# RogerBW's Blog

 The Weekly Challenge 202: Consecutive Valley 05 February 2023 I’ve been doing the Weekly Challenges. The latest involved various sorts of list filtering. (Note that this closes today.) Task 1: Consecutive Odds You are given an array of integers. Write a script to print `1` if there are `THREE` consecutive odds in the given array otherwise print `0`. For every language except Perl, I turn that to true/false, because why not? The logic is the same across all languages. Raku: ``````sub consecutiveodds(@a) { `````` Set up the odds counter. `````` my \$i = 0; `````` For each number in the sequence, `````` for (@a) -> \$v { `````` if it's odd, `````` if (\$v % 2 == 1) { `````` I have one more odd-in-sequence than I did last time. `````` \$i++; `````` If I have three odds-in-sequence, `````` if (\$i >= 3) { `````` stop here. `````` return True; } `````` If it isn't odd, `````` } else { `````` I have no odds-in-sequence. `````` \$i = 0; } } `````` And if I never got to three having run off the end of the list, return the other result. `````` return False; } `````` Another approach would be to mod2 the whole list, then take three-element windows onto it and check them individually, but I find the above pretty straightforward and reasonably efficient. Task 2: Widest Valley Given a profile as a list of altitudes, return the leftmost widest valley. A valley is defined as a subarray of the profile consisting of two parts: the first part is non-increasing and the second part is non-decreasing. Either part can be empty. The logic is a bit less obvious than some of these problems have been. Here's my Rust version, which falls into three stages. ``````fn widestvalley(a: Vec) -> Vec { `````` First, look through the list and combine consecutive multiples into `(value, count)` tuples. (Well, if I were writing only in Rust I'd probably use tuples; here for portability I just have separate vectors `av` and `ac`.) `````` let mut av = Vec::new(); let mut ac = Vec::new(); let mut l = 0; for &v in &a { if v == l { *ac.last_mut().unwrap() += 1; } else { av.push(v); ac.push(1); l = v; } } `````` Then I go looking for valley starts and ends - i.e. isolated highest points. The first and last distinct values always count as these; so does any value higher than both its neighbours. (Beware of languages that don't have short-circuiting evaluation, such as Lua and PostScript, because the offset indices won't be valid.) The start value of such a point is the index in `a` of the first such value in its group; the end value is the index of the last. (`c` is therefore the index into `a` that we're artificially incrementing.) `````` let mut s = Vec::new(); let mut e = Vec::new(); let mut c = 0; for i in 0..=av.len() - 1 { if i == 0 || i == av.len() - 1 || (av[i - 1] < av[i] && av[i] > av[i + 1]) { s.push(c); e.push(c + ac[i] - 1); } c += ac[i]; } `````` Finally, each valley runs from an `s`-value to the next `e`-value (i.e. including the full width of both peaks). Look through each of these, and if it's wider than we've had before, make it the output. `````` let mut out = Vec::new(); for i in 0..=s.len() - 2 { if e[i + 1] - s[i] + 1 > out.len() { out = a[s[i] ..= e[i+1]].to_vec(); } } out } `````` Full code on github. Comments on this post are now closed. 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