Answer / 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;
}
}

Answer / paras malik

//I strongly feel that this signature is of java instead of
c++ with wrong signature name because function name
according to java conventions should start with small letter.

import java.util.HashMap;
import java.util.Set;

class A{



public static void main(String [] a){


int array[]={1,1,2,1,2,3,4,5,2,3,1};

array = GetFrequency(array);


for(int i =0;i<array.length;i++){

System.out.println(array[i] + "\n");

}

}



public static int[] GetFrequency(int [] array){

HashMap<Integer ,Integer > map = new HashMap<Integer,Integer>();



for(int i =0;i<array.length;i++){


if(map.get(array[i])==null) map.put(array[i],1);
else{
int k = map.get(array[i]);
map.put(array[i],k+1);

}

}

int a[] = new int[map.size()];

Set<Integer> set= map.keySet();
int i =0;
for(int s : set)
a[i++]=map.get(s);
return a;




}

}

Answer / rob lange

// I added a size input because I don't know how it would
// work without it. I also don't know what you're going
// to do without the returned size information, but whatever.
int* FindMax( int * list, int size )
{
// Check bad input conditions
if ( size == 0 )
return NULL;
else if ( size == 1 )
return list;

// Turn int pointer list into vector
int* curr = list;
std::vector< int > vList;
while ( size != 0 )
{
vList.push_back( *curr );
curr++;
size--;
}

// Sort
sort( vList.begin(), vList.end() );
vList.push_back( 0 ); // dummy tail node

// Find most frequent number
std::vector<int>::iterator i = vList.begin();
int curMaxLength = 0;
int tmpLength = 0;
std::vector<int> MostFreqNums;
while ( i+1 != vList.end() )
{
if ( *i == *(i+1) )
tmpLength++;
else
{
if ( tmpLength > curMaxLength )
{
MostFreqNums.clear();
MostFreqNums.push_back( *i );
curMaxLength = tmpLength;
}
else if ( tmpLength == curMaxLength )
{
MostFreqNums.push_back( *i );
}
tmpLength = 0;
}
++i;
}

// Convert vector list to int pointer array
int * ret = new int[ MostFreqNums.size() ];
i = MostFreqNums.begin();
for( int j = 0; i != MostFreqNums.end(); ++i, ++j )
{
ret[j] = *i;
}

return ret;
}

Seat Reservation prog for the theatre. Write a function for seat allocation for the movie tickets. Total no of seats available are 200. 20 in each row. Each row is referred by the Character, "A" for the first row and 'J' for the last. And each seat in a row is represented by the no. 1-20. So seat in diffrent rows would be represented as A1,A2....;B1,B2.....;........J1,J2... Each cell in the table represent either 0 or 1. 0 rep would seat is available , 1 would represent seat is reserved. Booking should start from the last row (J) to the first row(A). At the max 20 seats can be booked at a time. if seats are available, then print all the seat nos like "B2" i.e (2 row, 3 col) otherwise Print "Seats are not available." and we must book consecutive seats only.

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