Question

Question 4 (10 pts) This algorithm counts the number of duplicate pairs in a sequence. Analysis Worst-case time complexity of the algorithm. Input: a1, a2,...,an, where n is the length of the sequence. Output: count = the number of duplicate pairs. count := 0 For i = 1 to n-1 For j = i+1 to n If (a? = aj), count := count +1 End-for End-for Return count

          Question 4 (10 pts)
This algorithm counts the number of duplicate pairs in a sequence. Analysis Worst-case time
complexity of the algorithm.
Input: a1, a2,...,an, where n is the length of the sequence.
Output: count = the number of duplicate pairs.
count := 0
For i = 1 to n-1
For j = i+1 to n
If (a? = aj), count := count +1
End-for
End-for
Return count

Question 4 (10 pts)
This algorithm counts the number of duplicate pairs in a sequence. Analysis Worst-case time
complexity of the algorithm.
Input: a1, a2,...,an, where n is the length of the sequence.
Output: count = the number of duplicate pairs.
count := 0
For i = 1 to n-1
For j = i+1 to n
If (a? = aj), count := count +1
End-for
End-for
Return count

Added by Teresa M.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert K S

Step 1

Step 1: State the operation to count — each time the inner If (a_i = a_j) is executed we perform one comparison; count the number of such comparisons as the cost measure. Show more…

Show all steps

Thanks for your feedback!

Question 4 (10 pts): Counting Duplicate Pairs in a Sequence This algorithm counts the number of duplicate pairs in a sequence. Analysis: The worst-case time complexity of the algorithm is O(n^2). Input: a1, a2,...,an, where n is the length of the sequence. Output: count = the number of duplicate pairs. count := 0 For i = 1 to n-1 For j = i+1 to n If (ai = aj) count := count + 1 End-for End-for Return count