1 Fill in the blank in the following hash function: int HashMap::HashFunction(int k, int mapSize){ int h = 0; int tmp = k; while (tmp > 0) { h += tmp % 10; tmp = tmp/10; //here return h; } A (S) tmp +=1; B (H) h %= mapSize; C (R) if (h > 0) D (T) map[h++] = k; E (A) other
1 Fill in the blank in the following hash function: int HashMap::HashFunction(int k, int mapSize){ int h = 0; int tmp = k; while (tmp > 0) { h += tmp % 10; tmp = tmp/10; //here return h; } A (S) tmp +=1; B (H) h %= mapSize; C (R) if (h > 0) D (T) map[h++] = k; E (A) other
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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![Hash Maps:
1 Fill in the blank in the following hash function:
int HashMap::HashFunction(int k, int mapSize){
int h = 0;
int tmp = k;
while (tmp > 0) {
h += tmp % 10;
tmp = tmp/10;
}
//here
return h;
}
A (S) tmp +=1;
B (H) h %= mapSize;
C (R) if (h > 0)
D (T) map(h++] = k;
E (A) other
2. Assume you have a perfect hash function such that there are NEVER any collisions (woah! Amazing!). Fill in the blank, below,
in the findKey method that (solely) finds a particular key (a string, k) in the hashMap:
int findkey(string k, int mapSize) {
}
A (H) if (map[k] == key) return k;
B (U) for (int i = 0; i < mapSize; i++) { if (map[i]->key = k) return i; }
C (0) return map[k]%mapSize;
D (E) return hashFunk (k);
E (Y) other
3. Assume you've got a hashMap of size 10. Use the HashFunction, in 1, and linear probing to place the following keys into the
hashMap: 32,21,41,53,60,14
A (R)
1
2
3
5
6
7
8
9
21
32
41
60
53
14
в (L)
6
8
9
41
14
21
32
60
53
C (E)
1
2
3
4
5
6
7
8
53
14
41
60
21
32
D (I)
1
4
5
6
7
8
41
21
60
53
14
32
E (D) Other](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f42c95c-d285-4d7d-9c78-356ca988b170%2F0b92a384-8e8f-4f5b-ba57-f21377201b40%2Ff817cs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Hash Maps:
1 Fill in the blank in the following hash function:
int HashMap::HashFunction(int k, int mapSize){
int h = 0;
int tmp = k;
while (tmp > 0) {
h += tmp % 10;
tmp = tmp/10;
}
//here
return h;
}
A (S) tmp +=1;
B (H) h %= mapSize;
C (R) if (h > 0)
D (T) map(h++] = k;
E (A) other
2. Assume you have a perfect hash function such that there are NEVER any collisions (woah! Amazing!). Fill in the blank, below,
in the findKey method that (solely) finds a particular key (a string, k) in the hashMap:
int findkey(string k, int mapSize) {
}
A (H) if (map[k] == key) return k;
B (U) for (int i = 0; i < mapSize; i++) { if (map[i]->key = k) return i; }
C (0) return map[k]%mapSize;
D (E) return hashFunk (k);
E (Y) other
3. Assume you've got a hashMap of size 10. Use the HashFunction, in 1, and linear probing to place the following keys into the
hashMap: 32,21,41,53,60,14
A (R)
1
2
3
5
6
7
8
9
21
32
41
60
53
14
в (L)
6
8
9
41
14
21
32
60
53
C (E)
1
2
3
4
5
6
7
8
53
14
41
60
21
32
D (I)
1
4
5
6
7
8
41
21
60
53
14
32
E (D) Other
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