RogerBW's Blog

I’ve been doing the Weekly Challenges. The latest involved various sorts of array hunting. (Note that this ends today.)

Task 1: Maximal Square

You are given an m x n binary matrix with 0 and 1 only.

Write a script to find the largest square containing only 1's and return it’s area.

So basically it's a search. Start at each point, then accrete layers o a square: first the point itself, then the three squares below and to the right of it to make a 2×2, then the five more that make 3×3, etc.

In JavaScript:

function maximalsquare(a) {

Set up the maximum value latch and the array sizes.

    let mx = 0;
    const boundy = a.length;
    const boundx = a[0].length;

Iterate over each starting point.

    for (let y = 0; y < boundy; y++) {
        for (let x = 0; x < boundx; x++) {

size is the last square-size we've found.

            let size = 0;
            while (true) {

Build a list of points that need to be included in the next size of square: horizontally including the corner

                let tests = [];
                for (let xx = x; xx <= x + size; xx++) {
                    tests.push([y + size, xx]);
                }

and vertically.

                if (size > 0) {
                    for (let yy = y; yy < y + size; yy++) {
                        tests.push([yy, x + size]);
                    }
                }

If any of those is a zero, the previous size value is what we've got. (In most of these languages I have an any filter, but min works when I don't.)

                if (Math.min(...tests.map(i => a[i[0]][i[1]])) == 0) {
                    break;
                }

Otherwise, add one to size and make sure that won't spill outside the grid.

                size += 1;
                if (x + size >= boundx || y + size >= boundy) {
                    break;
                }
            }

I've been working in linear size, but we have to store the count of points.

            mx = Math.max(mx, size * size);
        }
    }
    return mx;
}

I like this best in PostScript. Same algorithm, but the tests list can be an anonymous array.

/maximalsquare {
    0 dict begin
    /a exch def
    /mx 0 def
    /boundy a length def
    /boundx a 0 get length def
    0 1 boundy 1 sub {
        /y exch def
        0 1 boundx 1 sub {
            /x exch def
            /size 0 def
            {
                [
                    x 1 x size add {
                        [ exch y size add exch ]
                    } for
                    size 0 gt {
                        y 1 y size add {
                            [ exch x size add ]
                        } for
                    } if
                ]
                {
                    /i exch def
                    a i 0 get get i 1 get get
                } map
                { 0 eq } any {
                    exit
                } if
                /size size 1 add def
                x size add boundx ge
                y size add boundy ge or {
                    exit
                } if
            } loop
            /mx mx size dup mul max def
        } for
    } for
    mx
    end
} bind def

Task 2: Right Interval

You are given an array of @intervals, where $intervals[i] = [starti, endi] and each starti is unique.

The right interval for an interval i is an interval j such that startj >= endi and startj is minimized. Please note that i may equal j.

Write a script to return an array of right interval indices for each interval i. If no right interval exists for interval i, then put -1 at index i.

This turned out rather easier than task 1.

Raku:

sub rightinterval(@a) {

Build a list of the i values.

    my @ss = @a.map({$_[0]});

Sort a list of indices by the i values. (There's a term for this, sorting the indices rather than the thing itself, but I can never remember what it is.)

    my @si = (0 .. @a.end).sort({@ss[$^a] <=> @ss[$^b]});
    my @out;

For each interval set,

    for @a -> @l {

Find the indices of intervals that are validly to the right of it.

        my @ix = @si.grep({@ss[$_] >= @l[1]});

Since they're sorted, push the first one (if any).

        if (@ix.elems > 0) {
            @out.push(@ix[0]);
        } else {
            @out.push(-1);
        }
    }
    @out;
}

Full code on github.