Show that any array of integers x[1n] can be sorted in O (n + M) time, where

role="math" localid="1659938331794" M=maxxi-minxiii

For small M, this is linear time: why doesn’t the Ω(nlogn) lower bound apply in this case?

Short Answer

Expert verified

The Ωnlogn lower bound does not apply because this is not a comparison-based sort, and also because the information beforehand about the range of the elements in the array is given.

Step by step solution

01

Explain Array

There are different kind of Array. 1’s, 2’s, 3rd dimensional and multi-dimensional array’s. In the above question there will be sorted array technique through we can combine single sorted array in k/n elements. To check that see below detail related to combine theory of array.

02

Show that any array of integers x[1…n] can be sorted in  O (n + M)  time

Consider the any array of integers x [1...n] can be sorted in O (n + M) time where,

M=maxxi-minxiii

Counting sorting takes O(n) times to find maxXi and minXi , and then the size can be established as,

S=maxxi-minxi+1=M+1

The array for counting is Cminxi,,maxxi. Travers the array x [1...n] and for each occurrence of x [i] , the corresponding Cxi is incremented by 1. If C [i] is greater than zero, then the element i is the output of the array. The total time complexity is On+S=On+M.

Since counting sort is not based on comparison, it is not subject to lower bound Ωnlognlimit.

Therefore,The Ωnlognlower bound does not apply because this is not a comparison-based sort, and also because the information beforehand about the range of the elements in the array is given.

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Most popular questions from this chapter

Show that, if c is a positive real number, then g(n) = 1 + c + c2 + · · · + cn is:

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