</sup> smallest product is -6.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>-10<sup>5</sup> &lt;= nums1[i], nums2[j] &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= k &lt;= nums1.length * nums2.length</code></li>
+	<li><code>nums1</code> and <code>nums2</code> are sorted.</li>
+</ul>

2150-kth-smallest-product-of-two-sorted-arrays/solution.py

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+class Solution:
+    def f(self, nums2: List[int], x1: int, v: int) -> int:
+        if x1 > 0:
+            return bisect_right(nums2, v // x1)
+        elif x1 < 0:
+            return len(nums2) - bisect_left(nums2, -(-v // x1))
+        else:
+            return len(nums2) if v >= 0 else 0
+
+    def kthSmallestProduct(
+        self, nums1: List[int], nums2: List[int], k: int
+    ) -> int:
+        n1 = len(nums1)
+        left, right = -(10**10), 10**10
+        while left <= right:
+            mid = (left + right) // 2
+            count = 0
+            for i in range(n1):
+                count += self.f(nums2, nums1[i], mid)
+            if count < k:
+                left = mid + 1
+            else:
+                right = mid - 1
+        return left