Which XXX and YYY correctly complete the code to find the maximum score? Choices are in the form XXX/YYY. int[] scores = {43, 24, 58, 92, 68, 72); int maxScore; maxScore- scores[e]; for (XXX) { if (num > maxScore) { YYY; int scores: num/ maxScore = num scores != 0/num = maxScore int num: scores / maxScore = num num < scores/ num - maxScore
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- function result = result = ''; 1 tokens_to_str_code(token_mat, time_unit) 3 for i =1:size(token_mat,1) 4 5 if(token_mat (i,1) if(token_mat(i,2) > 4*time_unit) result = strcat(result,"-"); elseif((token_mat(i,2) > time_unit) && (token_mat(i,2) 8*time_unit) result = strcat(result,"/"); elseif((token_mat(i,2) 11 == 0) 12 13 14 4*time_unit) && (token_mat(i,2) < 8*time_unit)) 15 result strcat(result," "); %3D 16 end 17 end 18 end 19 end 20 21 Check Test Expected Got tokens %3D [1 4; е 3; 1 3; е 1; 1 3; е 1; 1 1; ө 1; 1 1 ]; time_unit = .5; disp( tokens_to_str_code( tokens, time_unit )) tokens = [ 0 4; 1 1457; 0 463; 1 1457; 0 463; 1 497; 0 1423; 1 1457; ---/..--.. --.---/..--.. о 463; 1 1457; 0 463; 1 1457; ө 3343; 1 497; ө 463; 1 497; 0 463; 1 1457; 0 1423; 1 1457; 0 463; 1 497; 0 463; 1 497; e 379 ]; time_unit = 240; disp( tokens_to_str_code( tokens, time_unit ))a) FindMinIterative public int FindMin(int[] arr) { int x = arr[0]; for(int i = 1; i < arr.Length; i++) { if(arr[i]< x) x = arr[i]; } return x; } b) FindMinRecursive public int FindMin(int[] arr, int length) { if(length == 1) return arr[0]; return Math.Min(arr[length - 1], Find(arr, length - 1)); } What is the Big-O for this functions. Could you explain the recurisive more in details ?2. code locations Consider the following C code and the corresponding assembly code: void baz(int a, int * p) { pushq %rbp movq %rsp,%rbp subq $16,%rsp int c; C = a-*p; movl %ecx,16(%rbp) movq %rdx, 24(%rbp) movq 24(%rbp), %rax movl (%rax),%eax movl 16(%rbp),%edx subl %eax,%edx movl %edx,%eax movl %eax,-4(%rbp) cmpl $0, -4(%rbp) je L2 if (c) { int d = c*2; movl -4(%rbp),%eax addl %eax,%eax movl %eax,-8(%rbp) *p -= d; movl 24(%rbp),%rax movl (%rax),%eax subl -8(%rbp),%eax movl %eax,%edx movq 24(%rbp), %rax movl %edx, (%rax) jmp L4: (continued on next page)
- void baz(int a, int * p) { pushq %rbp movq %rsp,%rbp subq $16,%rsp int c; c = a-*p; movl %ecx,16(%rbp) movq %rdx, 24(%rbp) movq 24(%rbp), %rax movl (%rax),%eax movl 16(%rbp),%edx subl %eax,%edx movl %edx,%eax movl %eax,-4(%rbp) cmpl $0, -4(%rbp) je L2 if (c) { int d = c*2; movl -4(%rbp),%eax addl %eax,%eax movl %eax,-8(%rbp) *p -= d; movl 24(%rbp),%rax movl (%rax),%eax subl -8(%rbp),%eax movl %eax,%edx movq 24(%rbp), %rax movl %edx, (%rax) jmp L4: } (continued on next page)3. Card Flipper: You walk into a room, and see a row of n cards. Each one has a number x; written on it, where i ranges from 1 to n. However, initially all the cards are face down. Your goal is to find a local minimum: that is, a card i whose number is less than or equal to those of its neighbors, xj-1 = X; <= Xj+1. The first and last cards can also be local minima, and they only have one neighbor to compare to. There can be many local minima, but you are only responsible for finding one of them. Obviously you can solve this problem by turning over all n cards, and scanning through them. However, show that you can find such a minimum by turning over only O(log n) cards.Given the following pseudocode: function fun2(n) { var outer_count = 0; var inner_count %3D for (i=0; i 0, as an expression of normal arithmetic and n, what is the value of outer_count that counts the number of times the outer loop executes? Enter an expression Fun.2.inner: Assuming n E N and n > 0, as an expression of normal arithmetic and n, what is the value of inner_count that counts the number of times the inner loop executes? Enter an expression
