# Splits strings by sum of letter values (a=1,b=2...) from random import choice # The core of the algo tries all solutions by enumerating i in 0..2**(len(x)-1) & allocating each team to set1 if bit k of i equals 1 or set 2 otherwise. Simple but slow. You can do better w dynamic programming see my other codes cnv=" abcdefghijklmnopqrstuvwxyz" guild=lambda v:(lambda x,n,s:(lambda h1:[[cnv[a[1]] for a in x if a not in h1],[cnv[a[1]] for a in h1]])(min(([e for e in x if i>>e[0]&1] for i in range (2**n//2)),key=lambda l:abs(len(l)-n/2)+n*abs(sum(e[1] for e in l)-s))))(*(lambda w:(list(enumerate(w)),len(w),sum(w)/2))([cnv.index(x) for x in v])) #for g in ['testcase','pjaaaaaa','what',"ab","aapaaaaj","venezuelabeaverchese","".join(choice(cnv[1:]) for i in range(12))]: # print(2**len(g)//2,"loops for",*(lambda v,f:(g,f(g),"\n",v[0],f(v[0]),"\n",v[1],f(v[1])))(guild(g),lambda s:sum(cnv.index(e) for e in s))) # Factorial fact=lambda i:int(i<2) or i*fact(i-1) # Al the ways to do n by adding positive integers combos=lambda n,p=1:[] if n

>i)&1])((lambda l,r:[min(([i,(l+1)*abs(sum(x[j] for j in r if (i>>j)&1)-t)+sum(((i>>k)&1) for k in r)] for i in range(1<>e[0])&1] for i in range (1<

Oneliners Compilation