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earnstyle [38]
3 years ago
9

Should the federal government have bug bounty programs? Why or why not?

Computers and Technology
2 answers:
ratelena [41]3 years ago
5 0
Yes they should
Hope this hkoed
gogolik [260]3 years ago
4 0

I think this question is supposed to be an opinion question but Im not sure

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Write a program that reads whitespace delimited strings (words) and an integer (freq). Then, the program outputs the strings fro
Lesechka [4]

Answer:

See explaination for the code

Explanation:

def wordsOfFrequency(words, freq):

d = {}

res = []

for i in range(len(words)):

if(words[i].lower() in d):

d[words[i].lower()] = d[words[i].lower()] + 1

else:

d[words[i].lower()] = 1

for word in words:

if d[word.lower()]==freq:

res.append(word)

return res

Note:

First a dictionary is created to keep the count of the lowercase form of each word.

Then, using another for loop, each word count is matched with the freq, if it matches, the word is appended to the result list res.

Finally the res list is appended.

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3 years ago
3. How are you able to create photographs differently than 100 years ago?
Agata [3.3K]

Answer:

it willbe black and white

Explanation:

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Which type of work is an electrical engineer most likely to do?
yKpoI14uk [10]

Answer:

B

Explanation:

design and analyze electrical devices

7 0
3 years ago
How many generations of computer languages have there been since the middle of the 20th century
frozen [14]
Four computer languages i think
5 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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