fs(n) = 1 + fs([n/2]) when n > 1, f;(1) = 1 %3D
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Find its time complexity using Big-O Notation.
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- helllpFigure 2 void sort_Algo (int anArray[], int n) { int i, j, idx; for (i = 0; i anArray[ idx ]) idx = j; Swap (anArray[ idx ], anArray[ i ]); } } int main() { const int n = 8; int anArray[n] { -9, 14, 52, 106, 22, -1, 99}; sort_Algo(anArray, n); } Name the sorting algorithm exercised by the sort_Algo() function. * Your answer Determine if anArray[] is to be sorted into ascending, or descending order. Your answer Trace the sorting algorithm in Figure 2 as it sorts anArray[]. (Answer the question according to the format, eg. Pass 0:3, 3, 3; Pass 1: 3, 3, 3; Pass2: 3, 3, 3;) * Your answer8
- Code: #include <bits/stdc++.h> using namespace std; void BUBBLE(int A[],int N){ for(int k=0;k<N-1;++k){ for(int ptr=0;ptr<N-k-1;++ptr){ if(A[ptr]>A[ptr+1]){ int temp = A[ptr]; A[ptr] = A[ptr+1]; A[ptr+1]=temp; } } }} //function to print the arrayvoid printArray(int arr[],int n){ int i; for(i=0;i<n;i++) cout<<arr[i]<<" "; cout<<endl;} //driver function to test the modulesint main(){ int arr[] ={15,17,5,3,25,66,14,7,59,100}; int n=sizeof(arr)/sizeof(arr[0]); cout<<"\nOriginal array: "; printArray(arr,n); cout<<"\n\nOutput of Bubble sort are shown below:\n"; BUBBLE(arr,n); printArray(arr, n); return 0;} Q: Remove the Function from the above codef= AB+C %D BH [1] [2]11 - The code segment below has time complexity? for (int i=0; i#include<bits/stdc++.h> using namespace std; void bubbleSort(int arr[], int n) { for (int i = 0; i < n - 1; i++) { for (int j = 0; j < n - i - 1; j++) { if (arr[j] > arr[j + 1]) { int temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; } } } } int binarySearch(int arr[], int l, int r, int x, int& comp) { comp++; if (r >= l) { int mid = l + (r - l) / 2; if (arr[mid] == x) { return mid; } if (arr[mid] > x) { return binarySearch(arr, l, mid - 1, x, comp); } return binarySearch(arr, mid + 1, r, x, comp); } return -1; } int main() { int Num[8192]; srand(time(NULL)); for (int i = 0; i < 8192; i++) { Num[i] = rand() % 10001; } clock_t starting_time = clock(); bubbleSort(Num, 8192); clock_t ending_time = clock(); clock_t result =…Solution Floating point representation: It is defined as the representation of floating numbers. It includes sign bit, exponent, and mantissa bits. Based on precision it has 2 types. 1. For IEEE 754 single-precision floating-point numbers, what is the exponent of a denormalized floating-point number in decimal? Solution: In IEEE 754 single-precision, exponent bits are 8. Therefore exponent = 2 ^(n-1) -1 = 2^(8-1) -1 = 127 OPTION D 2. For IEEE 754 single-precision floating-point numbers, how many bits for mantissa? Solution: In IEEE 754 single-precision, there are 23 bits for mantissa. sign = 1 bit exponent = 8 bits mantissa = 23 bits 3. For IEEE 754 single-precision floating-point numbers, which of the following is an example of NAN? Solution: In IEEE 754 single-precision, NAN is a special value where all exponents bits are 1's and the mantissa is non zero. a. 1 111 1 111 0000 0000 1101 0000 0000 0000 Exponent is not all 1's. Not a NAN b. 0 111 1 111 1000 0000…CFG: Example 1 • Draw the CFG for the following code: int f(int n){ } int m = n* n; if (n < 0) else return 0; return m;SEE MORE QUESTIONS
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