RogerBW's Blog

The Weekly Challenge 218: Maximum Matrix 28 May 2023

I’ve been doing the Weekly Challenges. The latest involved list manipulation and bit flipping. (Note that this is open until 28 May 2023.)

Task 1: Maximum Product

You are given a list of 3 or more integers.

Write a script to find the 3 integers whose product is the maximum and return it.

The simple version of this would just be to take the three highest. But that would be dull. Thus example 4, in which some of the integers are negative.

I end up sorting the list and then taking four combined slices of it to get the products. Raku:

sub maximumproduct(@lst) {
    my @l = sort {$^a <=> $^b}, @lst;
    my $b = @l.elems;
    my @t;

i is the number of integers to take from the left side (lowest values).

    for (0..3) -> $i {
        my $p = 1;

Multiply by each of those:

        if ($i > 0) {
            for (0..$i-1) -> $j {
                $p *= @l[$j];

Then by 3 - i numbers from the right side (highest values).

        if ($i < 3) {
            for ($b-3+$i..$b-1) -> $j {
                $p *= @l[$j];

Store that result.


And return the highest one we got.

    return max(@t);

Task 2: Matrix Score

You are given a m x n binary matrix i.e. having only 1 and 0.

You are allowed to make as many moves as you want to get the highest score.

A move can be either toggling each value in a row or column.

To get the score, convert the each row binary to dec and return the sum.

Like last week's problem, this doesn't need to be an exhaustive search. Given the choice in a sequence of (1, ...) or (0, ...), the (1, ...) will always produce a larger value, whatever the state of any digits after it.

Therefore every row must start with a 1; then for columns other than the first, the majority of digits should be 1.


fn matrixscore(matrix0: Vec<Vec<u8>>) -> u32 {
    let mut matrix = matrix0;

Check the first digit of each row. If it's a 0, flip the row.

    for i in 0..matrix.len() {
        if matrix[i][0] == 0 {
            for j in 0..matrix[i].len() {
                matrix[i][j] = 1 - matrix[i][j];

Now for each column after the first,

    let t = matrix.len() / 2;
    for i in 1..matrix[0].len() {

count the zeroes,

        let mut c = 0;
        for j in 0..matrix.len() {
            if matrix[j][i] == 0 {
                c += 1;

and if they're more than half the entries, flip the column.

        if c > t {
            for j in 0..matrix.len() {
                matrix[j][i] = 1 - matrix[j][i];

Then read out the results, converting each row to a number.

    let mut tot = 0;
    for m in matrix {
        let mut p = 0;
        for n in m {
            p *= 2;
            if n == 1 {
                p += 1;
        tot += p;

Turns out matrix is a keyword in PostScript…

Full code on github.

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