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
# | User | Rating |
---|---|---|
1 | jiangly | 3898 |
2 | tourist | 3840 |
3 | orzdevinwang | 3706 |
4 | ksun48 | 3691 |
5 | jqdai0815 | 3682 |
6 | ecnerwala | 3525 |
7 | gamegame | 3477 |
8 | Benq | 3468 |
9 | Ormlis | 3381 |
10 | maroonrk | 3379 |
# | User | Contrib. |
---|---|---|
1 | cry | 168 |
2 | -is-this-fft- | 165 |
3 | Dominater069 | 161 |
4 | Um_nik | 159 |
4 | atcoder_official | 159 |
6 | djm03178 | 157 |
7 | adamant | 153 |
8 | luogu_official | 150 |
9 | awoo | 149 |
10 | TheScrasse | 146 |
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
Name |
---|
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