![[Acuña] Give sortints Big-Oh order if its input was already sorted, where the expression bounds the number of swaps. Include](http://img.homeworklib.com/questions/907d7bb0-9859-11ea-9e6a-27cc2c4477a2.png?x-oss-process=image/resize,w_560)
Homework Answers
Given an sorting algorithm and input is already sorted.
--- If the input is already sorted. Then no swapping will happen.
So the number of swaps = 0
but irrespective of the sorted array, outer for loop will iterate n times when n is the size of the input array.
So based on swappings f(n) = 1
but if you consider the outer for loop also then T(n) = O(n)
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[Acuña] Give sortints' Big-Oh order if its input was already sorted, where the expression bounds the...
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public class CountIt {
public long SnippetNestedLoop(long n) {
long i, j, x = 0; i=0; x++; while(i<n){ x++;
//i<n
// SomeStatement // j = 0;
// j < n
// SomeStatement // j++;
// Can you explain why is this here? // i++;
// Can you explain why is this here? Ans: i < n
} }
}
x++; return...
What is the Big-Oh order of the following code fragment? The fragment is parametrized on the variable N. Assume that you are measuring the number of times j is decremented. public static void sort(Comparable[] a) { int N-a.length; for (int i = 1; i < N;i++) { for (int j = i; j > && less(a[5], a[j-1]); j--) //measure j -- exch(a, j, j-1); O(nlogn) O O(n^2) Q(n) Does not exist.
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Determine the Big 0 provide the order (Big O) of the execution
speed also determine the exact execution speed.
public class CountIt {
public long SnippetNestedLoop(long n) {
long i, j, x = 0; i=0; x++; while(i<n){ x++;
//i<n
// SomeStatement // j = 0;
// j < n
// SomeStatement // j++;
// Can you explain why is this here? // i++;
// Can you explain why is this here? Ans: i < n
} }
}
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