1 votes 1 votes Consider inserting the key 10,22,31,4,15,28,17,88,59 using open addressing technique into hash table of length m=11 with hash function h(k)= k mod 11 what is no of collisions encountered? A) 7 B) 10 C)12 D) none of these given ans is d is it right? ADITYA CHAURASIYA 5 asked Dec 7, 2017 ADITYA CHAURASIYA 5 231 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Ashwin Kulkarni commented Dec 7, 2017 reply Follow Share NO Ans should be option A 22 88 4 15 28 17 59 31 10 Here, 10, 22, 31, 4, 28 has no collisions, But 15 -> 1 collision with 4 17 - > one collision with 28 88 -> one collision with 22 59 -> coliision with 4,15,28,17 Hence total = 7 0 votes 0 votes Kalpataru Bose commented Dec 9, 2017 reply Follow Share How do you know how the collisions have been resolved by linear probing ? 0 votes 0 votes Please log in or register to add a comment.