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
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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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 = 3givesmid = 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) + 1i.e,left = 3, right = 3and 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