The process of converting information from a form/questionnaire is referred to as data preparation. This process follows a four-
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Answer:
Here is the C++ program:
#include<iostream> //to use input output functions
#include <algorithm> //to use tolower() function
using namespace std; //to identify objects like cin cout
int main() { //start of main function
string text; // to hold text input
cout<<"Enter a line of text: "; // prompts user to enter a string
getline(cin,text); // reads the input string (text) from user
transform(text.begin(), text.end(), text.begin(), ::tolower); //converts the text into lower case
int letter[26] = {0}; //to hold the letters
int i; // used as a loop variable
int words=0; // to hold the count of words
for(i = 0; i< text.size();i++){ // iterates through the text
if(isalpha(text[i])&&(text[i+1]=='.'||text[i+1]==','||text[i+1]==' ')) //checks if the character at i-th index of text is a letter, and checks if i+1 index position of text is a period, comma or a space
words++; // adds 1 to the count of words
if(isalpha(text[i])) // if the character at the i-th index of text is a letter
letter[text[i]-'a']++; } // counts the occurrences of each letter
char j = text[text.size()-1]; // sets j to the last character of text
if(j != '.' && j!= ' '&& j!=',' &&j!= '\0') //if the last character is not a period or empty space or comma or end of lone
words++; //add 1 to the count of words
cout<<"Number of words: "<<words<<endl; //display the number of words in the text
for(i=0; i<26; i++) { //iterates 25 times
if(letter[i]>0) //if letter at index i is greater than 0
cout<<(char)('a'+i)<<" : "<<letter[i]<<endl; }} //displays each letters and its number of occurrences in the text
Explanation:
The program is explained in the attached document with an example.
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.
