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Katena32 [7]
3 years ago
6

X333: countElements Given an array of integers and an integer target, return a count of the number of times the target value occ

urs in the array. Examples: countElements((25, 42, 18, 25, 38, 96, 25, 25〉, 25) -> 4 countElements (110, 20, 30, 40, 50, 99)-> θ 1 public int countElements (int[] nums, int target) 2 {34 }5
Computers and Technology
1 answer:
inn [45]3 years ago
5 0

Answer:

The method written in java is as follows:

public static int countElements(int[] nums, int target){

    int count=0;

    for(int i =0;i<nums.length;i++){

        if(nums[i] == target){

            count++;

        }

    }

    return count;

}

Explanation:

The method definition is not clear in the question you posted; so, I'll answer the question from scratch

This defines the method

public static int countElements(int[] nums, int target){

This initializes count to 0

    int count=0;

This iterates through the array

    for(int i =0;i<nums.length;i++){

This checks if the current array element is same as the target

        if(nums[i] == target){

If yes, the count variable is incremented by 1

            count++;

        }

    }

This returns the count

    return count;

}

See attachment for complete program which includes the main

Download txt
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g 1. Write a program that asks the user for a number greater than 5 and prints all values between 1 and n that are multiples of
a_sh-v [17]

PROGRAM 1

#include <iostream>  

using namespace std;

int main() {

   int n;

   // user input taken for n

   cout<<"Enter any number greater than five: ";

   cin>>n;

   

   cout<<"The multiples of 5 between 1 to "<<n<<" are shown below."<<endl;

   

   // displaying multiples of 5 between 1 to n

   for(int h=1; h<=n; h++)

   {

       // if number is completely divisible by 5, remainder will be 0

       if( (h%5)==0)

           cout<<h<<endl;

   }

 

   return 0;

}

OUTPUT

Enter any number greater than five: 22

The multiples of 5 between 1 to 22 are shown below.

5

10

15

20

The program does not implements any input validation since this is not mentioned in the question.

User input is taken. All the multiples of 5 computed inside a for loop, and displayed to the console.

PROGRAM 2

#include <iostream>

using namespace std;

int main() {

   // variables to hold respective values

   int n;

   int sum=0;

   double avg=0;

   

   do

   {

       // user input taken for n

       cout<<"Enter any number between 1 and 100: ";

       cin>>n;

       if(n<1 || n>100)

           cout<<"Invalid number. Enter valid number."<<endl;

       cout<<""<<endl;

       

   }while(n<1 || n>100);

   

   // computing sum and average of numbers from 1 to n

   for(int h=1; h<=n; h++)

   {

       sum = sum+h;

   }

   

   avg=avg+(sum/n);

   

   cout<<"Sum of all the numbers from 1 to "<<n<<" is "<<sum<<endl;

   

   cout<<""<<endl;

   cout<<"Average of all the numbers from 1 to "<<n<<" is "<<avg<<endl;

   

   return 0;

}

OUTPUT

Enter any number between 1 and 100: 111

Invalid number. Enter valid number.

Enter any number between 1 and 100: 0

Invalid number. Enter valid number.

Enter any number between 1 and 100: 23

Sum of all the numbers from 1 to 23 is 276

Average of all the numbers from 1 to 23 is 12

The program not implements any input validation since this is mentioned in the question.

User input is taken inside do-while loop till valid input is obtained.

The sum of all numbers from 1 to n computed inside a for loop, average computed outside for loop.

Both values are displayed to the console.

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