def bigoh1(n: int) -> int: res = 6 for i in range(n): res = res +i for i in range (n * n): res = res - i return res
def bigoh1(n: int) -> int: res = 6 for i in range(n): res = res +i for i in range (n * n): res = res - i return res
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Transcribed Image Text:(b)
def bigoh2(n: int) -> int:
res = 0
for i in range(n):
for j in range(i, n // 2):
res += (i * j)
return res

Transcribed Image Text:Give the worst-case big-oh time efficiency bound for each of the following
functions.
Requirements:
- Your big-Oh bound must be as TIGHT as possible, e.g., if 0(n) is the correct
answer, then O(n^2) will NOT be given marks because it is not tight enough.
- Your big-Oh bound must be SIMPLIFIED, e.g., answers such as O(n-1) or 0(3n)
will NOT ben given marks because they are not simplified enough.
Please write your answer for each question on the appropriate TODO Line.
(a)
def bigoh1(n: int) -> int:
res = 0
for i in range(n):
res = res + i
for i in range (n * n):
res = res - i
return res
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