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
Related questions
Question
![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
5
6
7
8
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?
○ 1, 2, 3, 4, 4,6
none of the others
O 0, 1, 2, 3, 5,5
O 4, 4, 5, 2,0
O 0, 0, 2, 3, 4, 4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feae7f5f2-1eca-45c9-9469-9617d2cd7b6f%2F63eef232-5d93-4f31-94c5-6e39d50ed6b4%2F4lcnnm_processed.png&w=3840&q=75)
Transcribed Image Text: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
5
6
7
8
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?
○ 1, 2, 3, 4, 4,6
none of the others
O 0, 1, 2, 3, 5,5
O 4, 4, 5, 2,0
O 0, 0, 2, 3, 4, 4
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