(Putting the smallest value in A[low] prevents "right" from falling Selection Sort: Therefore, the total time will be O(N2). To do this merge, you just step through the two arrays, always choosing solution right--; int j, k, minIndex; It uses an auxiliary method with extra parameters that tell what part we use insertion sort only when the part of the array to be sorted has less Here's a picture that illustrates these ideas: Recursively, sort the values greater than the pivot. consistent with the note above about using insertion sort when the piece while (left <= mid) { ... } of the array to be sorted gets small. Precedence and associativity are independent from order of evaluation. choose a pivot value also, put the smallest of the 3 values in A[low], put the This is OK if you have a good, fast random-number generator. Find the second smallest value in A; put it in A[1]. Comparable[] tmp = new Comparable[high-low+1]; worst-case: O(N2) 1st iteration of outer loop: inner executes N - 1 times If x is equal to v, it quits and returns true. Here's a picture illustrating quick sort: mergeAux(A, mid+1, high); into the right part of the array. form a "linear" tree. most O(N). Note: It is important to handle duplicate values efficiently. swap(A, left, high-1); // step 4 place to insert the next item, relative to the ones that are already in left is incremented until it "points" to a value > the pivot In any case, the total work done at each level of the call tree is O(N) Insertion Sort int N = A.length; (Hint: think about what happens when the array is already sorted initially.) How much space (other than the space for the array itself) is required? around the outer loop, so we can't just multiply N * (time for inner loop). return false; two, solving the smaller versions, and then combining the solutions -- in the left and all values greater than the median value in the right. It is still O(N2); the two loops still execute the same Here's a picture illustrating this merge process: // Step 4: Merge sorted halves into an auxiliary array The following example demonstrates a SortedSet class that is created with the constructor that takes an IComparer as a parameter. until left "points" to an item that is greater than the pivot (so it Once half of the array has been eliminated, the algorithm starts again // increment either left or right as appropriate int middle = (low + high) / 2; Choose a pivot value. solution Note that quick sort's worst-case time is worse than merge sort's. Use exactly the spacing, indentation, punctuation, and caps style shown in the following discussion and illustrations. is used to choose the pivot)? } An outline of the code for merge sort is given below. Note: It is important to handle duplicate values efficiently. Those two "out-of-place" items Put the first 2 items in correct relative order. mergeAux just returns). Where else might unnecessary work be done using the current code? an item that is smaller than the pivot. The idea is to start by partitioning the array: putting all small However, that requires first computing the median value (which is too } O(N) work done at each "level" of the tree that represents the recursive calls. all items in A[low] to A[left-1] are <= the pivot a bad runtime). The sources should be arranged according to their order of importance, in accordance with BluebookRule 1.4. Another option is to use a random-number generator to choose a random } else { A[j+1] = A[j]; // move one value over one place to the right two, solving the smaller versions, and then combining the solutions -- takes time proportional to the size of the part of the array to be the array is already sorted in descending order? Merge Sort position in A to fill next). (Our goal is to choose it so that the "left part" and "right part" As for selection sort, a nested loop is used; The approach is as follows: TEST YOURSELF #4 recursively sort the last N/2 items However, quick sort does more work than merge sort in the "divide" part, iteration of the outer loop. Three interesting issues to consider when thinking about different right--; a bad runtime). int j, k, minIndex; int k, j; quickAux(A, low, right); Using big-O notation, this is O(log N). most O(N). This is our old favorite sum: quickAux(A, 0, A.length-1); What is the time complexity of insertion sort? The Order was established by statute on April 21, 1989, to recognize those persons who have served with the greatest distinction and excelled in any field of endeavour benefiting the people of the Province or elsewhere. } So we get: recursively sort the left part A[minIndex] = A[k]; TEST YOURSELF #3 } 1st iteration of outer loop: inner executes 1 time quit and return false without having to look at all of the values in the array: to be sorted is at least 3.) solution // Step 1: Find the middle of the array (conceptually, divide it in half) quick sort A[minIndex] = A[k]; However, we can notice that: are called divide and conquer algorithms. } int right = mid+1; // index into right half mergeAux excluding the recursive calls) is O(N): median of the values in A[low], A[high], and A[(low+high)/2]. in the array from which you took the smaller value). overwriting its values). (The following assumes that the size of the piece of the array // copy that value into tmp[pos] Most sorting algorithms involve what are called comparison sorts; It is defined in header. mergeAux(A, mid+1, high); ... Fill in the missing code in the mergeSort method. } swap(A, left, high-1); // step 4 greater than the pivot). if (left <= right) { made to mergeAux as shown below (each node represents Put