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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;) 2Part a. Make a diagram of memory for point one when n == 2. int expo(int x, int n) { int r, t; if (n == 0) r = 1; else if (n == 1) r = x; else { t = expo (x, n / 2); r = t * t; if (n % 2 == 1) r *= t: } //-- -point one return r; int main(void) { int i; i = expo(4, 5); return 0; Part b. Using divide-and-conquer recursion to solve a problem similar to Prob- lem 3 is somewaiät tricky but a good test of programming skill. Write a function definition that provides a divide-and-conquer implementation of the following interface: double max_adj_sum(const double *a, int lo, int hi); // REQUIRES: hi - lo >= 2. // PROMISES: Return value is the largest sum that can be made by adding // Elements a[lo] ... a[hi -1] exist. two elements among a[lo] ... a[hi-1] with adjacent inderes. Hint: It makes sense to have two base cases, not just one.Question 4. Find the time complexity of the following code. 1. for (i = n; i >= 1; i/= 2): { } LA, 6B for j = m to i { ans += (i * j) }
- Q5. What is the time complexity of following code? int a for (i = 0; ii; j--) { a = a + i + j; } 0; } Q6. What is the time complexity of following code? int i, j, k = 0; for (i = n/2; i <= n; i++) { for (j = 2; j <= n; j = j * 2) { k=k+n/2; } }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.Q5. Find the time complexity of the following code int b = 0; for (j = 0; jj; i--) { b=b+j+i;
- void deleteRange( int from, int to) { int i, j = 0; for (i = 0; i < counter; i++) { if (i <= from - 1 || i >= to + 1) { A[j] = A[i]; j++; } } for (int i = 0; i < j; i++) cout << A[i] << " "; } Above method deletes range of elements from an array. Please explain the logic of above code in simple english (algorithm and comments).void deleteRange( int from, int to) {int i, j = 0;for (i = 0; i < counter; i++) {if (i <= from - 1 || i >= to + 1) {A[j] = A[i];j++;}}for (int i = 0; i < j; i++)cout << A[i] << " ";} Above method deletes range of elements from an array. Consider array A[] globally declared and counter is its size. Please explain the logic of above code in simple english (algorithm and comments).Draw the activation frame diagram for the following recursive function “factorial()”Code:#include<stdio.h>int factorial(int n); int main(void){int s; s=factorial(5); printf("%d",s); return 0;}/*Compute n! using a recursive definitionPre: n >= 0*/ intfactorial(int n){int ans;if (n == 0) ans = 1; elseans = n * factorial(n - 1);return (ans);}
- sum= 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.Consider the following function: public void bSearch(int[] A, int value, int start, int end) { if (end value) { return bSearch(A, value, start, mid); } else if (value > A[mid]) { return bSearch(A, value, mid+1, end); } else { return mid; } } The mutation program P' changes the part as shown below. } else if (value > A[mid]) { return bSearch(A, value, mid+2, end); } else { return mid; } A test input is given t1: {A = [2, 3, 5, 7, 10, 11], value = 6, start=0, end=5}. Which statement below best describes the mutant with regards to the test input? O a. t1 resolves mutant P O b. The mutant P' with regards to test input t1 is live.inhinclude Rsing namespace std; int maxResult( ) int maxVal for (int i { for (int j = 0; j <= n 0; i <= n; i += a) i; j += b) %3D %3D float z = (float)(n (i + j)) / (float)(c); if (floor(z) { int x = int y ceil(z)) i / a; j/ b; maxVal = max(maxVal, x + y (int)z); return maxVal; } int main() { max cout << maxResult( ); return 0; 1 } Input Compilation failed due to fellowing ons main.cpp:7:23: error: 'n' was not declared in this scope 7| for (int i = 0; i <= n; i a) main.cpp:7:31: error: 'a was not declared in this scope for (int i = 0; i <= n; i t= a) %3D main.cpp:9:36: error: b' was not declared in this scope | 6 for (int j = 0; j <= n - i; j - b) %3D main.cpp:11:45: error: 'c was not declared in this scope