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juin [17]
3 years ago
8

An array is sorted (in ascending order) if each element of the array is less than or equal to the next element .

Computers and Technology
1 answer:
Aleks04 [339]3 years ago
8 0

Answer:

// The program below checks if an array is sorted or not

// Program is written in C++ Programming Language.

// Comments are used for explanatory purpose

//Program Starts here

#include<iostream>

using namespace std;

//Function to check if the array is sorted starts here

int isSorted(int arr[], int count)

{

// Check if arrays has one or no elements

if (count == 1 || count == 0) {

return 1;

}

else

{

// Check first two elements

if(arr[0] >= arr[1])

{

return 0; // Not sorted

}

else

{ // Check other elements

int check = 0;

for(int I = 1; I<count; I++){

if (arr[I-1] > arr[I]) { // Equal number are allowed

check++; // Not sorted

}

}

}

if(check==0)

return isSorted(arr, count - 1); //Sorted

else

return 0; // Not sorted

}

// Main Method starts here

int main()

{

int count;

cin<<count;

int arr[count];

for(int I = 1; I<=count; I++)

{

cin>>arr[I-1];

}

int n = sizeof(arr) / sizeof(arr[0]);

if (isSorted(arr, n))

cout << "Array is sorted";

else

cout << "Array is not sorted";

}

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Answer:

The correct code for this question:

g=float(input("Enter your English test grade:")) #take input from user.

#check conditions

if (g>=100 and g<=90):

print ("A")

#g greater then equal to 100 and less then equal to 90.

if (g>=89 and g<=80):

print("B")

#g greater then equal to 89 and less then equal to 80.

if (g>=79 and g<=70):

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if (g>=69 and g<=65):

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if(g<=64):

print("F")

#g less then equal to 64.

else:

print ("Not a grade")

#not a grade or fail.

Explanation:

In this program, we use to take a value from the user and check the value from the various conditions. To check all the condition we use if-else statement and AND operator that check to the range to together.

If -else is a conditional operator. In that, If block is used to check the true part and else part takes false value, and AND is a logical operator that check the two range together  

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<u>Explanation</u>:

<u>Here's what the question entails in clearer detail;</u>

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The ArrayList class contains a trim method that resizes the internal array to exactly the capacity. The trim method is intended
andreev551 [17]

Answer:

public void trimToSize() {

modCount++;

if (size < elementData.length) {

elementData = (size == 0)

? EMPTY_ELEMENTDATA

: Arrays.copyOf(elementData, size);

}

}

Now, the running time for copyOf() function is O(N) where N is the size/capacity of the ArrayList. Hence, the time complexity of trimToSize() function is O(N).

Hence, the running time of building an N-item ArrayList is O(N^2).

Please find the sample code below.

CODE

==================

import java.util.ArrayList;

public class Driver {

  public static void main(String[] args) throws Exception{

      int N = 100000;

      ArrayList<Integer> arr = new ArrayList<>(N);

     

      long startTime = System.currentTimeMillis();

      for(int i=0; i<N; i++) {

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      long endTime = System.currentTimeMillis();

      double time = (endTime - startTime)/1000;

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  }

}

Explanation:

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