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The characters of the string K R P C S N Y T J M are inserted into a hash table of size 10 using hash function
h(x) = ((ord(x) - ord(A) + 1)) mod 10
If linear probing is used to resolve collisions, then the following insertion causes collision
h(x) = ((ord(x) - ord(A) + 1)) mod 10
If linear probing is used to resolve collisions, then the following insertion causes collision
- a)Y
- b)C
- c)M
- d)P
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The characters of the string K R P C S N Y T J M are inserted into a h...
The hash table with size 10 will have index from 0 to 9. hash function = h(x) = ((ord(x) - ord(A) + 1)) mod 10 So for string K R P C S N Y T J M: K will be inserted at index : (11-1+1) mod 10 = 1 R at index: (18-1+1) mod 10 = 8 P at index: (16-1+1) mod 10 = 6 C at index: (3-1+1) mod 10 = 3 S at index: (19-1+1) mod 10 = 9 N at index: (14-1+1) mod 10 = 4 Y at index (25-1+1) mod 10 = 5 T at index (20-1+1) mod 10 = 0 J at index (10-1+1) mod 10 = 0 // first collision occurs. M at index (13-1+1) mod 10 = 3 //second collision occurs. Only J and M are causing the collision. Final Hash table will be:
0 T
1 K
2 J
3 C
4 N
5 Y
6 P
7 M
8 R
9 S
So, option (C) is correct.
0 T
1 K
2 J
3 C
4 N
5 Y
6 P
7 M
8 R
9 S
So, option (C) is correct.
Most Upvoted Answer
The characters of the string K R P C S N Y T J M are inserted into a h...
Collision occurs when two or more elements are hashed to the same index in the hash table. In this case, the hash function h(x) = ((ord(x) - ord(A) + 1)) mod 10 is used to determine the index for each character. Let's analyze the given string and the hash function to understand why option 'C' causes a collision.
Given string: K R P C S N Y T J M
Hash function: h(x) = ((ord(x) - ord(A) + 1)) mod 10
Step 1: Calculate the index for each character using the hash function.
Character 'K':
ord('K') = 75
h('K') = ((75 - 65 + 1) mod 10) = 1
Character 'R':
ord('R') = 82
h('R') = ((82 - 65 + 1) mod 10) = 8
Character 'P':
ord('P') = 80
h('P') = ((80 - 65 + 1) mod 10) = 6
Character 'C':
ord('C') = 67
h('C') = ((67 - 65 + 1) mod 10) = 3
Character 'S':
ord('S') = 83
h('S') = ((83 - 65 + 1) mod 10) = 9
Character 'N':
ord('N') = 78
h('N') = ((78 - 65 + 1) mod 10) = 4
Character 'Y':
ord('Y') = 89
h('Y') = ((89 - 65 + 1) mod 10) = 5
Character 'T':
ord('T') = 84
h('T') = ((84 - 65 + 1) mod 10) = 4
Character 'J':
ord('J') = 74
h('J') = ((74 - 65 + 1) mod 10) = 0
Character 'M':
ord('M') = 77
h('M') = ((77 - 65 + 1) mod 10) = 3
Step 2: Check for collision.
Looking at the calculated indices, we can see that characters 'N' and 'T' both hash to index 4. This means that there is a collision between these two characters.
Therefore, the correct answer is option 'C' because inserting the character 'M' causes a collision with the character 'N' at index 4.
Given string: K R P C S N Y T J M
Hash function: h(x) = ((ord(x) - ord(A) + 1)) mod 10
Step 1: Calculate the index for each character using the hash function.
Character 'K':
ord('K') = 75
h('K') = ((75 - 65 + 1) mod 10) = 1
Character 'R':
ord('R') = 82
h('R') = ((82 - 65 + 1) mod 10) = 8
Character 'P':
ord('P') = 80
h('P') = ((80 - 65 + 1) mod 10) = 6
Character 'C':
ord('C') = 67
h('C') = ((67 - 65 + 1) mod 10) = 3
Character 'S':
ord('S') = 83
h('S') = ((83 - 65 + 1) mod 10) = 9
Character 'N':
ord('N') = 78
h('N') = ((78 - 65 + 1) mod 10) = 4
Character 'Y':
ord('Y') = 89
h('Y') = ((89 - 65 + 1) mod 10) = 5
Character 'T':
ord('T') = 84
h('T') = ((84 - 65 + 1) mod 10) = 4
Character 'J':
ord('J') = 74
h('J') = ((74 - 65 + 1) mod 10) = 0
Character 'M':
ord('M') = 77
h('M') = ((77 - 65 + 1) mod 10) = 3
Step 2: Check for collision.
Looking at the calculated indices, we can see that characters 'N' and 'T' both hash to index 4. This means that there is a collision between these two characters.
Therefore, the correct answer is option 'C' because inserting the character 'M' causes a collision with the character 'N' at index 4.
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Community Answer
The characters of the string K R P C S N Y T J M are inserted into a h...
The hash table with size 10 will have index from 0 to 9. hash function = h(x) = ((ord(x) - ord(A) + 1)) mod 10 So for string K R P C S N Y T J M: K will be inserted at index : (11-1+1) mod 10 = 1 R at index: (18-1+1) mod 10 = 8 P at index: (16-1+1) mod 10 = 6 C at index: (3-1+1) mod 10 = 3 S at index: (19-1+1) mod 10 = 9 N at index: (14-1+1) mod 10 = 4 Y at index (25-1+1) mod 10 = 5 T at index (20-1+1) mod 10 = 0 J at index (10-1+1) mod 10 = 0 // first collision occurs. M at index (13-1+1) mod 10 = 3 //second collision occurs. Only J and M are causing the collision. Final Hash table will be:
0 T
1 K
2 J
3 C
4 N
5 Y
6 P
7 M
8 R
9 S
So, option (C) is correct.
0 T
1 K
2 J
3 C
4 N
5 Y
6 P
7 M
8 R
9 S
So, option (C) is correct.
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The characters of the string K R P C S N Y T J M are inserted into a hash table of size 10 using hash functionh(x) = ((ord(x) - ord(A) + 1)) mod 10If linear probing is used to resolve collisions, then the following insertion causes collisiona)Yb)Cc)Md)PCorrect answer is option 'C'. Can you explain this answer?
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