10. One major thing to observe in above table that any entry of t say t[ i ] is dependent upon the entry t[i-1] and t[j] where j < i. Since the table t is getting filled in ascending order of indices, hence for calculating t[i], the dependent values t[i-1] and t[j] where j < i are already calculated, hence the above algorithm is based on dynamic programming, where present index value for t is calculated using previous entries.
11. Here outermost for loop run for n iterations while inner "while" loop run for i-1 iterations in worst case.
Hence the statements in innermost loop will run for O(n2) iterations in worst case.
Hence the worst case time complexity will be O(n2).
Please comment for any clarification.
For questions 10-12, refer to the following iterative code computes values for the table t. 1 pub...
(10') 6. For each of the following code blocks, write the best (tightest) big-o time complexity i) for (int i = 0; ǐ < n/2; i++) for (int j -0: ni j++) count++ i) for (int í = 0; i < n; i++) for (int ni j0 - for (int k j k ni kt+) count++ İİİ) for (int í ー 0; i < n; i++) for(int j = n; j > 0; j--) for (int k = 0; k...
Question 3: Given the following two code fragments [2 Marks] (i)Find T(n), the time complexity (as operations count) in the worst case? (ii)Express the growth rate of the function in asymptotic notation in the closest bound possible. (iii)Prove that T(n) is Big O (g(n)) by the definition of Big O (iv)Prove that T(n) is (g(n)) by using limits
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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.
Which big-O expression best characterizes the worst case time complexity of the following code? public static int foo(int N) ( int count = 0; int i1; while (i <N) C for (int j = 1; j < N; j=j+2) { count++ i=i+2; return count; A. O(log log N) B. O(log N2) C. O(N log N) D. O(N2)
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find complexity Problem 1 Find out the computational complexity (Big-Oh notation) of the code snippet: Code 1: for (int i = n; i > 0; i /= 2) { for (int j = 1; j < n; j *= 2) { for (int k = 0; k < n; k += 2) { // constant number of operations here } } } Code 2: Hint: Lecture Note 5, Page 7-8 void f(int n) { if (n...
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