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lord [1]
3 years ago
5

Which function works best when you need to remove an element at a specific index in a list?

Computers and Technology
2 answers:
Finger [1]3 years ago
4 0

Answer:

pop() function.

Explanation:

The pop() function is used in the format list.pop(integer), where the integer represents the index of the element of a list that is to be removed. If nothing is placed within the parentheses of pop(), the function will remove the last item of the list as default.

*This is for Python.

Hope this helps :)

AfilCa [17]3 years ago
3 0

Answer:

the poop() function

Explanation:

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5.23 LAB: Contains the character
torisob [31]

Answer:

In C++:

#include<iostream>

#include<vector>

using namespace std;

int main() {

int len;

cout<<"Length: ";  cin>>len;

string inpt;

vector<string> vect;

for(int i =0;i<len;i++){

  cin>>inpt;

  vect.push_back(inpt); }

char ch;

cout<<"Input char: ";  cin>>ch;  

for(int i =0;i<len;i++){

  size_t found = vect.at(i).find(ch);  

      if (found != string::npos){

          cout<<vect.at(i)<<" ";

          i++;

      }

}  

return 0;

}

Explanation:

This declares the length of vector as integer

int len;

This prompts the user for length

cout<<"Length: ";  cin>>len;

This declares input as string

string inpt;

This declares string vector

vector<string> vect;

The following iteration gets input into the vector

for(int i =0;i<len;i++){

  cin>>inpt;

  vect.push_back(inpt); }

This declares ch as character

char ch;

This prompts the user for character

cout<<"Input char: ";  cin>>ch;  

The following iterates through the vector

for(int i =0;i<len;i++){

This checks if vector element contains the character

  size_t found = vect.at(i).find(ch);  

If found:

      if (found != string::npos){

Print out the vector element

          cout<<vect.at(i)<<" ";

And move to the next vector element

          i++;

      }

}  

7 0
2 years ago
Requests to access specific cylinders on a disk drive arrive in this order: 24, 20, 4, 40, 6, 38, and 12, and the seek arm is in
GarryVolchara [31]

Answer:

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The time required to seek using Shortest Seek Time First algorithm is 360 milliseconds.

The time required to seek using LOOK (initialing moving upwards) algorithm is 348 milliseconds.

Explanation:

Part a,b : First-come, first-served:

The order of the cylinders is as 10, 22, 20, 2, 40, 6, 38

10 + 12 + 2 + 18 + 38 + 34 + 32 = 146 cylinders = 876 milliseconds.

ii) Shortest Seek Time First:

The order of the cylinders is as 20, 22, 10, 6, 2, 39, 40

0 + 2 + 12 + 4 + 4 + 36 + 2 = 60 cylinders = 360 milliseconds.

iii) LOOK (initialing moving upwards):

The order of the cylinders is as 20, 22, 38, 40, 10, 6, 2

0 + 2 + 16 + 2 + 30 + 4 + 4 = 58 cylinders = 348 milliseconds.

7 0
4 years ago
Consider the following code segment. int[][] arr = {{3, 2, 1}, {4, 3, 5}}; for (int row = 0; row &lt; arr.length; row++) { for (
ludmilkaskok [199]

Answer:

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Condition two - 2 times

Explanation:

Given

The above code segment

Required

Determine the number of times each print statement is executed

For condition one:

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<em />

For the given data array, this condition is true only once, when

row = 1  and  col = 2

i.e.

<em>if(arr[row][col] >= arr[row][col - 1]) </em>

=> <em>arr[1][2] >= arr[1][2 - 1]</em>

=> <em>arr[1][2] >= arr[1][1]</em>

<em>=> 5 >= 3 ---- True</em>

<em />

The statement is false for other elements of the array

Hence, Condition one is printed once

<em />

<em />

For condition two:

The if condition required to print the statement is:  <em>if (arr[row][col] % 2 == 0) </em>

<em />

The condition checks if the array element is divisible by 2.

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row = 0  and  col = 1

row = 1  and  col = 0

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<em>if (arr[row][col] % 2 == 0) </em>

<em>When </em>row = 0  and  col = 1

<em>=>arr[0][1] % 2 == 0</em>

<em>=>2 % 2 == 0 --- True</em>

<em />

<em>When </em>row = 1  and  col = 0

<em>=>arr[1][0] % 2 == 0</em>

<em>=> 4 % 2 == 0 --- True</em>

<em />

<em />

The statement is false for other elements of the array

Hence, Condition two is printed twice

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3 years ago
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yuradex [85]
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