Sorting
October 26, 2023JNU 2018-2 PhD CS
October 26, 2023Sorting
| Question 94 |
How many comparisons are needed to sort an array of length 5 if a straight selection sort is used and array is already in the opposite order?
| 1 | |
| 5 | |
| 10 | |
| 20 |
Question 94 Explanation:
Consider the array: 5 4 3 2 1
1st iteration will compare 4 numbers with the 5
2nd iteration will compare 3 numbers with the 4
3rd iteration will compare 2 numbers with the 3
4th iteration i will compare 1 number with the 2
So, the total number of comparisons is 4 + 3 + 2 + 1 = 10
It can be viewed as the sum of the sequence of the first (n-1) numbers starting by 1
S = ((1 + (n-1) )*(n-1) )/2
S = 10
1st iteration will compare 4 numbers with the 5
2nd iteration will compare 3 numbers with the 4
3rd iteration will compare 2 numbers with the 3
4th iteration i will compare 1 number with the 2
So, the total number of comparisons is 4 + 3 + 2 + 1 = 10
It can be viewed as the sum of the sequence of the first (n-1) numbers starting by 1
S = ((1 + (n-1) )*(n-1) )/2
S = 10
Correct Answer: C
Question 94 Explanation:
Consider the array: 5 4 3 2 1
1st iteration will compare 4 numbers with the 5
2nd iteration will compare 3 numbers with the 4
3rd iteration will compare 2 numbers with the 3
4th iteration i will compare 1 number with the 2
So, the total number of comparisons is 4 + 3 + 2 + 1 = 10
It can be viewed as the sum of the sequence of the first (n-1) numbers starting by 1
S = ((1 + (n-1) )*(n-1) )/2
S = 10
1st iteration will compare 4 numbers with the 5
2nd iteration will compare 3 numbers with the 4
3rd iteration will compare 2 numbers with the 3
4th iteration i will compare 1 number with the 2
So, the total number of comparisons is 4 + 3 + 2 + 1 = 10
It can be viewed as the sum of the sequence of the first (n-1) numbers starting by 1
S = ((1 + (n-1) )*(n-1) )/2
S = 10
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