2 votes 2 votes Let A,B,C,D,E are sorted sequences having length 70,74,80,85,102 respectively.They are merged into a single sequence by merging together two sequences at a time.The minimum number of comparisons that will be needed by algorithm in best case for going merging is _________. Algorithms merge-sort algorithms numerical-answers + – VS asked Jan 30, 2018 • retagged Jul 8, 2022 by Lakshman Bhaiya VS 4.3k views answer comment Share Follow See all 13 Comments See all 13 13 Comments reply Show 10 previous comments VS commented Jan 31, 2018 reply Follow Share @akash.dinkar12 But, correct ans given has 337 . I don't know how ! 0 votes 0 votes akash.dinkar12 commented Jan 31, 2018 reply Follow Share @VS Don't worry our answer is perfect!!! 1 votes 1 votes $ruthi commented Jan 31, 2018 i edited by $ruthi Jan 31, 2018 reply Follow Share i got 379 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes for minimum:- (70, 102)-:70 (172, 74)-:74 (246, 80)-:80 (246, 85)-:85 ans=309 kalra05 answered Dec 17, 2019 kalra05 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Given set {70,74,80,85,102} Merge (70,74) = 70 comparisons Merge (80,144) = 80 comparisons Merge (85,224) = 85 comparisons Merge (102,309) = 102 comparisons Total no. of comparisons=70+80+85+102 = 337 I hope this helps!! Jayan_78 answered Dec 14, 2019 Jayan_78 comment Share Follow See all 0 reply Please log in or register to add a comment.