- Q3: Interplanetary Spaceflight Milan Tusk is the richest person in the universe. After devoting decades of his life to further our space exploration technologies, he’s finally ready to retire. Being a space enthusiast, the first thing he wants to do is visit n planets p1, p2, …, pn, in this order. He’s currently on planet p0. Milan knows that the distance between planets pi and pi + 1 (for 0 ≤ i < n) is d[i]light years. His spaceship uses 1 tonne of fossil fuels per light year. He starts with a full tank and can fill up his tank at any of the n planets (but he must not run out in between two planets). There’s a huge cost to set up the spaceship for refuelling. Due to financial constraints (he’s not THAT rich), he can fill up his tank at most ktimes. In order to save money and make his spaceship lighter, Milan is looking for the smallest possible fuel tank that enables him to complete his space travel and reach planet pn. What is the smallest tank capacity that enables him to do so?…If you have the following statements: int i j; int ctr = 0; int myArray[2][3]; for (i=0; i<3; i++) for (j=0; j<2; j++) { myArray[ j ][ i] = ctr; ++ctr; } The value of the element myArray[1][2] after executing the above statements:CCC '13 J1 - Next in line Canadian Computing Competition: 2013 Stage 1, Junior #1 You know a family with three children. Their ages form an arithmetic sequence: the difference in ages between the middle child and youngest child is the same as the difference in ages between the oldest child and the middle child. For example, their ages could be 5, 10 and 15, since both adjacent pairs have a difference of 5 years. Given the ages of the youngest and middle children, what is the age of the oldest child? Input Specification The input consists of two integers, each on a separate line. The first line is the age Y of the youngest child (0Language: C Pascal’s triangle is a triangular array, useful for calculating the binomial coefficients, n k , that are used in expanding binomials raised to powers, combinatorics and probability theory. 0 0 1 0 1 1 2 0 2 1 2 2 3 0 3 1 3 2 3 3 4 0 4 1 4 2 4 3 4 4 Evaluating the values of the binomial coefficients, you get the following pattern, 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 The number of the entries in each row is increased by one, as we move down. Each number in the triangle, is constructed by adding the number above it and to the left, with the number above it and to the right. The blank entries as treated as 0. Using the recursion, implement the function that computes the Pascal’s triangle. PrC++ A robot is initially located at position (0; 0) in a grid [?5; 5] [?5; 5]. The robot can move randomly in any of the directions: up, down, left, right. The robot can only move one step at a time. For each move, print the direction of the move and the current position of the robot. If the robot makes a circle, which means it moves back to the original place, print "Back to the origin!" to the console and stop the program. If it reaches the boundary of the grid, print \Hit the boundary!" to the console and stop the program. A successful run of your code may look like:Down (0,-1)Down (0,-2)Up (0,-1)Left (-1,-1)Left (-2,-1)Up (-2,0)Left (-3,0)Left (-4,0)Left (-5,0)Hit the boundary! or Left (-1,0)Down (-1,-1)Right (0,-1)Up (0,0)Back to the origin! About: This program is to give you practice using the control ow, the random number generator, and output formatting. You may use <iomanip> to format your output. You may NOT use #include "stdafx.h".Q3: Superheroes Supervillains are tired of Toronto condo rental prices, so they are leaving Toronto for Mississauga. Luckily, we have valiant superheroes that can deal with them. The superhero () has a name , an intelligence score , and a strength score .ni-th0 = gin[x] + s[x] Detective Zingaro has asked for your help. For each of the supervillains, tell him the name of the superhero that should deal with that supervillain. Note: A superhero can be assigned to multiple supervillains (or none at all). Note: whenever there are multiple superheroes that satisfy the given requirements for a supervillain, report the one whose name is lexicographically smallest (i.e. the one that’s the smallest according to Python’s ordering of strings). It’s guaranteed that superheroes have distinct names. Hint: Tuples of multiple elements may be helpful here. In python, you can compare two tuples and . If and are different, the result is the same as comparing and . If and are equal, the result is the same…SEE MORE QUESTIONS