Find the complexity of the below program: (A) function(int n) { if (n==1) return; for (int i=1; i<=n; i++) { for (int j=1; j<=n; j++) { printf("*"): break; }
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- The fibonacci sequence of numbers are as follows: 0,1,1,2,3,5,8,13,21,34,... where F0=0, F1=1, F2=1, F3=2 and so on. The recursive function definition is F0=0 F1=1 Fn=F(n-2)+F(n-1) (if n>1) Write a recursive function, fib, to implement this definition in MATLAB. The function will receive a single input (integer), n, and return one integer alue that is the nth Fibonacci number.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>> }}Compute for the time complexity (in milliseconds) of each letter of the given program segment below, then get the total time complexity. int main( ) { int a, b, c, d; for (a=0;a<10;a++) for (b=0;b<10;b++) for (c=0;c<10;c++) for (d=0;d<10;d++) { If ((a==0)&&(b==0)&&(c==0)&&(d==0)) { a=1, b=2,c=3,d=4; } cout<<a<<b<<c<<d; } }
- Information is present in the screenshot and below. Based on that need help in solving the code for this problem in python. The time complexity has to be as less as possible (nlogn or n at best, no n^2). Apply dynamic programming. Do not use recursion/memoization. Make sure ALL test cases return expected outputs. Sample Input 04 412340123 Sample Output 013715 Explanation 0You are given the sequence 1,2,3,4.F0 = A0 = 1F1 = A1 + F0 = 2 + 1 = 3F2 = A2 + F1 + F0 = 3 + 3 + 1 = 7F3 = A3 + F2 + F1 + F0 = 4 + 7 + 3 + 1 = 15 The actual code def solve(k,a): MOD = 1000000007 # compute and return answer here q, n = list(map(int,input().rstrip().split(" ")))a = [int(input().rstrip()) for i in range(n)]outs = []for i in range(q): k = int(input().rstrip()) outs.append(solve(k,a))print("{}".format("\n".join(list(map(str,outs)))))Determine the time complexity function of the program snippet below, the proof must be in detail step by step until you get the time complexity function. int f1(int n) {if (n <= 1) return n;return 2 * f1(n/2);}int f2(int n) {if (n <= 1) return n;return f2(n/2) + f2(n/2);}Information is present in the screenshot and below. Based on that need help in solving the code for this problem in python. The time complexity has to be as less as possible (nlogn or n at best, no n2) and runtime should be under 1 second. Apply dynamic programming. Do not use recursion. Make sure ALL test cases return expected outputs. Output FormatFor each query, output one line containing an integer Dn. The number must be output mod (109 + 7). Sample Input 0:501234 Sample Output 0:10129 Edge Case: (to test against runtime and timeout error)300000 (number of queries)300000300000300000300000...and so on until there are 300,000 of them The actual code: def solve(n): MOD = 1000000007 # compute and return answer here q = int(input().rstrip()) outs = [] for i in range(q): n = int(input().rstrip()) outs.append(solve(n)) print("{}".format("\n".join(list(map(str,outs)))))
- In C program We want to emulate that, too! Using a recursive function, make it so that we print the number we are currently in on a new line. When we encounter a number n divisible by 5, we print "YEHEY" and jump 7 steps forward. However when we encounter a number divisible by 8, we print "OHNO" and go back 6 steps. If it is divisible by both, we print "LUCKY" and jump 11 steps. Otherwise just move 1 step. Start at n=1 and continue doing this until you reach a number greater than or equal to the number entered by the user. Instructions: In the code editor, you are provided with a main() function that must ask the user for a number and calls the snake_ladder() function. This snake_ladder() function is a recursive function which should perform the functionality explained in the problem description above. This function is declared but is initially empty. Input 1. Stopping number Description Input the number to which the recursive function must stop Constraints The input is a…In C program We want to emulate that, too! Using a recursive function, make it so that we print the number we are currently in on a new line. When we encounter a number n divisible by 5, we print "YEHEY" and jump 7 steps forward. However when we encounter a number divisible by 8, we print "OHNO" and go back 6 steps. If it is divisible by both, we print "LUCKY" and jump 11 steps. Otherwise just move 1 step. Start at n=1 and continue doing this until you reach a number greater than or equal to the number entered by the user. Instructions: In the code editor, you are provided with a main() function that must ask the user for a number and calls the snake_ladder() function.This snake_ladder() function is a recursive function which should perform the functionality explained in the problem description above.This function is declared but is initially empty.Input 1. Stopping number Description Input the number to which the recursive function must stop Constraints The input is a positive integer…The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b b) def f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a c) def f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] d) def f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return a
- Write a recursive Fibonacci function that computes the Fibonacci number for input n and prints each value computed in the series at each recursive step in the sequence from 0 to n, all in one line on the console. (In python)Information is present in the screenshot and below. Based on that need help in solving the code for this problem in python. The time complexity has to be as less as possible (nlogn or n at best, no n^2). Apply dynamic programming. Do not use recursion. Make sure ALL test cases return expected outputs. Sample Input 0:5 20011014 Sample Output 0:1 1 Explanation 0:The best way to jump from square 1 to square 5 is by jumping 4 squares. That requires only 1 jump.In fact, that is the only solution, thus the second output is 1. Sample Input 1:6 3010100123 Sample Output 1:2 2 Explanation 1The best sequence is to jump 2 squares to square 3 then 3 squares to end in square 6.This is one solution. The other solution is to jump from 1 -> 3 -> 5 -> 6. These are the only two solutions, therefore the second output is 2. Sample Input 2:7 2000110037 Sample Output 2:-1 0 Explanation 2:It is impossible to reach square 7 with the specified jump values. The actual…C programing Draw the recursion tree to find out the value of f(5)int f(int n){int ans;int i;if(n<3)return n;ans = f(n/2);for(i=0; i<n; i++)ans += f(i);return ans;}