Wrie a function which returns the most frequent number in a
list of integers. Handle the case of more than one number
which meets this criterion.
public static int[] GetFrequency(int[] list)

Answer Posted / sudha chinnappa

I somehow feel that the code can be made more efficient. I
will leave this job to someone else


#include<iostream>
#include<conio.h> //for getch
using namespace std; //for cout

int main()
{
int arr[100];

cout << "Enter the No. of integers\n";
cin >> n ;

cout << "enter the numbers\n";
for(i=0; i<n; i++)
cin >> arr[i];

//Sort the numbers
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if( arr[i] > arr[j] )
{
arr[j] = arr[i] + arr[j];
arr[i] = arr[j] - arr[i];
arr[j] = arr[j] - arr[i];
}
}
}

struct node
{
int num;
int frq;
node *nxt;
};

struct node *first = NULL, *temp, *temp1,*temp2;

i=0;
while(i<n)
{
temp = new node;
j=1;
while(arr[i] == arr[i+1])
{j++; i++;}

temp->num = arr[i];
temp->frq = j;
temp->nxt = NULL;

if (first == NULL)
first = temp;
else
{
temp1 = first;
while(temp1->nxt != NULL)
temp1 = temp1->nxt;

temp1->nxt = temp;
}
i++;
}

temp1 = first;
int large = first->frq;

temp1 = first;
temp2 = first->nxt;
while(temp1->nxt != NULL)
{
if(temp1->frq < temp2->frq)
large =temp2->frq;
temp1 = temp1->nxt;
temp2 = temp2->nxt;
}

cout << "\nHighest frequency numbers:\n\n";
cout << "Number Frequency\n";
temp1 = first;
while(temp1 != NULL)
{
if(temp1->frq == large)
cout << temp1->num << " "<< temp1->frq
<< endl;
temp1 = temp1->nxt;
}
}

Is This Answer Correct ?    10 Yes 5 No



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