A -> 5, 4, 2, 1 B -> 6, 5, 4, 2, 1 k = 3,
ignoring duplicates, the answer is 9 (a=5, b=4) but doesn't the algorithm below give 8 (a=2, b=6)? On Oct 6, 9:06 pm, ligerdave <david.c...@gmail.com> wrote: > use pointers and lengths of two arrays. depends on what K is, if K> > m*n/2, you reverse the pointers. therefore, the worst case is either > O(m) when length of m is shorter or O(n) when length of n is > shorter, > > make the pointers pointing to the first elements in both arrays. > > A) > 4,3,2,2,1 > ^ > > B) > 5,3,2,1 > ^ > > compare them to find out which one is larger, here 5 is larger than 4. > by definition, you know 5 would be bigger than any elements in array > A, and sum of 5 with kth element of array A (here, kth <= A.length) > will be the one(kth largest sum(a+b) overall) you are looking for. > > if k>A.length, shift the pointer of B one number to the right and > repeat the same process. > > like i said, if the k> m*n/2, start from small > > On Oct 6, 6:34 am, sourav <souravs...@gmail.com> wrote: > > > > > you are given 2 arrays sorted in decreasing order of size m and n > > respectively. > > > Input: a number k <= m*n and >= 1 > > > Output: the kth largest sum(a+b) possible. where > > a (any element from array 1) > > b (any element from array 2) > > > The Brute force approach will take O(n*n). can anyone find a better > > logic. thnkx in advance. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
