1 1 vote Given a set of $n$ points: $S=\left\{P_{1}, P_{2}, P_{3}, P_{4} \ldots P_{n}\right\}$, where $P_{1}=\left(x_{i}, y_{j}\right)$. Finding the pair of points that has the smallest distance among all pairs that can be solved in $O$ (nlogn) time using Divide and Conquer Greedy Technique Dynamic programming All of these Algorithms algorithms time-complexity test-series + – nikkey123 844 views answer comment Share Follow Print See all 3 Comments 3 3 Comments reply Anu007 commented Jan 3, 2018 reply Follow flag by greedy it will O(nlogn) , by dynamic may be in O(n) 0 0 replyShare nikkey123 commented Jan 3, 2018 reply Follow flag can u explain me how u have done it 0 0 replyShare rahul sharma 5 commented Jan 4, 2018 reply Follow flag Divide and conquer? What is the answer? Ref: https://www.cs.cmu.edu/~ckingsf/bioinfo-lectures/closepoints.pdf 0 0 replyShare Please log in or register to add a comment.
0 0 votes using greedy approach we can solve in O(nlogn). ie. by using Kruskal's algo $ruthi answered Jan 7, 2018 $ruthi comment Share Follow 0 reply Please log in or register to add a comment.
