Q1. 7 and 8 subpart only Q1. Analyze the time complexity for following code segments/ functions.i.1. int count =0;2. for (int i = n; i> 0; i /= 2)3.ii.for (i=1; i<=n; i++)for (j=1; j<=log(i); j++)PRINT "HELLO JUET !";for (int j = 0; j< i; j++)count += 1;4.iii.iv.void fun(int n, int arr[])1. {int i = 0if (n==1)return;2.for (int i=1; i<=n; i++){3.for(; i a[J+1] then,Set temp = a[J]Set a[J] = a[J+1]Set a[J+1] = temp3.4.for (int k-1; k<=n; k = k * 2)5.count++;6.}vii.vii.void function(int n){int count = 0;for (int i-n/2; i<=n; i++)for (int j=1; j+n/2<=n; j = j++)1. For I = 0 to n-22.min=I3.For J = I+1 to (n-1)4.If a[min] > a[J]for (int k=1; k<=n; k = k * 2)count++;5.min=JSet temp = a[min]Set a[min] = a[I]Set a[I] = temp6.7.8.

Q1. Analyze the time complexity for following code segments/ functions. i. 1. int count = 0; 2. for (int i n; i> 0; i /= 2) 3. ii. for (i=1; i<=n; i++) for (j=1; j<=log(i); j++) PRINT "HELLO JUET !" ; for (int j= 0; j a[J+1] then, Set temp = a[J] Set a[J] = a[J+1] Set a[J+1] = temp 2. 3. 4. 5. for (int k=1; k<=n; k = k * 2) count++; 6. vii. viii. void function(int n) 1. For I = 0 to n-2 2. min=I int count = 0; for (int i-n/2; i<=n; i++) for (int j=1; j+n/2<=n; j = j++) for (int k=1; k<=n; k = k * 2) 3. For J = I+1 to (n-1) 4. If a[ min] > a[J] 5. min=J count++; 6. Set temp = a[min] Set a[min] = a[I] Set a[I] = temp 7. 8.

Q1. Analyze the time complexity for following code segments/ functions. i. 1. int count = 0; 2. for (int i n; i> 0; i /= 2) 3. ii. for (i=1; i<=n; i++) for (j=1; j<=log(i); j++) PRINT "HELLO JUET !" ; for (int j= 0; j a[J+1] then, Set temp = a[J] Set a[J] = a[J+1] Set a[J+1] = temp 2. 3. 4. 5. for (int k=1; k<=n; k = k * 2) count++; 6. vii. viii. void function(int n) 1. For I = 0 to n-2 2. min=I int count = 0; for (int i-n/2; i<=n; i++) for (int j=1; j+n/2<=n; j = j++) for (int k=1; k<=n; k = k * 2) 3. For J = I+1 to (n-1) 4. If a[ min] > a[J] 5. min=J count++; 6. Set temp = a[min] Set a[min] = a[I] Set a[I] = temp 7. 8.

