Sorting Lower Bound. Show that any array of integers x[1...n] can be sorted in O(n + M) time, M = max.x; -min.x; i For small M, this is linear time: why doesn't the 2(n logn) lower bound apply in this case? where

Sorting Lower Bound. Show that any array of integers x[1...n] can be sorted in O(n + M) time, M = max.x; -min.x; i For small M, this is linear time: why doesn't the 2(n logn) lower bound apply in this case? where

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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where
Sorting Lower Bound. Show that any array of integers x[1...n] can be sorted in O(n + M) time,
M = max.x; -min.x;
i
For small M, this is linear time: why doesn't the (n logn) lower bound apply in this case?

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1. Let S be a set of n arbitrary but distinct numbers. (a) Give a deterministic algorithm to output the largest n³/4 of the numbers in S in sorted order. Your algorithm must use O(n) comparisons. (b) Show that any deterministic comparison-based algorithm that correctly outputs the median on inputs of n distinct elements must use at least n 1 comparisons. To be completely precise, the median is defined to be the [n/2]th smallest element.
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Problem 3. Given an array A[1 : n] of n positive integers (possibly with repetitions), your goal is to find whether there is a sub-array A[l : r] such that EAli) = n. i=l Example. Given A = up to n = 10. On the other hand, if the input array is A = [3, 2, 6, 8, 20, 2, 4| for n = 7, the answer is No since no sub-array of A adds up to n = 7. [13, 1, 2, 3, 4, 7, 2, 3, 8,9] for n = 10, the answer is Yes since elements in A[2 : 5] add Hint: Observe that if - A[i] = n, then A[i] = E-1 A[i] - - n; this may come handy!