D19 Answer

int numberDuplicates( int x[][NUMCOLS], int nrows )
{
  int r, c;     /* indexes to the current array element */
  int j;        /* index into array[] */
  int array[nrows*NUMCOLS];
  int duplicateCount = 0;
  int inRun = 0;

  /* Copy elements from x[][] to array[] */
  j = 0;
  for ( r=0; r<nrows; r++ )
    for ( c=0; c<NUMCOLS; c++ )
      array[ j++ ] = x[r][c];

  /* Sort the 1D array */
  selectionSort( array, nrows*NUMCOLS );
  
  /* Scan the 1D array for duplicates */
  for ( j=0; j<nrows*NUMCOLS-1; j++ )
  {
    if ( !inRun && array[j] == array[j+1] )
    {
      inRun = 1;
      duplicateCount += 2 ;
    }
    else if ( inRun && array[j] == array[j+1] )
      duplicateCount++ ;
    else 
      inRun = 0;
  }
  return duplicateCount;
}

Comment: Notice the tricky code dealing with the start of a run:

    if ( !inRun && array[j] == array[j+1] )

The start of a run is detected when the first two elements of it are found to be the same. The duplicateCount must be incremented by two.

In the middle of a run,

    else if ( inRun && array[j] == array[j+1] )

duplicateCount is incremented by only one for each new element in the run.

Of course, when not in a run, duplicateCount is not incremented at all.