island rainfall problem - more documentation

#include
#include
#include
#include
#include
#include
using namespace std;
int const island[] = { 54, 50, 54, 54, 52, 55, 51, 59, 50, 56, 52, 50 };
///////////////   Pos # 0   1   2   3   4   5   6   7   8   9  10  11
int const size = sizeof(island) / sizeof(int);
int accu = 0;
//adapted from STL
template
ForwardIterator max_element_last(ForwardIterator scanner, ForwardIterator const end) {
ForwardIterator ret = scanner;
if (scanner == end)
return ret;//empty range, with zero element!
while (++scanner != end)
if (*ret <= *scanner) //"=" means find LAST
ret = scanner;
return ret;
}
//print height and address of a column
void print1(int const* const pos, char const * const label) {
//int const height = *pos;
printf("%s=%d/%d ", label, *pos, pos - island);
}
void printAll(int const* const L, int const* const l, int const* const h,
int const* const H) {
if (l < h) {
print1(L, "wallL");
print1(l, "ptr");
printf("  ");
print1(h, "ptr");
print1(H, "wallH");
} else {
print1(H, "wallH");
print1(h, "ptr");
printf("  ");
print1(l, "ptr");
print1(L, "wallL");
}
printf("%d=Accu\n", accu);
}
//Rule: move the lo-side pointer only
void onePassAlgo(){
int*loptr; //moving pointer, moving-inward.
int*wallLo, *wallHi; //latest walls
int*h;

//1st we ASSUME the first left side wall will be lower than the first right side wall
wallLo = loptr = const_cast (island);
wallHi = h = const_cast (island) + size - 1;
//2nd, we validate that assumption
if (*wallLo > *wallHi) {
std::swap(wallLo, wallHi);
std::swap(loptr, h);
}
// now lo is confirmed lower than the hi side
printAll(wallLo,loptr,h,wallHi);
printf("All pointers initialized (incl. 2 walls\n");
while (loptr != h) {
if (*loptr > *wallHi) {
wallLo = wallHi;
wallHi = loptr;
std::swap(loptr, h);
//printf("new wallHi:");
} else if (*loptr >= *wallLo) {//see the >=
wallLo = loptr;
//printf("wallLo updated:");
} else {
assert (*loptr < *wallLo);
accu += (*wallLo - *loptr);
printf("adding %d liter of water at Pos_%d (%d=A\n", *wallLo - *loptr,
loptr - island, accu);
}
printAll(wallLo,loptr,h,wallHi);
// only by moving the loptr (not h) can we confidently accumulate water
if (loptr < h)
++loptr; //lo side is on the left, move loptr right
else
--loptr; //lo side is on the right, move loptr left
}
}
void twoPassAlgo() {//less convoluted
int const* const peak = max_element_last(island, island + size);
printf("highest peak (last if multiple) is %d, at Pos %d\n", *peak, peak
- island);
//(island, island + size, ostream_iterator (cout, " "));

//forward scan towards peak
int* pos = const_cast (island); //left edge of island
int* wall = pos;
for (++pos; pos < peak; ++pos) {
if (*wall > *pos) {
accu += *wall - *pos; // accumulate water
printf("adding %d liter of water at Pos#%d (T=%d)\n", *wall - *pos,
pos - island, accu);
continue;
}
//ALL new walls must match or exceed previous wall.
printf("found new wall of %d^ at Pos#%d\n", *pos, pos - island);
wall = pos;
}
cout << "^^^ end of fwd scan ; beginning backward scan vvv\n";
//backward scan
pos = const_cast (island) + size - 1;
wall = pos;
for (--pos; pos > peak; --pos) {
if (*wall > *pos) {
accu += *wall - *pos; // accumulate water
printf("adding %d liter of water at Pos#%d (T=%d)\n", *wall - *pos,
pos - island, accu);
continue;
}
//Note all new walls must match or exceed previous wall.
printf("found new wall of %d^ at Pos#%d\n", *pos, pos - island);
wall = pos;
}
}
int main(int argc, char *argv[]) {
twoPassAlgo();
accu = 0;
cout<<"-----------------------------\n";
onePassAlgo();
}
/*
 Requirement -- a one-dimentional island is completely covered with columns of bricks.
 If  between Column 
 A(height 9) and Column B(10) all columns are lower, then we get a basin to
 collect rainfall. Watermark height (absolute) will be 9.  We can easily calculate the
 amount of water. If I give you all the column heights, give me total rainfall collected.
 Code showcasing
 - stl algo over raw array
 - array/pointer manipulation
 - array initialization
 - array size detection
 - std::max_element modified
 - std::swap
 */