Could someone please show how to find the order of growth (big O) for the swim and sink functions on binary heaps? I need to check my work. Thank you!
1.) Swim
private void swim(int k) { while (k > 1 && less(k/2, k)) { exch(k, k/2); k = k/2; } }
2.) Sink
private void sink(int k) { while (2*k <= N) { int j = 2*k; if (j < N && less(j, j+1)) j++; if (!less(k, j)) break; exch(k, j); k = j; } }
Could someone please show how to find the order of growth (big O) for the swim...
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.
They NAME sc 162- lec. 18 (Big quiz 1. Arrange the following functions in order of increasing rate of growth. Also, identify any functions with the SAME rate of growth by putting then below the others. a) sn, 44log n, 10n log n, 500, 2n, 28, 3n b) n', n +2 nlog2 n, n! ne log, n, n n n'. 4", n, na, 2 2. Use the Big-o notation to estimate the time complexity for the following segments/methods. (Assume all...
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Show the Big O Complexity of the following functions and loop constructions: (Please show work and explain) a. f(n) = 2n + (blog(n+1)) b. f(n) = n * (log(n-1))/2 c. int sum = 0; for (int i=0; i<n; i++) sum++; for (int j=n; j>0; j /= 2) sum++; d. int sum = 0; for (int i=n; i>0; i--) for (int j=i; j<n; j *= 2) sum++;
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 } } } x++; return...
Show how to get the big-Oh for the following code: void CountSort (int A[N], int range) { // assume 0 <= A[i] < range for any element A[i] int *pi = new int[range]; for ( int i = 0; i < N; i++ ) pi[A[i]]++; for ( int j = 0; j < range; j++ ) for ( int k = 1; k <= pi[j]; k++ ) cout << j << endl; }
For each of the below code snippets, identify the bounding function (the big O) of the number of times the indicated line is run (justify your answer briefly): int i = 1: while (i < n) { printf ("Insert difficult work here!") i = i + i: } for(i = 0: i < n: i++) { for (j = 0: j < n: j++) { for (k = 0: k < n: k++) { if(i==j && j==k) arr[i] [j] [k]...
Provide a "big oh" run-time analysis for each of the following. When a value of “n” is used, it is the size of the input. 4.) void problem 40 cin n min max for (int i min i n, i++) for (int j- 1: j< max, j++) tota while (total n tota total 2 total 5.) void problem 50 cin n; for (int i 0: i n, i++) for (int j 0; j i2; j++) for (int k 0; k...
Describe the order of magnitude of the following code section using Big(O) notation j = 1; While (j < N) { j = j * 2); } Can someone give me a more Clearer answer please.
[Acuña] Give sortints' Big-Oh order if its input was already sorted, where the expression bounds the number of swaps. Include a short explanation public static void sortints (int[] data) { for(int i = 1; i < data.length; i++) { int j = i; while(j > 0 && data[j-1] > data[j]) { swap (data[j], data[j-1]) //f(n) counts these j--;