Bartleby Related Questions Icon
Related questions
Question
Transcribed Image Text:Given an array of numbers X1 = {x1, x2, ..., n } an exchanged pair in X is a pair xi, xj such that i < j and
xį > x¡ . Note that an element x; can be part of up to n - 1 exchanged pairs, and that the maximal possible
number of exchanged pairs in X is n(n − 1)/2, which is achieved if the array is sorted in descending order.
Give a divide-and-conquer algorithm that counts the number of exchanged pairs in X in O(nlogn) time.
SAVE
AI-Generated Solution
info
AI-generated content may present inaccurate or offensive content that does not represent bartleby’s views.
bartleby
Unlock instant AI solutions
Tap the button
to generate a solution
to generate a solution
Click the button to generate
a solution
a solution
Knowledge Booster
Background pattern image
Similar questions
- I need help with this pleasearrow_forwardLets say we have an array of n items that we need to sort. Lets say that there is a sorting algorithm that has the property that in every step, no item moves more than log n places in the array in any step of the algorithm. Show that this algorithm runs in time at least Ω(n2 / log n) in the worst case.arrow_forward1. Asymptotics. Given an array A of n integers, you'd like to output a two-dimensional n x n array B in which B[i, j] = max {A[i], A[i + 1],..., A[j]} for each i < j. For i j the value of B[i, j] can be left as is. for i = 1,2, η for j = i + 1, n " Compute the maximum of the entries A[i], A[i + 1], , A[j]. Store the maximum value in B[i, j]. (a) Find a function of such that the running time of the algorithm is O(f(n)), and clearly explain why. (b) For the same function f argue that the running time of the algorithm is also (f(n)). (This establishes an asymptotically tight bound (f(n)).) (c) Design and analyze a faster algorithm for this problem. You should give an algorithm with running O(g(n)), where lim→∞ g(n)/f(n) = 0.arrow_forward
- During each iteration of Quick Sort algorithm, the first element of array is selected as a pivot. The algorithm for Quick Sort is given below. Modify it in such a way that last element of array should be selected as a pivot at each iteration. Also explain the advantages.arrow_forwardFind a sorted matrix: The input is a real number x and a matrix A[1..n, 1..m] of nm real numbers, with each row A[i, 1..m] and column A[1..n, j] sorted. The objective is to locate the largest array entry A[i, j] that is less than or equal to x, or to report that all components of A are greater than x. Create and test an iterative method that analyses as few matrix elements as feasible. Be cautious if you assume a simple binary search would fix the problem. Find a sorted matrix: The input is a real number x and a matrix A[1..n, 1..m] of nm real numbers, with each row A[i, 1..m] and column A[1..n, j] sorted. The objective is to locate the largest array entry A[i, j] that is less than or equal to x, or to report that all components of A are greater than x. Create and test an iterative method that analyses as few matrix elements as feasible. Be cautious if you assume a simple binary search would fix the problem.arrow_forwardLet A be an array of n integers: (a) Write a pseudo code for a divide-and-conquer algorithm that computes the MAX and MIN values in array A . (b) Write the recurrence for this algorithm. (c) Solve the recurrence using the substitution method.arrow_forward
- The algorithm below searches for the maximum in an input array A. Assume A is a random sequence containing n distinguishable real numbers, what is the probability for line #4 to be executed exactly once? (The phrase "random sequence" here means the n numbers can appear in any order with equal chance.) mymax(A, n) { 1: max = A[1]; 2: for i = 2:n 3: if A[i]> max 4: max = A[i); 5: end; 6: end; 7: return max; }arrow_forwardCan someone help me with this algorithm? No coding neededarrow_forward15. Consider the problem of finding the first position in which an array b occurs as a subsequence of an array a. Write two nested loops: let result undefined for (let i = 0; i < a.length - b.length; i++) { for (let j = 0; j < b.length; j++) { if (a[i+j] = b[j]) . . . } } Complete with labeled break and continue statements.arrow_forward
arrow_back_ios
arrow_forward_ios