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter15: Recursion

Section: Chapter Questions

Problem 8SA

See similar textbooks

Similar questions

Compare the complexities of the following two pieces of code using the cyclomatic complexity metrics. What conclusions can you draw about relative complexity of the code? [6 marks] Code A Code B int i = 1; int j = 1; while (i <= 5) { int i = 2; Playacard(i); if (j =i); if (playhaswon(i)) j++; else i++; i++; else System.out.println(j); i--; i++; }
For function log, write the missing base case condition and the recursive call. This function computes the log of n to the base b. As an example: log 8 to the base 2 equals 3 since 8 = 2*2*2. We can find this by dividing 8 by 2 until we reach 1, and we count the number of divisions we make. You should assume that n is exactly b to some integer power. Examples: log(2, 4) -> 2 and log(10, 100) -> 2 public int log(int b, int n ) { if <<Missing base case condition>> { return 0; } else { return <<Missing a Recursive case action>> }}
solve the problem in java:Consider the are n=2 subjects and needed =[4,5]answered questions, to pass.The student has answered =[2,4] questions in the two subjects so far, and can answer another q=1 questions across all subjects combined. The best outcome is to answer an additional question in the second subject on order to pass it, as 2 more answers required to pass the first subject. the max number of subject s that can be passed is 1. the function must return an integer that can represent the max number of subjects that can be passed. public static int maxNumsub(List <Integer> answered, List<Integer> needed, int q){//write code here}
Assume that you were given N cents (N is an integer) and you were asked to break up the N cents into coins consisting of 1 cent, 2 cents and 5 cents. Write a dynamicprogramming based recursive algorithm, which returns the smallest (optimal) number of coins needed to solve this problem. For example, if your algorithm is called A, and N = 13, then A(N) = A(13) returns 4, since 5+5+2+1 = 13 used the smallest (optimal) number of coins. In contrast, 5+5+1+1+1 is not an optimal answer.
Write a recursive function that, given a sequence of comparable values, returns the count of elements where the current element is less than the following ( next ) element in the given sequence. See the examples given below. def count_ordered ( seq ) : """ Input : A sequence of comparable elements Output : The number of elements that are less than the following element in the sequence Example : >>> count_ordered ( [ 1 , 2 , 3 , 4 , 5 , 6 ] ) 5 >>> count_ordered ( ( 1 , 12, 7.3 , -2,4 ) ) 2 >>> count_ordered ( 'Python' ) 2 >>> count_ordered ( [ 6 ] ) 0 >>> count_ordered ( [ ] ) 0 """ In the first example above , count_ordered ( [ 1,2,3,4,5,6 ] )the returned answer is 5 because for all the first 5 numbers the current number is less than the next number. In the second example above, count_ordered ( ( 1,12,7.3 , -2,4 ) )the…
in, java Assume a function g(x) is defined as follows where x is an int parameter:g(x) = g(x - 1) * g (x - 3) if x is even and x > 3 = g(x - 2) if x is odd and x > 3 = x otherwiseWrite a recursive method to compute g.
Create a recursive function with arguments N and K in python 3 that does the following specifications:Using numbers ranging from 0 to 15 (inclusive), create all possible lists which sum up to K and have a length of N. Duplicated numbers are allowed as long as it fulfills the conditions above (this means [0,0,1], [0,1,0] and [1,0,0] are all correct outputs if K=1 and N=3) The function returns the number of created lists which follow the specifications mentioned. For example, given K=23 and N=2, the expected output of the function is 8. Constraints:1) The time complexity must be O(n) or faster. Indicate the time complexity of created code as proof.(Redundant due to constraint above, but still it is important to note that the function must be able to finish in 9 seconds or less when N=10)2) You are not allowed to use external libraries or itertools.
Implement a program that contains the following two functions: 1. A function that accepts a list of numbers as an argument, the function should recursively calculate the sum of all the numbers in the list and returns that value2. A function that uses recursion to raise a number to a power. The function should accept two arguments: the number to be raised, and the exponent. Assume the exponent is a positive integer
(Using the programming language C, answer the following question) Write recursive C functions to do the following: Example: int summ(int n) returns the summation of integers from 0 to n. Here is the code. int summ(int n) { if(n < 0) return -1; if(n == 0) return 0; return n + summ(n - 1);} 1) If n is less than 0, the expression int bunnyEars(int n) returns -1; otherwise, it returns the number of ears for n bunnies. int factorial(int n) returns -1 if n is less than 0, otherwise returns the factorial of n. If n is greater than or equal to 0, the expression int fib(int n) returns -1; otherwise, it returns fib of n - 1 + fib of n - 2. (This will run quite slowly if you test it with more than roughly 50 digits.). int posPow(int base, int exp) returns -1 if exp is less than 0, otherwise returns base raised to the power exp. Write a main() that tests all the other functions thoroughly.
1: |R|←|P| Reserve space for |P| = 13 values.2: x ← n3: for i ← 0 ...(|P| − 1) do4: c ← x div Pi Number of multiplicands Pi in x.5: Ri ← c6: x ← x − c · Pi7: end for8: return RA Java programmer could implement Algorithm by first modelling the primitivenumbers with the enumeration type RomanNumeral. Each enum constant (I, IV, ..., M) is declared with its decimal value, which can be accessed with the function getValue().
Please explain Crime Wave more clearly for me please there's a picture of the decription and also please comment on each line if possible on the code below to for me to have a better understanding of what each line is doing. It is in C++. Please also give the space and time complexity and the reason why. #include <stdio.h>#include <string.h>#include <math.h>#define eps 1e-6double W[105][105];int N, M;int mx[105], my[105]; // match arrdouble lx[105], ly[105]; // label arrint x[105], y[105]; // used arrint hungary(int nd) { int i; x[nd] = 1; for(i = 1; i <= M; i++) { if(y[i] == 0 && fabs(W[nd][i]-lx[nd]-ly[i]) < eps) { y[i] = 1; if(my[i] == 0 || hungary(my[i])) { my[i] = nd; return 1; } } } return 0;}double KM() { int i, j, k; double d; memset(mx, 0, sizeof(mx)); memset(my, 0, sizeof(my)); memset(lx, 0, sizeof(lx)); memset(ly, 0, sizeof(ly));…
Consider the following recursive function: def rec1(aList,first,last): if first==last: return aList[first] else: return aList[last]*rec1(aList,first,last-1) Show the output of the following call (show your work and all recursive calls): print(rec1([1,2,3,4,5,6,7],3,5))? describe the task of the rec1 function.