Answer Posted / 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;
}

A suduco given & u hv 2 check if it is incomplete(blanks left),or correct or incorrect

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Ask the user to input three positive integers M, N and q. Make the 2 dimensional array of integers with size MxN, where all the elements of I (I = 1,…,M) line will be members of geometrical progression with first element equal to the number of line (I) and denominator q.

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write a program to perform generic sort in arrays?

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Given a table of the form: Product Sold on A 1/1/1980 B 1/1/1980 C 1/1/1980 A 1/1/1980 B 1/1/1980 C 2/1/1980 A 2/1/1980 There are 30 products and 10,000 records of such type. Also the month period during which sales happened is given to u. Write the program to display the result as: Product Month No. of copies A January 12 A February 15 A March 27 B January 54 B February 15 B March 10 C January 37

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