Trace counting sort on the following list L of unordered keys which lie within the range [10, 18]. L[1:10]: 16 11 18 13 11 12 15 15 18 16

O(N) is the order of growth execution time of the index-based add operation when using the ABList class, assuming a list size of N.

Heapsort has heapified an array to: 76 62 40 32 30 and is about to start the second for loop. What is the array after each loop iteration? i= 4: Ex: 86, 75, 30 i = 3: i = 2: i= 1:

Heapsort has heapified an array to: 63 50 37 26 89 and is about to start the second for loop. What is the array after each loop iteration? i = 4: Ex: 86, 75, 30 i = 3: i = 2: i = 1:

The add method of the sorted array-based list implementation takes O(N) time because searching in the array takes O(N) time. True False

Please help me quickly I will vote your answer

Trace counting sort on the following list L of unordered keys which lie withinthe range [10, 18].L[1:10]: 16 11 18 13 11 12 15 15 18 16

Consider the following list: int list[] = (14, 18, 29, 35, 44, 49, 55, 58, 66, 71, 75, 80, 89, 95, 98) %3D When performing a binary search for 75, after the first comparison, the search is restricted to O list[8].. .list[14] O list [5]...list[11] O list[0].. .list[7] O list [0]. . .list[6]

True or false: Merge sort can be implemented using O(1) auxiliary space.

In Java, Apply replacement selection sort to the following list assuming array size M = 3; 27 47 35 7 67 21 32 18 24 20 12 8 Show the content of each sorted file.

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"

Create three problem instances of size n is around 10, representing the best-case, worst-case and average-case for Radix Sort respectively. Show how the Radix Sort works on the best-case, and explain why each of them represents the best-case, worst-case and average-case respectively. psudocode for Radixsort attached

Trace counting sort on the following list L of unordered keys which lie withinthe range [10, 18].L[1:10]: 16 11 18 13 11 12 15 15 18 16