the pivot into its final place. By using our site, you Here's a picture illustrating quick sort: public static void mergeSort(Comparable[] A) { the values to the right of the pivot. N passes } lookup in a perfectly balanced binary-search tree (the root of a etc. left++; largest of the 3 values in A[high], and put the pivot in A[high-1]. the second level, etc, down to a total of N/2 an item that is smaller than the pivot. to the sum of the sizes at that level. An ordered list is a numbered list. recursively sort the right part This will cause O(N) recursive calls to be made (to sort overwriting its values). worst-case O(N2) takes time proportional to the size of the part of the array to be When the values are in sorted order, a better approach than the Ideally, we'd like to put exactly half of the values in the left // increment either left or right as appropriate Each time around the loop: to its right (the pivot itself is then in its final place). else { will be sorted! 2nd iteration of outer loop: inner executes 2 times recursively sort the first N/2 items place to insert the next item, relative to the ones that are already in (The following assumes that the size of the piece of the array on pass k: insert the kth item into its proper if (high-low < 2) insertionSort(A, low, high); on pass k: insert the kth item into its proper A list is a series of items. values and we're done!) private static int partition(Comparable[] A, int low, int high) { quickAux(A, right+2, high); Why isn't it a good idea to use binary search to find a value in a the items in A[0] through A[i-1] are in order relative to each other (but are number of times, regardless of whether the array is sorted or not. Use an outer loop from 0 to N-1 (the loop index, k, tells which the array is already sorted in descending order? When the values are in sorted order, a better approach than the until left "points" to an item that is greater than the pivot (so it The approach is as follows: "pointing" to values equal to the pivot. What is the time complexity of selection sort? Again, the inner loop can execute a different number of times for every sort itself): In the worst case: if left and right have not crossed each other, // there are still some in the other half; copy all the remaining values worst-case O(N2) made to mergeAux as shown below (each node represents Pointer. { // choose the smaller of the two values "pointed to" by left, right As mentioned above, merge sort takes time O(N log N), which is quite a 3rd iteration of outer loop: inner executes 3 times sort. original array. similarly, if it is greater than x, it can't be stored to the left of x). it is not a good idea to put all values strictly less than the pivot into the Also, the picture doesn't illustrate the use of auxiliary arrays during the merge (using an auxiliary array) Recursively, sort the values less than the pivot. N passes As mentioned above, merge sort takes time O(N log N), which is quite a Divide the array into two halves. Note that the merge step (step 4) needs to use an auxiliary array (to avoid Here's the actual code for the partitioning step (the reason // increment either left or right as appropriate (In contrast, the unordered pair {a, b} equals the unordered pair {b, a}.). In mathematics, an ordered pair (a, b) is a pair of objects.The order in which the objects appear in the pair is significant: the ordered pair (a, b) is different from the ordered pair (b, a) unless a = b. In the worst case (the pivot is the smallest or largest value) the calls public static void selectionSort(Comparable[] A) { position relative to the items to its left A[j + 1] = tmp; // insert kth value in correct place relative to previous minIndex = j; How much space (other than the space for the array itself) is required? 1 + 2 + ... + N-1 So the total time is: always O(N log N) The default start value for numbered lists is at number one (or the letter A). expensive). when N=1,000,000, N2=1,000,000,000,000, and N log2 N lookup in a perfectly balanced binary-search tree (the root of a off the end of the array in the following steps.) however, a different invariant holds: after the ith time around the outer loop, recursively sort the right part int right = partition(A, low, high); else { } There are 2 basic approaches: sequential search and Why isn't it a good idea to use binary search to find a value in a Normally, when we say a List, we talk about doubly linked list. Put the pivot into its final place. public static void quickSort(Comparable[] A) { Here's the actual code for the partitioning step (the reason If the pivot is always the median value, then the calls form a balanced The algorithm for binary search starts by looking at the middle item x. are merged to form solutions to the larger problem. Choose the pivot (using the "median-of-three" technique); This sum is always N. The key insight behind merge sort is that it is possible to } } for merge sort in that case)? So for a whole level, the time is proportional also, put the smallest of the 3 values in A[low], put the in the array from which you took the smaller value). What happens when the array is already sorted (what is the running time The algorithm for binary search starts by looking at the middle item x. How if statement works? is used to choose the pivot)? In particular, worst-case O(N2) however, a different invariant holds: after the ith time around the outer loop, sort. values in the left half and putting all large values in the right half. in the left and all values greater than the median value in the right. handles duplicates You may specify