0 votes 0 votes What is the number of comparisons required to extract 45th element of the min heap? Algorithms algorithms binary-heap time-complexity + – bts1jimin asked Sep 10, 2018 bts1jimin 1.8k views answer comment Share Follow See all 15 Comments See all 15 15 Comments reply Show 12 previous comments Shaik Masthan commented Sep 10, 2018 reply Follow Share it is bonded ===> O(1) https://gateoverflow.in/1110/gate2003-23 0 votes 0 votes neerajyadav commented Dec 4, 2018 reply Follow Share for the 45th minimumm in heap total uumber of comparisons =sum of first 44 natural numbers=44*45/2=990 1 votes 1 votes Somtirtha commented Dec 10, 2019 reply Follow Share why how ? 0 votes 0 votes Please log in or register to add a comment.