BS-7. Find out how many times array has been rotated

We have solved the problem, and we have gone from brute force and ended with the most optimal solution. Every approach's code has been written in the video itself. Also, we have covered the algorithm with intuition.

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0:00 Introduction of Course

[takeUforward]

Please comment understood and give us a like if you got everything :)

[pranavmisra]

Learning platforms appear, reach their peak and disappear, but takeuforward is going to be evergreen. The quality, the hardwork and the amazing way he teaches is unmatched.

[vaishnavejp]

used a different approach. it goes to the side that is unsorted(as that will have the pivot element). it keeps doing that until nums[low] < nums[mid]. the moment we find that, we find our pivot value.

int findKRotation(vector<int> &nums) {
// Code Here

if((nums[0] < nums[nums.size()-1]) || (nums.size() == 1)) return 0;

int low = 0, high = nums.size()-1;

while(low <= high) {

int mid = (low+high)/2;

if(nums[low] > nums[mid]) high = mid - 1;
else low = mid + 1;

if(nums[low] < nums[mid]) return low;
}

return 0;
}

[dilipkumarbk]

Don't know how to thank you for your contribution to the community. I just did this problem by my self that I was able to use the previous question finding the minimum in a rotated sorted array.
When I saw the question.
I just remembered the line,
"DON'T DIRECTLY JUMP TO THE SOLUTION, FIRST GIVE IT A TRY!"
Thanks a lot striver bhaiya. I am feeling confident now, that I can understand the pattern in the problem.

[cinime]

Understood! Wonderful explanation as always, thank you very much for your effort!!

[Manasidas]

Help me a lot thank you sir It is the biggest YouTube channel I follow every time on the channel. I learn lot .

[SuvradipDasPhotographyOfficial]

Understood, awesome striver keep going, eagerly waiting for remaining videos

[manavsingh]

Thank you Striver...Understood everything

[inderjeet]

Bro we need a playlist on Bit manipulation also . Please consider this request.

[KCOYASH]

understood :) , please continue such great videos

[gurpritsingh]

a tricky thing to spot here is whether the rotation is clockwise or anti-clockwise, in this problem we considered anti-clockwise rotation.
the final rotated position can take a different no of steps depending upon the type of rotation.

[Gaurav_Tripathi_]

understood, wonderful explaination Bhaiya..!!

[visase]

Thanks Striver. Adding my thoughts for (duplicates) .
If we apply the previous logic of high-- or low++ as the mid and low/high values are equal, we will end up getting the minimum element but that does not gaurentee the no of times array has been rotated .
Examples:
array=[1, 1, 2, 1, 1],
orignal array = [1, 1, 1, 1, 2].
The correct answer is 3 (as the element at 0th index has been moved to 3rd index [1, 1, 2, 1, 1]).
But if we apply the previous logic, the answer would come as 0 as 0th index is the minimum element.

To upsolve this, we can do the following :
Keep reducing high, untill [high-1] > [high] (2>1).
Once you attain this point, high(3rd index) will be the answer.

[Anshydv]

bhaiya aap, , sbko knight and gaurdian badge dil ke rahoge, Leetcode pe ❤❤