Search an element in a sorted and rotated array | GeeksforGeeks

This video is contributed by Harshit Jain.

[akshaydwivedi]

you are just making a simple algo sophisticated by the way you explain

[tcobvaibhavwable]

can we have more vedios on DE Shaw interview preperation

[amanjaiswal]

7:28 high < low can any one explain what does it actually mean?

[vaishnavigoteti]

Great Job Geeksforgeeks! this video is certainly not for the beginners, but for intermediate level programmers. The algorithm was very clearly demonstrated.

[sainathpatil]

in the solution on last line return search(arr, l, mid - 1, key); -- is it required ?

[pman-codes]

How can you pass 0 and n-1 as low and high and compare them with the mid where the array is pivoted in the first place?

[shwetadk]

Giving an example would definitely be better but your explaination is very good, much better than just reading the text. Thanks. Keep making such videos.

[raghavchadha]

clear it with some examples ...reading algo we will do it ourselves, ,

[satyamgupta]

Why are we checking from first element to key element ?
How ??

[subham-raj]

*This is what happens when you memorize*

[akankshagupta]

You just read the algorithm that's it, there is no use of any of your videos. This is very disappointing from your side "geeksforgeeks"

[saikumarmadhadi]

finding pivot will itself take O(n) in worst case, so i guess linear search will also be a considerable option

[ragas_]

GfG should just give up this stuff. Extremely confusing

[spicytuna]

finding pivot code is too complex. prone for an error. if traversing every element is O(n), use easier method to get a pivot.

[alperozdamar]

So disappointing. Without explanation reading code line by one doesn't make any sense. Everybody can do it! You need to add clear explanations and examples.

[hemant_malav]

Isse Accha aap Hindi me explain kare to jada acche se samaj aayega

[karimfarhouti]

while the array is sorted we don't have to do any search i think, here's a simple fun
int fun(int arr[], size_t t_size, unsigned int n)
{
if (n <= t_size / 2)return n==arr[(t_size / 2 + n)-1]?n:-1;
else if (n == t_size) return n==arr[t_size/2]?n:-1;
else return n == arr[(n - (t_size / 2)) - 1] ? n : -1;
}