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34kurt
3 years ago
14

When can you access the BIOS setup utility?

Computers and Technology
1 answer:
torisob [31]3 years ago
8 0

Turn off the computer and wait five seconds.

Turn on the computer, and then immediately press the Esc key repeatedly until the Startup Menu opens.

Press F10 to open the BIOS Setup Utility.

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Write measure five safety measures can be taken to stay safe in social media​
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Read 2 more answers
Write a java program to read elements in an array from user and count total number of duplicate elements in array.
Anna11 [10]

Answer:

Explanation:

The following code is written in Java and is a function/method that takes in an int array as a parameter. The type of array can be changed. The function then creates a counter and loops through each element in the array comparing each one, whenever one element is found to be a duplicate it increases the counter by 1 and moves on to the next element in the array. Finally, it prints out the number of duplicates.

public static int countDuplicate (int[] arr) {

               int count = 0;

               for(int i = 0; i < arr.length; i++) {

                       for(int j = i + 1; j < arr.length; j++) {

                               if(arr[i] == arr[j])

                                       count++;

                       }

               }

               return count;

       }

8 0
3 years ago
You are asked to simulate a binary search algorithm on an array of random values.An array is the list of similar type of element
Alex Ar [27]

Answer:

Explanation:

Problem statement:

to simulate a binary search algorithm on an array of random values.

Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.

Input/output description

Input:

Size of array: 4

Enter array:10  20 30 40

Enter element to be searched:40

The Output will look like this:

Element is present at index 3

Algorithm and Flowchart:

We basically ignore half of the elements just after one comparison.

Compare x with the middle element.

If x matches with middle element, we return the mid index.

Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.

Else (x is smaller) recur for the left half.

The Flowchart can be seen in the first attached image below:

Program listing:

// C++ program to implement recursive Binary Search

#include <bits/stdc++.h>

using namespace std;

// A recursive binary search function. It returns

// location of x in given array arr[l..r] is present,

// otherwise -1

int binarySearch(int arr[], int l, int r, int x)

{

   if (r >= l) {

       int mid = l + (r - l) / 2;

       // If the element is present at the middle

       // itself

       if (arr[mid] == x)

           return mid;

       // If element is smaller than mid, then

       // it can only be present in left subarray

       if (arr[mid] > x)

           return binarySearch(arr, l, mid - 1, x);

       // Else the element can only be present

       // in right subarray

       return binarySearch(arr, mid + 1, r, x);

   }

   // We reach here when element is not

   // present in array

   return -1;

}

int main(void)

{ int n,x;

cout<<"Size of array:\n";

cin >> n;

int arr[n];

cout<<"Enter array:\n";

for (int i = 0; i < n; ++i)

{ cin >> arr[i]; }

cout<<"Enter element to be searched:\n";

cin>>x;

int result = binarySearch(arr, 0, n - 1, x);

   (result == -1) ? cout << "Element is not present in array"

                  : cout << "Element is present at index " << result;

   return 0;

}

The Sample test run of the program can be seen in the second attached image below.

Time(sec) :

0

Memory(MB) :

3.3752604177856

The Output:

Size of array:4

Enter array:10  20 30 40

Enter element to be searched:40

Element is present at index 3

Conclusions:

Time Complexity:

The time complexity of Binary Search can be written as

T(n) = T(n/2) + c  

The above recurrence can be solved either using Recurrence T ree method or Master method. It falls in case II of Master Method and solution of the recurrence is Theta(Logn).

Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.

8 0
3 years ago
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