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Mrac [35]
2 years ago
8

Given an array of integers a, your task is to calculate the digits that occur the most number of times in the array. Return the

array of these digits in ascending order.
Computers and Technology
1 answer:
Annette [7]2 years ago
3 0

Answer:

I hope this should be helpful

Explanation:

In Javascript

function solution(numbers) {

   let hashmap = {};

   let ans = [];

   for (let i = 0; i < numbers.length; i++) {

     // check wether it double or single digit

     if (hasOneDigit(numbers[i])) {

       // check if it exists in hashmap

       // if exist add number of occurence

       // check if the occurence is greater than 2

       // add  if > 2 to the answer list

       if (numbers[i] in hashmap) {

         hashmap[numbers[i]] = hashmap[numbers[i]] + 1;

 

         if (hashmap[numbers[i]] = 2 && !ans.includes(numbers[i])) {

           // max_freq = hashmap[numbers[i]]

           ans.push(numbers[i]);

         }

         // else if (max_freq == hashmap[numbers[i]])

         //     ans = min(ans, numbers[i])

       } else {

         hashmap[numbers[i]] = 1;

       }

     } else {

       // change number to string

       number = numbers[i].toString();

       // loop to iterate every single element

       for (let j = 0; j < number.length; j += 1) {

 

         if (+number.charAt(j) in hashmap) {

           hashmap[+number.charAt(j)] = hashmap[+number.charAt(j)] + 1;

 

           if (hashmap[+number.charAt(j)] = 2 && !ans.includes(+number.charAt(j))) {

     

             ans.push(+number.charAt(j));

           }

       

         } else {

             hashmap[+number.charAt(j)] = 1

         }

       }

     }

   }

     return ans.sort();

 }

 

 function hasOneDigit(val) {

   return String(Math.abs(val)).charAt(0) == val;

 }

 console.log(solution([2, 1,42,44, 2, 3,32,7777, 3 , -1]));

 

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Using the program below, explain what the output will be at LINE A. 1 #include 2#include 3#include 4 5 int value - 128; 6 7 int
slega [8]

Answer:

This is the complete correct program:

#include <stdio.h>

#include<sys/types.h>

#include<unistd.h>

int value = 128;

int main()

{

  pid_t pid;

  pid=fork();

  if (pid==0) /* child process */

  {

  value +=8;

  return 0; }

  else if (pid > 0) {/* parent process */

 wait (NULL);

 printf ("PARENT: value =%d\n" ,value); /* LINEA */

 return 0;

}

}

The output of the LINE A is:

PARENT: value = 128

Explanation:

The fork() function used in the program creates a new process and this process is the child process. The child process is same as the original process having its own address space or memory.

In the child process the value of pid is 0. So the if condition checks if pid==0. Then the child process adds 8 to the value of its variable according to the following statement  

value +=8;

Now the original process has value = 128. In else if part the parents process has the value of pid greater than zero and this portion of the program is of the parent process :

else if (pid > 0)

{ wait (NULL);

printf ("PARENT: value =%d\n" ,value);

return 0; }

So the value 128 is printed at the end in the output.

wait(NULL) is used to wait for the child process to terminate so the parent process waits untill child process completes.

So the conclusion is that even if the value of the variable pid is changed in the child process but it will not affect the value in the variable of the parent process.

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Answer:

showProduct(int,double)

for example: showProduct(10,10.5) is the correct answer even showProduct(10,10.0) is also correct but showProduct(10.0,10.5) or showProduct(10,10) or showProduct(10.0,10) are wrong calls.

Explanation:

The code is

  1.       <em>public static void showProduct (int num1, double num2){</em>
  2. <em>       int product;</em>
  3. <em>       product = num1*(int)num2;</em>
  4. <em>       System.out.println("The product is "+product);</em>
  5. <em>       }</em>

showProduct is function which asks for two arguments whenever it is called, first one is integer and second one is of type double which is nothing but decimal point numbers. Generally, in programming languages, 10 is treated as integer but 10.0 is treated as decimal point number, but in real life they are same.

If showProduct( 10,10.0) is called the output will be 'The product is 100'.

Strange fact is that, if you enter showProduct(10,10.5) the output will remain same as 'The product is 100'. This happens because in the 3rd line of code,which is <em>product=num1*(int)num2</em>, (int) is placed before num2 which makes num2 as of type integer, which means whatever the value of num2 two is given, numbers after decimal is erased and only the integer part is used there.

This is necessary in JAVA and many other programming languages as you <u>cannot</u><u> multiply two different datatypes</u> (here one is int and another is double). Either both of them should be of type int or both should be of type double.

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