have question about scratch as you can see in this photo , it request to write a algorithm and Implement the algorithm using scratch what i need is a photo of algorithm and photo of scr
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i have question about scratch as you can see in this photo , it request to write a
what i need is a photo of algorithm and photo of scratch implementation
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- scheme: Q7: No Repeats Implement no-repeats, which takes a list of numbers s as input and returns a list that has all of the unique elements of s in the order that they first appear, but no repeats. For example, (no-repeats (list 5 4 5 4 2 2)) evaluates to (5 4 2). Hints: To test if two numbers are equal, use the = procedure. To test if two numbers are not equal, use the not procedure in combination with =. You may find it helpful to use the filter procedure. (define (no-repeats s) 'YOUR-CODE-HERE ) ;;; Tests (no-repeats (list 5 4 5 4 2 2)) ; expect (5 4 2)* allSame returns true if all of the elements in list have the same value. * allSame returns false if any two elements in list have different values. * The array may be empty and it may contain duplicate values. * * Your solution should contain at most one loop. You may not use recursion. * Your solution must not call any other functions. * Here are some examples (using "==" informally): * * * * * * * * true == allSame (new double[] { }) true == allSame(new double[] {11}) true == allSame (new double[] { 11, 11, 11, 11 }) false == allSame(new double[] { 11, 11, 11, 22 }) false == allSame (new double[] { 11, 11, 22, 11 }) true == allSame (new double[] { 22, 22, 22, 22 }) * */ public static boolean allSame (double[] list) { return StdRandom.bernoulli(); //TODO: fix thisSelect which of the following statements about searching for items in a list is incorrect. Group of answer choices 1. In an unsorted list, if the item is not present in the list, we will need to check every single item. In a sorted list, if we start the search at the smallest item, at most we would only need to check all items smaller than the target. If we get to an item larger than the target we can end our search early. This would make searching faster in the sorted list. 2. Sorting a list only makes sense for primitive types (ints, doubles) and Strings but not objects. For this reason we cannot sort a list of objects, and will always have to check each object in a list when we search for a target object. 3. In an unsorted list you need to check each item, to see if it is the item you want. If the list is sorted you can look at the middle item and then you can direct your search to the relevant portion of the list (higher or lower). This is called binary search. This will…
- Prob 5 Given a list of x objects, create a program that performs the following operation: -Get the first and last object and place them both at the end of the list. Example: QWERT becomes WERQT Find the worst case time complexity if the solution will be implemented using a dequeue. Give both the EQUATION AND THE BIG-O NOTATION.Complete the following recursive function that returns the sum of all the numbers in a list that are positive and even numbers. Replace the bold text with the correct expression. #lang racket (define (positiveEvenNums 1st) (if (null? 1st) (if (THE CAR OF THE LIST IS POSITIVE AND EVEN) (+ (car 1st) (positiveEvenNums (cdr 1st))) (positiveEvenNums (cdr 1st))))) (positiveEvenNums '(-3 8 4 3 -2 0 5))Correct answer will be upvoted else Multiple Downvoted. Don't submit random answer. Computer science. anglers have recently gotten back from a fishing excursion. The I-th angler has gotten a fish of weight man-made intelligence. Anglers will flaunt the fish they got to one another. To do as such, they initially pick a request where they show their fish (every angler shows his fish precisely once, in this way, officially, the request for showing fish is a stage of integers from 1 to n). Then, at that point, they show the fish they discovered by the picked request. At the point when an angler shows his fish, he may either become glad, become dismal, or stay content. Assume an angler shows a fish of weight x, and the most extreme load of a formerly shown fish is y (y=0 if that angler is quick to show his fish). Then, at that point: in the event that x≥2y, the angler becomes cheerful; in the event that 2x≤y, the angler becomes miserable; in the event that none of these two…
