int L = 0, R = 0;
while(L < R){
if(v[L] == x) cout << "I found it";
if(v[R] == x) cout << "I found it";
L++, R--;
}
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3382 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 155 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 134 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
int L = 0, R = 0;
while(L < R){
if(v[L] == x) cout << "I found it";
if(v[R] == x) cout << "I found it";
L++, R--;
}
| Name |
|---|



Actually you could have done something similar on a graph, and it will reduce the time complexity by a lot if it's a very big graph with a high branching factor. This has its own name — "Bidirectional Search".
I have a more superior idea
Would you please explain it?
it's an nlog(n) search method, whereas binary search is only log(n). Basically it's n times better than binary search
Big L if
RAND_MAXis 32767 andxis located on index 32768That's why you should buy a lottery ticket.