Certainly having a smaller volume is necessary for a box to fit in another box, but it is not sufficient. E.g., a box of size 1 x 1 x 1 will not fit in a box of size 2 x 2 x 1/2.
Dave On Sep 21, 1:16 pm, rajess <rajeshrules...@yahoo.com> wrote: > find the volume of boxes as v=l*b*h > sort boxes in volumes in descending order and this is the way to > insert boxes one into another > > On Sep 21, 7:55 pm, Rashmi Shrivastava <rash...@gmail.com> wrote: > > > > > If there are n number of boxes and each with different dimensions and your > > job is to insert one box having lesser dimension than that to another. > > Consider size of boxes as, > > b1->s1(h1,l1,w1) > > b2->s2(h2,l2,w2) > > . > > . > > . > > bn->sn(hn,ln,wn)- Hide quoted text - > > - Show quoted text - -- 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.
