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Python, 91 85 83 bytes -6 bytes thanks to @Steffan -2 bytes thanks to @movatica f=lambda a,b,c:c and[(i,*j)for i,d in enumerate(a)for j in f(d,b,c-1)]or[[]]*(a==b) Attempt This Online!

Rust, 68 bytes |u:Vec<_>|u.iter().fold((0,0,0),|(u,v,w),&x|(u+w*(v!=x)as i8,x,1)).0 Attempt This Online! Non-Competing Rust, 63 bytes This version crashes on 0-length inputs so doesn't technically co …

><> (Fish), 30 bytes i0\ ;n$\.13$\ 1+>i:0(?/:{=?^$ Try it

Rust, 135 130 bytes -5 bytes thanks to @alpehalpha |v:Vec<i8>|v.iter().scan(0,|c,&a|Some((match a-*c{1=>vec![a],2=>vec![a-1,a],_=>vec![-1,a]},*c=a).0)).flatten().collect::<Vec<_>>() Attempt This Onlin …

Rust, 112 bytes |l:&[u8]|{(0..).zip(l).scan(0,|a,(b,c)|{*a=(*a).max(*c);Some((*a).min(*l[b..].iter().max()?)-c)}).sum::<u8>()}; Playground Link Basically min(max(..index),max(index..)-value

Rust, 150 bytes |m|f(m,1);fn f(k:&[i8],u:i8)->bool{let m=k.len()/2;k!=&[]&&(k==&[u]||f(&k[..m],u)&&(&k[..m]==&k[m..]||(f(&k[m..],-u)&&(0..m).all(|i|k[i]==-k[i+m]))))} Attempt This Online!

Rust, 64 bytes |x:&[_]|(0..x.len()).map(|i|x[i&i+1..=i].iter().sum()).collect() Attempt This Online! Uses XNOR's trick. Also uses blatant abuse of type annotations outside the code.

Python, 69 bytes lambda x:max(sum(x[a:b])for a in range(len(x))for b in range(len(x))) Attempt This Online! Python, 74 bytes Outputs the list lambda x:max((x[a:b]for a in range(len(x))for b in range( …

Given a list of lists of positive integers, output a subset of them so their union forms a continuous non-empty range with no numbers missing. For example, consider this input: [ [1, 2, 3, 5], [ …

Rust, 90 bytes fn f(a:&[i32],b:i32)->bool{a.len()<2||a[1]-a[0]<=b&&f(&a[1..],b-(a[0]-a[1]>b&&b>0)as i32)} Attempt This Online! Recursive approach

Python, 154 bytes lambda x:(z:=len(x))and next(i for i in permutations(range(z))if all(x[j]==i[j]or("RL"[j>i.index(i[j]+1)]==x[j])for j in range(z))) from itertools import* Attempt This Online! 0-ind …

Rust, 63 bytes |k:&[u8]|(0..k.len()).all(|z|(z..k.len()).any(|q|k[q-z]==k[q])) Attempt This Online!

Given a list, L, of sets of numbers like this: [[1], [0,2,3], [3,1], [1,2,3]] Output a single list of numbers such that 2 numbers A and B appear next to each other at least once if and only if B in L …

Python, 79 bytes lambda x,j=0:(i for i in count()if i-j or(j:=next(x))*0) from itertools import* Attempt This Online!

><> (Fish), 431 385 bytes -46 bytes by creating a new stack with l-2 elements so I can remove the 2- from many places. l[bb01. >rl2=?vl3=?vl2%?vl2-2,[r0501. 0v=1:< 0 < < < << / 1 …