Search element in a circular sorted array

[@wonderInNoWhere]

You are a legend. Even almost ten years later, your explanation is the best.

[@AruneshSrivastava]

Before watching this , i watched 2 other videos on similar issue , dint understand a damm thing , then i just see this video's link on suggestions and i smiled . The complexity of the problem doesn't bother you if you have a good teacher. Thanks man

[@findoc9282]

as I keep learing and comparing I realize how energy saving it can be with a good teacher/tutotial

[@excitingmonkey3970]

Man you are awesome! I was Leetcoding, got stuck in understanding the solution and searched for explanation. And here i am , mindblown by how simple you made me understand.

[@RaoVenu]

Thanks for the great tutorial. My implementation (assuming we already have implementations for findRotations and binary search in place) CircularArraySearch(A, n, x) { lowIndex = findRotations(A, n) // Index of lowest value in circular sorted array O(lgN) //Now we have two cases. The array to left of lowIndex is sorted. And the array to the right(including lowIndex is sorted) if (x == A[lowIndex]) return lowIndex // O(c) else if (x > A[lowIndex]) return BST(A, lowIndex + 1, n, x) // O(lgN) else return BST(A, 0, lowIndex - 1, x) //O(lgN) }

[@S__Arslaan]

Amazing!!!No other youtube channel matches your level of explaination!!!!Thanks mycodeschool

[@Aggarwal_Anshul]

Such an Efficient & Simple Solution to a tricky problem! Fantastic explanation BIG THANKS!

[@mycodeschool]

Thanks !

[@Guillermorosales]

Thank you very much!!! for this series of videos, you explained it very well.

[@Joyddep]

Exactly what I was looking for. Thank you for your excellent content!

[@StartDeutsch]

Hi! What do you use for writing? A tablet or what? Looks very good! Thank you

[@yunususmani3294]

Can't thanks enough...... You just saved my night :) Thanks a lot :)

[@SaumyaSharma007]

Thank you so much....The evergreen lectures 🌟👌

[@codingandmathvideos]

great tutorial! you simply rock!!! God bless you. You probably should adjust the way you calculate the middle index. If you add 2 big positive numbers or 2 big negative numbers for that matter, you can get overflow. Use mid = low + (high - low)/2 instead to get the index of the middle element.

[@AbhijithVMohan]

In typical system, overflow will only happen if the array has about 2**30 integers. In that case, the array itself will take ~4 GB space.

[@saiadityagedda]

He already mentioned about that in his initial lecture

[@S__Arslaan]

not if u use python

[@saileshverma1942]

Thank you for your excellent teaching

[@siddharthasharma9316]

as you mention that a circular sorted array with Distinct elements in description part of this problem , we can also skip equal sign- Array[mid] < Array[high] and same for Array[mid] > Array[low].

[@siddharthasharma9316]

and thanks for such nice explanation.

[@adityagupta6338]

Hey, thanks for this. It really helps. I want to just verify my approach is correct. How about we find the pivot using the previous tutorial, figure out which sorted array it belongs to (left of pivot or right of pivot) and apply binary search to whichever direction of pivot x belongs to? The time complexity is still 0 (log n), right?

[@allanbee5752]

Hi, Firstly, you are doing amazing job. Your teaching skills are superb!! Can you please share your algorithm approach to print all the permutations of a string? String length could be any length. e.g. abc, acb, bac,bca, cab,cba.

[@swapnanilsharma]

#include void swap (char *x, char *y) { char temp; temp = *x; *x = *y; *y = temp; } void res(char *A,int m,int n) { int i; static int num=1; if(m==n) { printf("[%d] ",num); printf("%s",A); printf(" "); num++; } else { for(i=m;i

[@nouralaaeldinmounirhashem8570]

Thanks for your great effort You are an excellent teacher

[@bharathrajakumar1823]

Simple and easy to understand. Thank you !!!

[@jalajkhajotiaiitr]

Thanks for your work sir, it is really helpful

[@bhoopendrasharma9474]

Excellent tutorial, Thanks a ton. I believe this will work for duplicates as well, especially for the example shown in the tutorial { 2, 2, 2, 2, 2, 0, 2, 2 }. I tried it and found working.

[@stym06]

best explanation ever.