- 4272871/Downloads/AP%20Computer%20Science%20PPincipl Daily Tabs Calculator E Student Technology.. A Clas ales How to reduce (sim. 2 Equivalent Fraction.. les Practice Exam 1 for the Spring ... 68 / 68 88% 138. The procedure Numoccurrences is intended to count and return the number of times targetWord appears in the list wordList. The procedure does not work as intended. PROCEDURE Numoccurences wordList, targetWord count FOR EACH word IN wordList count IF (Word = targetWord count count + 1 RETURN COunt For which of the following code segments will the call to Numoccurrences NOT return the intended value? Select two answers. (A) treeList + "birch", "maple", "birch" numoccurences treeList, "birch" (B) treeList "birch" "maple", "oak" numoccurences treeList, "maple" (C) treeList + "birch", "maple", "ak" numoccurences treeList, "oak" (D) treeList - "birch", "maple", "oak" numoccurences treeList, "spruce"Question 41: Write an algorithm that compares each odd element with each even element of a list consisting of first m natural numbers, ie. I. 2. 3. -.. m-, m. and for each comparison, puts the larger element into a new list A . Assume m is an even number, thus there are equal number of odd and even elements. At the end, algorithm should display number of times even number is larger than the odd number. 1.write a pseudocodeand a python code. Execute with K=10 and present the output. 2. What is the time complexity of this algorithm (worst)? 3. Extend the algorithm such that it searches for an odd-even pair which is sum is 10 and terminates when first such pair is found. 4. What is the time complexity (worst case) of the resulting algorithm?CodeW For fun X C Solved https://codeworkou... 臺亂 CodeWorkout X272: Recursion Programming Exercise: Is Reverse For function isReverse, write the two missing base case conditions. Given two strings, this function returns true if the two strings are identical, but are in reverse order. Otherwise it returns false. For example, if the inputs are "tac" and "cat", then the function should return true. Examples: isReverse("tac", "cat") -> true Your Answer: 1 public boolean isReverse(String s1, String s2) { 2. if > 3. 4. else if > return true; return false; 5. 6. else { String s1first = String s2last return s1first.equals (s2last) && 51. substring(0, 1); s2, substring(s2.length() 1); 7. 8. 6. isReverse(s1.substring(1), s2.substring(0, s2.length() 1)); { 12} 1:11AM 50°F Clear 12/4/2021
- Sorting ascends if a list is empty or all items except the last are less than or equal to their successors. Define isSorted to return True if a list is sorted and False otherwise. (Hint: Loop through a list of length 2 or more and compare pairs of items left to right, returning False if the first item is greater.) Main function output: def main(): Print (isSorted(lyst)) lyst = [1] print(isSorted(lyst)) Lyst list(range(10)) print(isSorted(lyst)) lyst[9]=3 print(isSorted(lyst)) %3D True FalseConsider the problem of sorting a list of 10 elements from smallest to largest. Which of the following statements must be true of the state of the list after five iterations of the outer loop of selectionsort? (Select all that apply) Question 2 options: A) The list is sorted. B) The first five elements are sorted. C) The last five elements are the largest in the list. D) The last five elements are sorted.Weighted Intervals Problem EXAMPLE:Suppose we have the following weighted intervals. start finish weight 11 14 4 10 13 5 9 11 5 11 14 3 8 9 2 To solve the problem, we start out by sorting the intervals by finish time. This gives us the following. start finish weight 8 9 2 9 11 5 10 13 5 11 14 4 11 14 3 Next we calculate the p-values. P(j) = k where interval k is the last interval prior to interval j in the list below which does not conflict with interval j, or P(j) = 0 if all intervals prior to interval j conflict with it. # start finish weight P 1 8 9 2 0 2 9 11 5 1 3 10 13 5 1 4 11 14 3 2 5 11 14 4 2 Finally, we use the formula OPT(j) = max{OPT(j-1), weight(j)+ OPT(P(j))} to fill in the OPT value for each j. (The…