### rum3r's blog

By rum3r, 7 weeks ago,

Can someone point out the error in this code -> PROBLEM
I am assuming there's something wrong with my binary search.
This is the implementation I normally use. In case if someone could suggest something better.

And also if someone can give me a link to the blogs explaining different implementation's of binary search, it would be very helpful. I think there were some blogs regarding this topic. I can't remember properly during round Div. 2 (703).

• -8

 » 7 weeks ago, # | ← Rev. 3 →   +6 There are twp ways of writing binary search that I know of. In the first, the current search range is represented by [lo, hi]. If m = (l + r) / 2, then we have two cases to continue to check: check towards smaller numbers: then we know the search range will be [lo, m — 1] (no need to include m again) check towards larger numbers: then we know the search range will be [m + 1, hi] As long as [lo, hi] is non-empty, we continue searching, so the condition in the while loop is while (lo <= hi).The other way is when the search range is represented by [lo, hi). For the two cases above, they can be represented as: check towards smaller numbers: new range will be [lo, m) check towards larger numbers: new range will be [m, hi) As long as [lo, hi) is non-empty, we continue searching, so the condition in the while loop is while (lo < hi).You can use whichever implementation that you wish, but you have to make sure that you set the boundaries to the correct values.
•  » » 7 weeks ago, # ^ |   +3 $Thanks For Replying!$ I think this is a typo $[m + 1, hi)$ $->$ $[m, hi)$ Now, I have a question let's say I want to find the max element $<= x$ . Now if I write something like this. int l = 0, h = n; while(h-l > 1) { int m = (l+h) >> 1; if(a[m] <= x) l = m; else h = m-1; } return l; Now, as far as the invariant and my small brain goes, $l$ should always hold element $<= x$ and $h$ should hold element $> x$. But as we are assigning $h = m-1$. There may be a case where m is at the boundary(first element $> x$). Then $h$ would have been assigned $<= x$ value which destroys the invariant. And also when to use $while(h>l)$ or $while(h-l>1)$ is there any difference ? $while(h>l)$ was not working so I used $while(h-l>1).$ My usual implementation looks like this. I modify $l$ and $h$ accordingly to the problem. Is there anything wrong in this ? int l = 0, h = n; while(h-l > 1) { int m = (l+h) >> 1; if(a[m] <= x) l = m; else h = m; } return l; Damn, this $latex..$
•  » » » 6 weeks ago, # ^ |   0 Yes, that was an error on my part. I have fixed it.The way that I usually do binary search uses an ans variable which is initially set at -1. If the current value of m is "valid" (<= x in this case), then we can set the variable value at this stage. By doing this, you can deal with the case that there is no solution within the search range. I'm not very familiar with setting both l and h to m, but it at least does not work conceptually with the idea that [l, h) is the current search space (however there could be another analogy where it would work, and it if you have used it before, then why doubt that it works). However, this is why I am unsure about the boundary condition. Thinking about it, it looks like it represents the interval (l, h), which matches the while loop condition as well. However, your starting values wouldn't work then.
 » 6 weeks ago, # | ← Rev. 2 →   0 Try this. Sort the second list (based on D). Create a list of prefix sums for the amount of gold you accumulate. Then binary search the second list for the index that is greater than A[i]. Once you find this, you know that you can get the gold from every station lower than this. Subtract one from the index and take the prefix sum of gold and output it. Check my submission for reference. Note that upper bound and lower bound are useful STL functions, and you don't need to implement a binary search if you are unsure about implementation.https://codeforces.com/contest/1184/submission/109112662