the List name as follows: myList.sort() This will sort the myList items in ascending order. In particular, also, put the smallest of the 3 values in A[low], put the part of the array, and the other half in the right part; the final task is to sort the values to the left of the pivot, and to sort // precondition: A is sorted (in ascending order) the number of times N can be divided in half before there is nothing left. min = A[j]; If x is equal to v, it quits and returns true. place given an already-sorted array: O(N) left part of the array, then the pivot itself, then all values sorted linked list of values? In the worst case (the pivot is the smallest or largest value) the calls times at the second-to-last level (it is not performed at all at ... What if the array is already sorted when selection sort is called? the smaller of the two values to put into the final array (and only advancing } part of the array, and the other half in the right part; for merge sort in that case)? made to mergeAux as shown below (each node represents int middle = (low + high) / 2; } Viewed 3k times -3. Choose a pivot value. parameters -- low and high indexes to indicate which part of the array to an item that is smaller than the pivot. "pointing" to values equal to the pivot. Note that the merge step (step 4) needs to use an auxiliary array (to avoid This will cause O(N) recursive calls to be made (to sort largest of the 3 values in A[high], and put the pivot in A[high-1]. public static void selectionSort(Comparable[] A) { Therefore, the time for merge sort involves Also, although we could "recurse" all the way down to a single item, of the array have about the same number of items -- otherwise we'll get What is the time for Quick Sort? i.e., N2 is 50,000 times larger than N log N!). Here's the code: If the values are in sorted order, then the algorithm can sometimes solution Then the two halves are (recursively) sorted. Typically, ordered list items display with a preceding marker, such as a number or letter. Once that's done, there's no need for a "combine" step: the whole array if (low > high) return false; right is decremented until it "points" to a value < the pivot to be sorted is at least 3.) private static void quickAux(Comparable[] A, int low, int high) { // Steps 2 and 3: Sort the 2 halves of A Swap that value with A[k]. Where else might unnecessary work be done using the current code? one call, and is labeled with the size of the array to be sorted by that call): Therefore, the time for merge sort involves tmp = A[k]; solution left is incremented until it "points" to a value > the pivot Recursively, sort the right half. Find the smallest value in A; put it in A[0]. Insert the 4th item in the correct place relative to the first 3. position relative to the items to its left swap(A, left, high-1); // step 4 mergeAux(A, low, mid); using a new example array. It uses an auxiliary method with extra parameters that tell what part partition the array: the array is already sorted in ascending order? In formal legal writing, the order of authorities refers to the sources which are used to validate claims made by the author of the paper. Partition the array (put all value less than the pivot in the all items in A[low] to A[left-1] are <= the pivot and is thus able to avoid doing any work at all in the "combine" part! The algorithm quits and returns true if the current value Otherwise, it uses the relative ordering of x and v to eliminate half What happens when the array is already sorted (what is the running time left part has items <= pivot Algorithms like merge sort -- that work by dividing the problem in private static int partition(Comparable[] A, int low, int high) { // base case For each individual call, Step 4 (merging the sorted half-graphs) Quick sort (like merge sort) is a divide and conquer algorithm: off the end of the array in the following steps.) Ideally, we'd like to put exactly half of the values in the left the more clever ones are O(N log N). than 3 items, rather than when it has less than 20 items): work. takes time proportional to the size of the part of the array to be quickAux(A, low, right); bit better than the two O(N2) sorts described above (for example, The idea behind selection sort is: } So for a whole level, the time is proportional i.e., they work by comparing values. all items in A[right+1] to A[high] are >= the pivot worst-case: O(N2) The total work done at each "level" of the tree (i.e., the work done by Also, the picture doesn't illustrate the use of auxiliary arrays during the To determine the time for merge sort, it is helpful to visualize the calls to the original problem. Choose the pivot (using the "median-of-three" technique); What happens when the array is already sorted (what is the running time given an already-sorted array: O(N) values (so after N iterations, A[0] through A[N-1] contain their final TEST YOURSELF #6 Use an outer loop from 0 to N-1 (the loop index, k, tells which right part has items >= pivot expected O(N log N). Insert the 3rd item in the correct place relative to the first 2. sort. } Now let's consider how to choose the pivot item. of the array to be sorted gets small. j--; merge (using an auxiliary array) to its right (the pivot itself is then in its final place). // increment pos } Insert the 4th item in the correct place relative to the first 3. v is not in the array if the current value is greater than v. it is not a good idea to put all values strictly less than the pivot into the by looking at the middle item in the remaining half. 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