Bro great video bruh! If the input is sorted: Two Pointers Binary Search If there are multiple sorted sequences or the input is sorted: Binary Search If the problem involves brackets or arrangements: Stack Queue If asked to find the longest, smallest, shortest, etc.: Backtracking ipdi entha problem enna method use panalam numsollunga bro it would be helpful
Nice hair cut. All the best for the tournament man, stay healthy. Try incorporating meditation as well. Happy to find a person whose thought process is similar to mine like coding, financial literacy, physical fitness. The content was very useful as always. Please try to improve audio, video and thumbnail as well. With the quality of content you provide you will reach much more audience. If possible make a video on freelancing. Cheers !
bro but sorted ah iruntha thaana binary use panna mudiyum if unsorted array question na then sort pannitu binary search panna TC athigam aagume ithukku enna solution
Yeah sometimes data set from the question has to be sorted but for the square root of an number the solution space is sorted right , there you can apply binary search , similarly in number comes only once ,there solution space indexes are sorted , so you gotta think in terms of solution space In short solution you are looking for the problem lies in a range then you can do binary search there and get your answer faster Kindly see the video again and try to understand this point its very important
Hi bro,I've been looking for a job in a top MNC in Accounts and Finance for one year, but I feel exhausted from the search. Could you make videos on how to select an MNC?
array = 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 8 8 target =7 step 1 : mid element start = 0 end =17 = 0+17/2 mid = 8 array of [8] = 3 which is lesser than target [7] so increase pannanum i mean right side pakanum. step 2 : 8+17/= 12 mid = 12 arr[12] = 5 which is not equal to target so again increase panrrom so apa 12 - 17 kulla than answer irukkum again right side pakurom. step 3: 12+17/2 = 14 mid = 14 arr[14] =16 again increase it step4 14+17/2 = 15 mid = 15 arr[15] = 7 target found
@@jaya_bhuvanesh okay bro If mid is even: If arr[mid] == arr[mid + 1], the odd element is on the right side. Otherwise, it's on the left side. If mid is odd: If arr[mid] == arr[mid - 1], the odd element is on the right side. Otherwise, it's on the left side. correct ah bro.
Thala ; 1. If mid is equals to it's previous element, Start_point= mid+1; else End_point = mid - 1; Thala mid element athoda previous element kooda pair sh irunthuchuna, apo athuku munnadi entha unpaired element um illa, so naama RIGHT move panniklam, apdi illana ithoda pair pinadi irukuna, apo ithuku munnadi tha antha un_paired element iruku. Ithu correct ah thala ? pathutu sollunga.
Bro great video bruh!
If the input is sorted:
Two Pointers
Binary Search
If there are multiple sorted sequences or the input is sorted:
Binary Search
If the problem involves brackets or arrangements:
Stack
Queue
If asked to find the longest, smallest, shortest, etc.:
Backtracking
ipdi entha problem enna method use panalam numsollunga bro it would be helpful
More or less correct but yeah we will cover these things in deep
Anna neenga sollikudukura method easya irruku continue pannunga and time and space complexity pathi oru clarify kodunga 👍👍
Nice hair cut. All the best for the tournament man, stay healthy. Try incorporating meditation as well. Happy to find a person whose thought process is similar to mine like coding, financial literacy, physical fitness. The content was very useful as always. Please try to improve audio, video and thumbnail as well. With the quality of content you provide you will reach much more audience. If possible make a video on freelancing. Cheers !
What tournament bro?
Clear explanation, keep sharing your knowledge.
Happy to meet you again jaya bhuvanesh
❤😊
Best of luck bro 👍
Welcome back ↩️ thala
Nice video bro . waiting for coding explanation 🙂
Bro ithemari example oda ella algorithm kum podunga bro❤
Dsa padikka aasa..but..enga eppadi start panna nu theriyala..dsa ku munnadi java padikava venama..
bro but sorted ah iruntha thaana binary use panna mudiyum if unsorted array question na then sort pannitu binary search panna TC athigam aagume ithukku enna solution
Yeah sometimes data set from the question has to be sorted but for the square root of an number the solution space is sorted right , there you can apply binary search , similarly in number comes only once ,there solution space indexes are sorted , so you gotta think in terms of solution space
In short solution you are looking for the problem lies in a range then you can do binary search there and get your answer faster
Kindly see the video again and try to understand this point its very important
Hii bro can you please let me know about scaler devops course?
Bro freshers epdi job vangurathu
vaa thalaiva vaa thalaiva vaa
System design learn panrathuku nalla youtube channel sollunga from scratch lenthu start panrathuku
Hi bro,I've been looking for a job in a top MNC in Accounts and Finance for one year, but I feel exhausted from the search. Could you make videos on how to select an MNC?
I have already added a video reg this , kindly check
@@jaya_bhuvanesh thanks a lot for ur response
I will check
There is any eligibility criteria to join Scaler
No
Java learn pannurathukku nalla platform solluga any TH-cam channel
From scratch la irunthu start pannurathukku
Check Ravindra Babu Ravula java playlist
@@jaya_bhuvanesh thankyou
But athi OOPS consept mattum thaa irukku
Bro can u do C programming please 🙏 i can understand your teaching
Check out my DSA playlist.
array = 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 8 8
target =7
step 1 : mid element
start = 0 end =17
= 0+17/2
mid = 8
array of [8] = 3 which is lesser than target [7] so increase pannanum i mean right side pakanum.
step 2 :
8+17/= 12
mid = 12
arr[12] = 5 which is not equal to target so again increase panrrom so apa 12 - 17 kulla than answer irukkum again right side pakurom.
step 3:
12+17/2 = 14
mid = 14
arr[14] =16 again increase it
step4
14+17/2 = 15
mid = 15
arr[15] = 7 target found
No here the question is to find odd one out element which occurs only once we dont know what that element is
@@jaya_bhuvanesh
okay bro
If mid is even:
If arr[mid] == arr[mid + 1], the odd element is on the right side.
Otherwise, it's on the left side.
If mid is odd:
If arr[mid] == arr[mid - 1], the odd element is on the right side.
Otherwise, it's on the left side. correct ah bro.
Right uhh vilayadumpothu shoe va pathukonga
Engineering aspirants can only get high package comparable to BSC BCA aspirants tell
No not like that!
Thala ;
1. If mid is equals to it's previous element,
Start_point= mid+1;
else
End_point = mid - 1;
Thala mid element athoda previous element kooda pair sh irunthuchuna, apo athuku munnadi entha unpaired element um illa, so naama RIGHT move panniklam, apdi illana ithoda pair pinadi irukuna, apo ithuku munnadi tha antha un_paired element iruku.
Ithu correct ah thala ?
pathutu sollunga.
You are thinking in the right way but think how index number will change based on the element occurrence
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