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Diffstat (limited to 'kth_smallest_product_sort_arr.py')
-rw-r--r-- | kth_smallest_product_sort_arr.py | 45 |
1 files changed, 45 insertions, 0 deletions
diff --git a/kth_smallest_product_sort_arr.py b/kth_smallest_product_sort_arr.py new file mode 100644 index 0000000..459a00c --- /dev/null +++ b/kth_smallest_product_sort_arr.py @@ -0,0 +1,45 @@ +# 2040. Kth Smallest Product of Two Sorted Arrays +import bisect +from math import ceil + +""" +given two sorted 0-indexed integer arrays 'nums1' and 'nums2' as well as an +integer 'k', return the k'th smallest product of 'nums1[i] nums2[i]' where +'0 <= i < nums1.length' and '0 <= j < nums2.length' +""" + + +class Solution(object): + def kthSmallestProduct(self, nums1, nums2, k): + """ + :type nums1: List[int] + :type nums2: List[int] + :type k: int + :rtype: int + """ + + def check(n): + total = 0 + for i in nums1: + if i > 0: + total += bisect.bisect(nums2, n // i) + if i < 0: + total += len(nums2) - bisect.bisect_left(nums2, ceil(n / i)) + if i == 0 and n >= 0: + total += len(nums2) + + start, end = -(10**10) - 1, 10**10 + 1 + while start + 1 < end: + mid = (start + end) // 2 + if check(mid) >= k: + end = mid + else: + start = mid + return start + 1 + + +if __name__ == "__main__": + obj = Solution() + print(obj.kthSmallestProduct(nums1=[2, 5], nums2=[3, 4], k=2)) + print(obj.kthSmallestProduct(nums1=[-4, -2, 0, 3], nums2=[2, 4], k=6)) + print(obj.kthSmallestProduct(nums1=[-2, -1, 0, 1, 2], nums2=[-3, -1, 2, 4, 5], k=3)) |