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NNADVOKAT [17]
3 years ago
7

If a binary search is applied to an array with 1024 elements, in the worst case, the main loop executes, approximately, _____.

Computers and Technology
1 answer:
trasher [3.6K]3 years ago
7 0

Answer:

10.

Explanation:

Binary search divides the array to be search each in half according to the value of the element.

The worst case time complexity of binary search is O(logN).

In this case the time complexity will come out to be log₂(1024)=10.

So the binary search can divide this array in half maximum of 10 times.

Hence the main loop will executes 10 times.

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How to write a function that counts the letters in a string in C?
stiv31 [10]

Answer:

Here is the C function that counts the letters in a string:

int LetterCount(char string[]){  //function to count letters in a string passed as parameter

 string[100];  // char type string with size 100

  int i, letters;  // letter variable stores the count for no. of letters in string

   i = letters = 0;  //initialize variables i and letters to 0

  while (string[i] != '\0')  { // loop iterates through the entire string until end of string is reached

    if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  { // if condition checks the letters in the string

     letters++;    }  // increments 1 to the count of letters variable each time a letter is found in the string

    i++;  }  //increments value of i to move one character forward in string

   printf("Number of Letters in this String = %d", letters);   // displays the number of letters in the string

   return 0; }                              

Explanation:

Here the question means that the function should count the letters in a string. So the letters are the alphabets from a to z and A to Z.

Here is the complete program:

#include <stdio.h>   // to use input output functions

int LetterCount(char string[]){  // method that takes a character string as parameter and counts the letters in the string

string[100];  

int i, letters;

i = letters = 0;

while (string[i] != '\0')   {

 if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z'))

   {letters++;  }

   i++; }

   printf("Number of Alphabets in this String = %d", letters);  

   return 0;}  

int main(){  // start of main function

  char string[100];  //declares a char array of string

   printf("Please Enter a String : ");   //prompts user to enter a string

  fgets(string,100,stdin);  //get the input string from user

   LetterCount(string); } // calls method to count letters in input string

I will explain this function with an example

Lets suppose the user input "abc3" string

string[100] = "abc3"

Now the function has a while loop that while (string[i] != '\0')  that checks if string character at i-th position is not '\0' which represents the end of the character string. As value of i = 0 so this means i is positioned at the first character of array i.e. 'a'

At first iteration:

i = 0

letter = 0

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )   if condition checks if the character at i-th position of string is a letter. As the character at 0-th position of string is 'a' which is a letter so this condition evaluates to true. So the statement letter++ inside if condition executes which increments the letter variable to 1. So the value of letter becomes 1. Next statement i++ increments the value of i to 1. So i becomes 1. Hence:

i = 1

letter = 1

At second iteration:

i = 1

letter = 1

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  

if condition checks if the character at i-th position of string is a letter. As the character at 1st position of string is 'b' which is a letter so this condition evaluates to true. So the statements letter++ inside if condition and i++ executes which increments these variables to 1. Hence:

i = 2

letter = 2

At third iteration:

i = 2

letter = 2

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  

if condition checks if the character at i-th position of string is a letter. As the character at 2nd position of string is 'c' which is a letter so this condition evaluates to true. So the statements letter++ inside if condition and i++ executes which increments these variables to 1. Hence:

i = 3

letter = 3

At fourth iteration:

i = 3

letter = 3

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )  

if condition checks if the character at i-th position of string is a letter. As the character at 3rd position of string is '3' which is not a letter but a digit so this condition evaluates to false. So the statement letter++ inside if condition does not execute. Now i++ executes which increments this variable to 1. Hence:

i = 4

letter = 3

Now the loop breaks because while (string[i] != '\0') condition valuates to false as it reaches the end of the string.

So the statement: printf("\n Number of Letters in this String = %d", letters); executes which prints the value of letters on the output screen. Hence the output is:

Number of Letters in this String = 3

3 0
3 years ago
Libby wrote an email to her friend. She pressed Shift and the number key 2 together to enter the email address. Which symbol did
exis [7]

When Libby wrote the email to her friend, she typed the '@' symbol. When pressing SHIFT and 2 together, it pastes this aforementioned symbol. However, there isn't any specific name for the symbol. As a matter as fact, there are several names that this symbol goes by.

The most famous name this symbol is called is the "at symbol" or the "at sign". In terms of a formal name, "commercial at" would be a good one.

Here's an example with the symbol:

[email protected]

This is essentially telling the email server where to send your email. From this, they'll know it's located at brainly.com! It's pretty neat.

4 0
3 years ago
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Jim wants to develop an app for a specific purpose that would run even when his computer is not connected to the Internet. What
Vadim26 [7]

Answer:

NOT mobile app. NOT Cloud-based app.

Explanation:

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You can access sites else by ip address.
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