Every time I call remove‐from‐end on a dynamic array list it takes O(1) time, except if I’m using half the space or less. In that case, it takes O(n) because it will shrink the array (by making a new smaller array and copying all the elements to the smaller space). Determine and prove the runtime of this algorithm for removal‐from‐end is O(1) amortized. [Note: you may not "prove" it by saying "this is the same as X", use more formal proof methods.]

You are given an array of integers nums and an integer target. Return the indices of two numbers in the array such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Implement a function in Java to solve this problem using a hash map, and explain the time and space complexity of your solution.

Write a bottom-up mergesort that makes use of the array's order by carrying out the following steps each time it needs to locate two arrays to merge: locate the first element in an array that is smaller than its predecessor, then locate the next, and finally merge them to form a sorted subarray. Consider the array size and the number of maximal ascending sequences in the array while analysing the running time of this method.

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.

Suppose you have an array-based implementation of a linked list. What would be the worst case asymtotic runtime of the following functions? Select the tightest. Please explain for each funtion. insert() search() delete() minimum() maximum()

There are two n-element arrays A and B the task is to permute them such that A[i]+b[i]>=k Can someone walk me through the code. I dont understand the iList and jList and why we have if j and i not in List we do List.append. Also at the end why we do if len(iList)+len(jList)==len(A)+len(B) ? def twoArrays (k, A, B): A.sort() B. sort() iList = [] jList = [] for i in range (len (A)): for j in range (len (B)): if(A[i]+B[j]>=k): if j not in jList and i not in iList: iList.append(i) jList.append(j) print (iList) print (jList) if(len (iList) + len(jList) ==len (A) + len(B)) return "YES" else: return "NO"