Hello, can someone explain me in this question BINARY SEARCH for the test case 3 1 2 my answer is giving answer 2 but the judge says it should be 0. Can someone explain me this test case. thanks in advance. 96599228
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 3985 |
2 | jiangly | 3814 |
3 | jqdai0815 | 3682 |
4 | Benq | 3529 |
5 | orzdevinwang | 3526 |
6 | ksun48 | 3517 |
7 | Radewoosh | 3410 |
8 | hos.lyric | 3399 |
9 | ecnerwala | 3392 |
9 | Um_nik | 3392 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 169 |
2 | maomao90 | 162 |
2 | Um_nik | 162 |
4 | atcoder_official | 161 |
5 | djm03178 | 158 |
6 | -is-this-fft- | 157 |
7 | adamant | 155 |
8 | awoo | 154 |
8 | Dominater069 | 154 |
10 | luogu_official | 150 |
Hello, can someone explain me in this question BINARY SEARCH for the test case 3 1 2 my answer is giving answer 2 but the judge says it should be 0. Can someone explain me this test case. thanks in advance. 96599228
Название |
---|
Their binary search is a bit different in the sense that they keep going even after finding x.
Here if we consider 1 3 2 and 2 1 3. Apply BS implementation.
So for 1 3 2
left = 0, right = 3
givesmid = 1
, since a[1] = 3(it's fixed) soleft = 1(mid) + 1 = 2
.Now we have
left = 2, right = 3
.Again calculating mid,
mid = 2
, Now no matter what you place 1 or 2 in position = 2, you will always satisfya[mid] <= x
, soleft = 2(mid) + 1
i.e,left = 3, right = 3
and now BS will terminate.So you see now if you do
a[left-1] == x
, it's not true. So the answer will be 0.got it !!!! thanks