Angelina Veni
Harus dikuasai :
- sequential search O(n)
- binary search O(log n )
Latihan:
To be added : source code masing2 algo, more problems, detailed explanation
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ka’ ada ngga yg lebih cepet dari binary search it.?
gini, ada array string m smpe indeks 5000. trus ada 3^12 string lain yg mau di cek apkah dia ada di array string m tadi. nah code nya ak runtime error (> 1 detik). pake binary search.
gmn k’.? ap ada cara lain.?
jawabannya cuma ya / tidak? kalo iya, masukin 3^12 string itu ke array of string lain, misal p. sort m dan p, ascending. menentukan ada / tidak kan bisa cuma sekali jalan jadinya.
yay. sip banget deh tu. hha. makasi makasi k’
k’ gmna dh.? ak malah bingung. hha ckck.
ak code nya gini k’ :
jd ada array n yg kita mau tau dia ada di array m atau nggak. kalau ada tulis. max it indeks tergede n trus 4617 it indeks tergede m.
i:=1; bottom:=1;
while i<=4617 do begin
j:=bottom;
while (j<=max) and (m[i]n[j]) do inc(j);
if m[i]=n[j] then begin
writeln(n[j]);
bottom:=j;
found:=true;
end;
inc(i);
end;
tp msh runtime error. malah lebih lama tu drpda pake binary search. gmn k’.?
itu karena kamu bottomnya nambahnya cuma kalo ketemu doang. kira2:
walkM = 1; walkN = 1;
while (walkM <= maxM) do begin
while ((walkN <= maxN) and (n[walkN] < m[walkM])) do inc(walkN);
if (n[walkN] == m[walkM]) then //n [walkN] ada di dalam array M.
end;
oo gitu. hmmm ngerti ngerti k’. yap. makasii bnget k
binary search O(n log n )?? veni kamu pendek deh
whooops typo. thanks su!