EBK DATA STRUCTURES AND ALGORITHMS IN C
4th Edition
ISBN: 9781285415017
Author: DROZDEK
Publisher: YUZU
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What is the best case complexity (in terms of the number of comparisons) for standard (non-early termination) bubble sort, sorting an array of n elements?
Write a version of bottom-up mergesort that takes advantage of order in the array by proceeding as follows each time it needs to find two arrays to merge: find a sorted subarray (by incrementing a pointer until finding an entry thatis smaller than its predecessor in the array), then find the next, then merge them. Analyze the running time of this algorithm in terms of the array size and the number of maximal increasing sequences in the array.
Write a version of bottom-up mergesort that takes advantage of order in the array by proceeding as follows each time it needs to find two arrays to merge: find a sorted subarray (by incrementing a pointer until finding an entry that is smaller than its predecessor in the array), then find the next, then merge them. Analyze the running time of this algorithm in terms of the array size and the number of maximal increasing sequences in the array.
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EBK DATA STRUCTURES AND ALGORITHMS IN C
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- The median m of a sequence of n elements is the element that would fall in the middle if the sequence was sorted. That is, e ≤ m for half the elements, and m ≤ e for the others. Clearly, one can obtain the median by sorting the sequence, but one can do quite a bit better with the following algorithm that finds the kth element of a sequence between a (inclusive) and b (exclusive). (For the median, use k = n/2, a = 0, and b = n.) select(k, a, b)Pick a pivot p in the subsequence between a and b.Partition the subsequence elements into three subsequences: the elements <p, =p, >p Let n1, n2, n3 be the sizes of each of these subsequences.if k < n1 return select(k, 0, n1).else if (k > n1 + n2) return select(k, n1 + n2, n).else return p. c++arrow_forwardImplement bucket sort with an array of lists data structure to sort i) a list ofstrings, ii) a list of floating point numbers and iii) a list of integers, which areuniformly distributed over a range, for an appropriate choice of the number ofbuckets.arrow_forwardWrite a program to implement insertion sort and also test this using an unsorted array in javaarrow_forward
- Trace the passes of insertion sort on the following lists:i) {H, K, M, N, P} ii) {P, N, M, K, H}Compare their performance in terms of the comparisons made.arrow_forwardExplain Quick sort algorithms with the following set of integers {50 , 70, 60, 90, 40, 80, 10, 20 , 30}arrow_forwardWrite a program in Java to implement the Insertion sort algorithm to sort a given set of elements and determine the time required to sort the elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted, and plot a graph of the time taken versus n. The elements can be read from a file or can be generated using the random number generatorarrow_forward
- There is a variation of the bubble sort algorithm called a gap sort that, rather than comparing neighboring elements each time through the list, compares elements that are some number i positions apart, where i is an integer less than n. For example, the first element would be compared to the (i + 1) element, the second element would be compared to the (i + 2) element, the nth element would be compared to the (n - i) element, and so on. A single iteration is completed when all of the elements that can be compared, have been compared. On thenext iteration, i is reduced by some number greater than 1 and the process continues until i is less than 1. Implement a gap sort.arrow_forwardConsider an array A of N values (N entered by the user). Write a C program that creates a second array B such that the i-th element of B, i.e., B[i] is equal to A[0] *A[1] * A[2] * ... A[i]; In other words:- B[0] = A[0]- B[1] = A[0] * A[1];- B[2] = A[0] * A[1] * A[2];- B[3] = A[0] * A[1] * A[2] * A[3]- ....etc.For this exercise, we would like to implement two solutions and compare their complexity /performance in terms of computation time.Solution 1: Implement blindly the following algorithms: for each element B[i], compute A[0] *A[1] * A[2] * .... *A[i] and save the result in B[i];To test the solution, create an array A of 1000 elements and initialize each of its elements with arandom value (instead of asking the user to . To do that, use the function rand, which, every timecalled, returns a random number Questions:- How long does your program take to produce the solution?- Analyze the solution above and explain why it is taking long time. What are the mainfactors that affect how long…arrow_forwardImplement bucket sort using an array of lists data structure to sort three lists of uniformly dispersed numbers: a list of integers, a list of floating point numbers, and a list of strings for the right number of buckets.arrow_forward
- The Bubble Sort procedure bubblesort(a1,a2,...,an: real numbers with n>1) for i:=1 to n-1 for j:=1 to n-i if aj>aj+1 then interchange aj and aj+1 _____________________________________________ Show how this algorithm works on the input sequence 3, 6, 1, 4, 2. What sequence do we have after the first pass (with i=1)? Make sure that you give the current state of the sequence after every pass.arrow_forwardA variation of the bubble sort algorithm known as a gap sort examines elements that are some number I positions apart, where I is an integer less than n, rather than neighbouring elements each time through the list. The first element, for instance, would be compared to the element I + 1), the second element, to the element I + 2), the nth element, to the element (n - I and so on. Once all the components that can be compared have been done so, the iteration is finished. I am decreased by a factor greater than 1 on the following iteration, and the procedure continues until I is less than 1. Apply the gap filter.arrow_forwardA gap sort is a version of the bubble sort method that compares items that are some number i places away, where i is an integer smaller than n, rather than neighbouring elements each time across the list. For example, the first element would be compared to the (i + 1) element, the second element to the (i + 2) element, the nth element to the (n - i) element, and so on. When all of the items that may be compared have been compared, a single iteration is accomplished. On the following iteration, i is reduced by a value bigger than one, and the procedure is repeated until i is less than one. Utilise a gap kind.arrow_forward
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