Binary Search in an Array | Java | ICSE ISC Computer
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- Опубликовано: 11 сен 2024
- What is Binary Search Technique ?
Binary search is an efficient algorithm for finding an item from a sorted list of items. It works by repeatedly dividing in half the portion of the list that could contain the item, until you've narrowed down the possible locations to just one.
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Kasam se sir 15 number pakke BINARY SEARCH to aayega hi ICSE 2024 MAI 😊
I wish it will come ❤
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Binary search was hardest to me but after this video it is now easiest to me
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WAP in Java to store 20 different number in SDA Now enter a number and search whether the number is present or not wearing binary search technique if the number is present then display the measure number is present otherwise display number is not present you have to give following trans index number
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Thank you!
Sir , why you have not use loop in this question ??
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Best for revision ❤
very nice video sir
Sir I have one doubt that if in question the elements of the array are not given then do we have to necessarily input it from the user or we create our own elements
nice explanation but a small suggestion is that you should have completed your example demonstration at the end of the video.
Sir can you please make a video on important programs for icse board exams 2020 .
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sir doubt h ek jo ye lst m apne pos likha h woh pos + 1 hoga kya ?? apne linear search m bola tha position nikalni h toh index +1 krna h ki ?...
Watch on 12/3/2024
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Bahut vdyaa sir .keep it up!!!
At 16:54 pos should be printed as pos +1 because pos is initialised as -1 in beginning.??? Alok bhaiya please explain....
Yes, you can print pos + 1 because indexes starts from 0
And pos taken as -1 in the beginning, to ensure if we don't get the result, pos will remain -1 and we can print element not found
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Sir aap indeed bahut acche ho aur aap bahut aachha padhate ho no doubt.maine aapki bahut saari videos dekhi hai and all videos clears my concept but sir mujhe ye binary search samajh nahi aa raha😢😢😢 will u help please 😢😢
Sir i have confusion after e=mid-1 we don't have to start loop to check whether the search number is lying less than mid or after mid+1?? Plz reply
sir, if(n==a[mid]) true ho jae,
to kya
pos=indexOf(mid);
likh sakte hain kya?
mid is already an index position.
You will just write pos = mid;
@@AmplifyLearning ok sir, thank you!
Thank you sir!! Understood everything 😀
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tq sir loved this session
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16:57 tough nhi hai sir
All the things are clear in this
Nice video
👍