Arrays

Arrays

Postby kate » Fri Jul 04, 2008 11:19 am

We have given the array A[N]. Make a program, which:
1. checks whether all numbers from the array A[N] are bigger than 4.
2. moves all even numbers from the array A[N] into the array B[k] (sorts the array B[k] in an ascending sequence and print it on the screen)
3. checks whether the array A[N] includes odd numbers.
kate
 
Posts: 91
Joined: Mon Apr 09, 2007 3:58 pm
Reputation: 2

Re: Arrays

Postby Guest » Sun Jul 01, 2018 5:04 pm

// A C++ program to print elements with count more than n/k
#include<iostream>
using namespace std;

// A structure to store an element and its current count
struct eleCount
{
int e; // Element
int c; // Count
};

// Prints elements with more than n/k occurrences in arr[] of
// size n. If there are no such elements, then it prints nothing.
void moreThanNdK(int arr[], int n, int k)
{
// k must be greater than 1 to get some output
if (k < 2)
return;

/* Step 1: Create a temporary array (contains element
and count) of size k-1. Initialize count of all
elements as 0 */
struct eleCount temp[k-1];
for (int i=0; i<k-1; i++)
temp[i].c = 0;

/* Step 2: Process all elements of input array */
for (int i = 0; i < n; i++)
{
int j;

/* If arr[i] is already present in
the element count array, then increment its count */
for (j=0; j<k-1; j++)
{
if (temp[j].e == arr[i])
{
temp[j].c += 1;
break;
}
}

/* If arr[i] is not present in temp[] */
if (j == k-1)
{
int l;

/* If there is position available in temp[], then place
arr[i] in the first available position and set count as 1*/
for (l=0; l<k-1; l++)
{
if (temp[l].c == 0)
{
temp[l].e = arr[i];
temp[l].c = 1;
break;
}
}

/* If all the position in the temp[] are filled, then
decrease count of every element by 1 */
if (l == k-1)
for (l=0; l<k; l++)
temp[l].c -= 1;
}
}

/*Step 3: Check actual counts of potential candidates in temp[]*/
for (int i=0; i<k-1; i++)
{
// Calculate actual count of elements
int ac = 0; // actual count
for (int j=0; j<n; j++)
if (arr[j] == temp[i].e)
ac++;

// If actual count is more than n/k, then print it
if (ac > n/k)
cout << "Number:" << temp[i].e
<< " Count:" << ac << endl;
}
}
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