for ( i = 1 ; i < n ; i * 2 ){ for ( j = 0 ; j < min( i , k ) ; j++){ sum++ } } } what is the time complexity
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for ( i = 1 ; i < n ; i * 2 ){
for ( j = 0 ; j < min( i , k ) ; j++){
sum++
}
}
}
what is the time complexity
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- I Evaluate the time complexity of following code: int a-0; for (i-0; i< N; i++){ aa+i;) for (i 0; i< N; i+){ a-a+i;) 2sum= 0; for (int i = 0; i 1) { sum++; i= 1/2; } = 2*log2 (n) We denote by Ta(n), Tb (n), Te(n) the running time of the three fragments. 1. Give evaluations for Ta(n), Tb (n), Te(n). 2. Is T(n) = O(Ta(n)) ? Answer YES or NO and justify your answer. 3. Is Te(n) = (Ta(n)) ? Answer YES or NO and justify your answer.In questions 4-10 estimate the Big O value by analyzing the code. Note the algorithms are written in English. Hint: you are interested in the number of operations for each algorithm.
- For each of the following program fragments: Give an analysis of the running time (Big-Oh). Justify your answer? A. public class GFG { // Linearly search x in att[]. if x is present then //return the index, otherwise return -1 static int search (int arr[], int n, intx) { int i; for (i = 0: i%3B for ( int i = 0; i < n; ++i ) { for ( int j= 0; j< n; ++j) { sum += f( j, 10 ); } } iii. int sum = 0; %3D Assume f(n, m) runs in O( n + m) time.java C++ Ackermann’s FunctionAckermann’s Function is a recursive mathematical algorithm that can be used to test how well a computer performs recursion. Write a function A(m, n) that solves Ackermann’s Function. Use the following logic in your function:If m = 0 then return n + 1If n = 0 then return A(m−1, 1) Otherwise, return A(m−1, A(m, n−1))Test your function in a driver program that displays the following values:A(0, 0) A(0, 1) A(1, 1) A(1, 2) A(1, 3) A(2, 2) A(3, 2) SAMPLE RUN #0: ./AckermannRF Hide Invisibles Highlight: Show Highlighted Only The·value·of·A(0,·0)=·1↵ The·value·of·A(0,·1)=·2↵ The·value·of·A(1,·1)=·3↵ The·value·of·A(1,·2)=·4↵ The·value·of·A(1,·3)=·5↵ The·value·of·A(2,·2)=·7↵ The·value·of·A(3,·2)=·29↵Explain the functionality of below recursive functions. static void fun1(int n) { int i = 0; if (n > 1) fun1(n - 1); for (i = 0; i<n; i++) System.out.print(" *"); } 2. int LIMIT = 1000; void fun2(int n) { if (n<=0)return; if (n>LIMIT)return; System.out.print( String.format("%d", n)); fun2(2*n); System.out.print( String.format("%d", n)) }sum = 0; for (k=1; k<=n; k*=2) for (j=1; j<k; j++) sun++ What is the time complexity of the above code snippet? Is it O(2n) or O(nlogn)For funX |C Solved xb Answer x+ CodeW X https://codeworko... 田) CodeWorkout X267: Recursion Programming Exercise: Cumulative Sum For function sumtok, write the missing recursive call. This function returns the sum of the values from1 to k. Examples: sumtok(5) -> 15 Your Answer: 1 public int sumtok(int k) { 2. } (0 => ) return 0; 3. } else { return > 6. { Check my answer! Reset Next exercise 1:09 AMQuestion 4: The function T(n) is recursively defined as follows: 1 if n = 1, 고 +. T(n -1) if n> 2. Prove that T(n) = 0(n log n).Find the time complexity of the following code int x = 0, i = n; while (i> 0) { x += i; i= 2; }Fun(n) { if (n ≤ 1){return;} else { Statement; Fun (√) return; } Assume "Statement" takes (1) time, Further assume that ʼn = 2m, then the time complexity of Fun() is:SEE MORE QUESTIONS
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