Count Number Of Duplicates Elements Present In Array

Count Number Of Duplicates Elements Present In Array

You need to count Number of duplicates elements present in an array that occurs two or more times.

For Example: 

Input: arr = 1,2,3,1,1,2,4,5,2,3

Output :      3  (1 present 3 times, 2 present 3 times , 3 present 2 times )

Count-Duplicates-In-Array

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
int arr[] = { 1,2,3,1,1,4,5,6,4,4,5,2,2,12};
        int size1 = sizeof(arr)/sizeof(arr[0]);
int count1 = 0,flag=0, temp = 0;;
int n = 0;

/*first sort array elements so that all duplicates will be next to each other*/

for (int i = 0; i<size1; i++)
{
for (int j = 0; j<size1 - i - 1; j++)
{
if (arr[j] > arr[j + 1])
{
temp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = temp;
}
}
}

       /*Print sorted array*/

for (int i = 0; i<size1; i++)
cout << arr[i] << "  ";
int i = 0;
while(i<size1)
{
flag = 0;
for (int j = i + 1; j<=size1; j++)
{
if (arr[i] == arr[j])
{
n++;
flag++;
if (flag ==1)
count1++;
continue;
/*if elements present the continue to next iteration inside nested loop*/
}
else
{
n++;
break;
}
}
i = n;
}
cout << "\n\n count value:-  " << count1;
getchar();

}

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Delete All Node From A Link List Greater Than A Given Value(X)

Delete All Node From Link List Greater Than A Given Value

Linked-List

Program:

void DeleteNode(node** head, int x)
{

    node *ptr = *head;
    node *prev,*temp,*temp1;
    if((*head== NULL))
    {
        printf("\n\nlist is empty");
    }
    else if(ptr->data >x)
    {
        temp = ptr;
        ptr=ptr->next;
        free(temp);
    }
    else
    {
         temp = ptr;
         prev = ptr;
         printf("\n data :  %d",temp->data);
        while(temp!=NULL)
        {
          
            if (temp->data >x)
            {
                temp1 = temp;
                temp=temp->next;
                prev->next = temp;
                free(temp1);
                temp =prev;

            }
            else
            {
                prev=temp;
                temp=temp->next;
            }
        }

    }
}

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Interview Question #1

What happens if you Call a free on a pointer twice?

Deallocating a memory area with free does not make the contents of the pointer NULL. 

Suppose that you have int *a = malloc (sizeof (int)) and a has 0xdeadbeef and you execute free (a) then after execution a still contains 0xdeadbeef but after the free call this memory address is no more reserved for you. 

Something like you have rented a flat with malloc used for some time, returned the flat by free then you might have a duplicate key for the flat, but it is not reserved for you.

Doing a free on an already freed memory will result in double free memory corruption.

Callling-free-twice-on-pointer

Answer Original Source: StackOverFlow

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