This is possible because the algorithm assumes that each of the n input elements is in the range[0, k], i.e. no element is smaller than 0 nor greater than k. The input is an array A indexed from 1to n, the output is stored in the array B also indexed from 1 ton, and the array C is used as atemporary container, indexed from 0 to k.1235678101112let C[0..k] be a new arrayfor i=0 to kC[i] = 0for j = 1 to A.lengthC[A[j]] = C[A[j]] + 1// C[i] now contains the number of elements equal to i.for i=1 to kC[i] = C[i] + C[i-1]// C[i] now contains the number of elements less than or equal to i.for j = A.length downto 1B[C[A[j]]]= A[j]C[A[j]] = C[A[j]] - 1Given an input array A = [5, 1, 2, 1, 5, 3], what is the status of C' at the end of the algorithm?○ 1, 2, 3, 4, 4,6none of the othersO 0, 1, 2, 3, 5,5O 4, 4, 5, 2,0O 0, 0, 2, 3, 4, 4

This is possible because the algorithm assumes that each of the n input elements is in the range [0, k], i.e. no element is smaller than 0 nor greater than k. The input is an array A indexed from 1 to n, the output is stored in the array B also indexed from 1 ton, and the array C' is used as a temporary container, indexed from 0 to k. 1 2 3 4 5 6 7 8 9 10 11 12 let C[0..k] be a new array for i = 0 to k C[i] = 0 for j = 1 to A.length C[A[j]] = C[A[j]] + 1 // C[i] now contains the number of elements equal to i. for i=1 to k C[i] = C[i] + C[i-1] // C[i] now contains the number of elements less than or equal to i. for j = A.length downto 1 B[C[A[j]]] = A[j] C[A[j]] = C[A[j]] − 1 - Given an input array A = [5, 1, 2, 1, 5, 3], what is the status of C' at the end of the algorithm? O 1,2,3,4,4,6 O none of the others O 0, 1, 2, 3, 5, 5 O4, 4, 5, 2,0 O 0, 0, 2, 3, 4, 4

This is possible because the algorithm assumes that each of the n input elements is in the range [0, k], i.e. no element is smaller than 0 nor greater than k. The input is an array A indexed from 1 to n, the output is stored in the array B also indexed from 1 ton, and the array C' is used as a temporary container, indexed from 0 to k. 1 2 3 4 5 6 7 8 9 10 11 12 let C[0..k] be a new array for i = 0 to k C[i] = 0 for j = 1 to A.length C[A[j]] = C[A[j]] + 1 // C[i] now contains the number of elements equal to i. for i=1 to k C[i] = C[i] + C[i-1] // C[i] now contains the number of elements less than or equal to i. for j = A.length downto 1 B[C[A[j]]] = A[j] C[A[j]] = C[A[j]] − 1 - Given an input array A = [5, 1, 2, 1, 5, 3], what is the status of C' at the end of the algorithm? O 1,2,3,4,4,6 O none of the others O 0, 1, 2, 3, 5, 5 O4, 4, 5, 2,0 O 0, 0, 2, 3, 4, 4

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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