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
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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 |
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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