# RogerBW's Blog

 The Weekly Challenge 248: Shortest Submatrix 24 December 2023 I’ve been doing the Weekly Challenges. The latest involved searching a string and adding matrix elements. (Note that this ends today.) Task 1: Shortest Distance You are given a string and a character in the given string. Write a script to return an array of integers of size same as length of the given string such that: distance[i] is the distance from index i to the closest occurence of the given character in the given string. The distance between two indices i and j is abs(i - j). So each instance of the target character maps to zero, its neighbours to 1, etc. There were several possible approaches to this, but I ended up building a list of the target character indices and then doing a BFS to find their neighbours. Rust, which is my usual language for the first solution of these problems, has a handy `match_indices` which lets me find all the matches at once. Most of the others have a "start your match at a particular point" function. For PostScript, I tried building a working string copy and cropping it down, but it felt cleaner to take a functional approach: take the input string as an array, enumerate it so that I have index values, filter for matching characters, then map the list into the queue format. In Scala, I'm on Scala 2.11 (Debian/stable), which only has a Queue type rather than a general-purpose double-ended queue. Still, it works. ``````sub shortestdistance(\$a, \$c) { my @q; my \$i = 0; `````` Find all instances of the target character and queue them with a zero value. `````` while (True) { with (\$a.index(\$c, \$i)) { @q.push((\$_, 0)); \$i = \$_ + 1; } else { last; } } `````` In most languages that was a loop as in this Perl, where I start the next string search one character after the previous one. (Raku returns a fail state of `Nil` if the character isn't found, rather than -1, but documentation on how to catch that cleanly is sparse.) `````` while (\$i >= 0) { my \$p = index(\$a, \$c, \$i); if (\$p > -1) { push @q, [\$p, 0]; \$i = \$p + 1; } else { \$i = -1; } } `````` In Rust it was: `````` for (i, _c) in a.match_indices(c) { q.push_back((i, 0)); } `````` while in PostScript (which only has "does this substring appear at the start of the string" and "does this substring appear at all" functions) I went functional: We're going to define a variable: ``````/q `````` Build a list of characters and indices: ``````a enumerate.array `````` Choose only those where the character matches the target: ``````{ 1 get c eq } filter `````` Use the index to build the queue entry: ``````{ 0 get [ exch 0 ] } map `````` and put that list into the variable: ``````def `````` Back to Raku. Fill the output list with an invalid valuue. `````` my \$invalid = \$a.chars + 1; my @out = \$invalid xx \$a.chars(); `````` Now do the BFS. If I find an entry that hasn't been initialised already… `````` while (@q.elems > 0) { my (\$i, \$v) = @q.shift; if (@out[\$i] == \$invalid) { `````` Fill it with the calculated value (and because this is BFS the lowest value will be the first time I've seen this cell). `````` @out[\$i] = \$v; `````` Now queue neighbouring cells with a value one higher than this. One of these will always be a cell we've already seen, and sometimes both. `````` if (\$i > 0) { @q.push((\$i - 1, \$v + 1)); } if (\$i < \$a.chars - 1) { @q.push((\$i + 1, \$v + 1)); } } } return @out; } `````` Another approach, which I didn't think of until writing this up: for each matching character, build a list `[..., -2, -1, 0, 1, 2, ...]` centred on that position. Then align them all and take the minimum value. But I was in a BFS sort of mood, so this is the one you get. Task 2: Submatrix Sum You are given a NxM matrix A of integers. Write a script to construct a (N-1)x(M-1) matrix B having elements that are the sum over the 2x2 submatrices of A, b[i,k] = a[i,k] + a[i,k+1] + a[i+1,k] + a[i+1,k+1] This is basically just nested loops, and looks the same in most languages. JavaScript: ``````function submatrixsum(a) { let out = []; for (let y = 0; y < a.length - 1; y++) { let row = []; for (let x = 0; x < a[y].length - 1; x++) { let s = 0; for (let ya = y; ya <= y + 1; ya++) { for (let xa = x; xa <= x + 1; xa++) { s += a[ya][xa]; } } row.push(s); } out.push(row); } return out; } `````` Full code on github. Comments on this post are now closed. If you have particular grounds for adding a late comment, comment on a more recent post quoting the URL of this one. 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