[@yuvarajravi7849]

Great tutorial. Other approach is just keep comparing first and last elements. Say if i=1, j=n-1, perform if a[i]

[@harshbaliyan5867]

O(n) complexity still

[@piyushjindal4955]

simple and nice explanation :)

[@vramali1]

I am finding it hard to understand the code. I think i need a lot of example use cases for each of the conditions applied in the code.

[@salemouail627]

Write it write it write it

[@nobody2937]

Thank you very much... Thank you ...

[@sawyerford2191]

How you were able to return the index value without returning it

[@saisudheermittapalli574]

Looks like you are not returning any value from CircularArraySearch() function in success case (when x is part of array)

[@unknownman1]

mid is returned

[@401varun]

Very nice.. this code is for ascending array rotations.. but doesn't work for descending array rotations... kindly generalize for both.. working for : {4,5,1,2,3} ---> initial array {1,2,3,4,5} rotated twice right . not working for : {5,4,8,7,6} --> initial array {8,7,6,5,4} rotated twice right. Thanks :)

[@kiranmahesh93]

same here

[@dangidelta]

Just change every '' and vice-versa. It should work just fine.

[@prs314]

have you considered the possibility of (low + high) overflowing in "int mid = (low + high) / 2"? Since int is signed the (low + high) will give an undefined behavior in C++ or a negative number in Java. I think the better way would be to cast them to unsigned int and shift. int mid = ((unsigned int) low + (unsigned int) high) >> 1;

[@adarshverma3372]

One of the variation can be achieved using below code: private static int searchNumberInCircularSortedArray(int []inputs, int numberToSearch, int startIndex, int endIndex) { if (startIndex > endIndex) return -1; int mid = (startIndex + endIndex) / 2; if (inputs[mid] == numberToSearch) return mid; if (inputs[mid] = numberToSearch) return searchNumberInCircularSortedArray(inputs, numberToSearch, mid + 1, endIndex); return searchNumberInCircularSortedArray(inputs, numberToSearch, startIndex, mid - 1); } else { if ((inputs[mid] > numberToSearch && numberToSearch

[@krushnnabaviskar2170]

Completed playlist

[@sdhupar]

If array has duplicate elements then we can still apply Binary Search

[@WhisperingWalnuts]

saurabh Dhupar yes, i tried the code by giving array with duplicate elements as input, and it works!!! Simple and great algo!!

[@artyzach]

but wouldn't it only show the index of the first occurrence of duplication?

[@dangidelta]

A recursive approach : int CircularlySorted(int Array[ ], int first, int last, int value){ if(first > last) return -1; int m = first + (last - first)/2; if(Array[m] == value) return m; else if(Array[m] = Array[first]){ if(Array[m] < value) return CircularlySorted(Array, m+1, last, value); return CircularlySorted(Array, first, m-1, value); } else if(Array[m] = Array[first]) return CircularlySorted(Array, first, m-1, value); return CircularlySorted(Array, m+1, last, value); } }

[@Sohanbadaya]

public int searchItem(int arr[], int item){ int low = 0; int high = arr.length-1; while(low arr[mid] && item

[@srkrohit]

+Sohan badaya The authenticity of this code is only valid if the segment is sorted otherwise not. "if(item > arr[mid] && item

[@Sohanbadaya]

+Rohit Kumar You are right Rohit. Thanks :)

[@omkarkulkarni3207]

// Search an element in rotated/circular array. public static int findElementInCircularArray(int[]a, int n, int x) { int pivot=findRotationCount(a,n); if(x==a[pivot]) { return pivot; } if(x> a[0]) { //Search in sub-array left of the pivot return binarySearchRecursive(a,0,pivot-1,x); } else { //Search in sub-array right of the pivot return binarySearchRecursive(a,pivot+1,n,x); } }

[@yashtailor1543]

love you @mycodeschool

[@mayuraggarwal5840]

This solution will not work if the element to be searched is present in unsorted part of the array.

[@Hack_Neuron_To_DSA]

Wrong!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!.

[@Hack_Neuron_To_DSA]

subscribe, share my channel and comment on them, then I would explain in depth how mine is correct

[@tejaskadam1889]

Hi please remove this videos this solution is not correct at all

[@psn999100]

Why not ? This exact implementation passed all test cases in Leetcode leetcode.com/problems/search-in-rotated-sorted-array/