First And Last Position Of An Element In A Sorted Array | FREE DSA Course in JAVA | Lecture 54

This question has been asked in the interview of Facebook, Google, and Microsoft and is a Leetcode problem number 34.
The question reads - Given an array of integer numbers sorted in ascending order, find the starting and ending position of a given target value.
if the target is not found in the array return [-1,-1].
You must write an algorithm with O(log n) runtime complexity.
Now a simple brute force approach will be to run a loop from the first index and return the index when the element is found first for a first position or starting position.
And, for the ending or last position run a loop from the last index.
But in this method, we will compromise with the time complexity given.
So one thing which is very clear is that we have to use the binary search algorithm for this program.
Let's see how a binary search approach with little changes can give the desired results as asked in this program.
Subscribe to our channel for regular updates on the dsa course and click on the bell icon to never miss an update from our dsa course.
Data Structures and Algorithms Free Course (Learn DSA Without Paise) Playlist - • Data Structures And Al...
For more information, fill this form: forms.gle/8eiUmM92Fx563Aen9
or call us at 8884881203
Facebook: / thetapacademy
Instagram: / tapacademy_online
Linkedin: / 73820805
Website: www.thetapacademy.com
#coding #dsa #dsacourse #java #javainterview #array #binarysearch

[@bhavanimohan6809]

You are such a genius made me understand very clearly when ever I have doubt on DSA I just immediately landed on your channel.

[@arjungillela9565]

Sir, You made it this simple, directly entered to the brain, Thank you

[@TAPACADEMY]

You are very welcome. Do subscribe to the channel.

[@krishnavar7219]

the way you explained this challenge...it is genius

[@bun_bun17]

you guys make the best effort, you are great

[@musicallymad9]

SUCH A WONDERFUL EXPLAINATION BY YOU SIR. THANKS A LOT SIR. THE WAY YOU EXPLAIN EVERY PROBLEM IS JUST AWESOME😍

[@UmaGowda1232]

Superb explaination sir 💥

[@anand1557]

very nice and simple explanation instead, easy to understand

[@aaviff1810]

Best explanation sir thank u so much❤️

[@sohamsonwane1534]

best explanation sir

[@Ashritha_purna]

Great lecture

[@garimachawla775]

what an explaination wowwwwww

[@BilalKhanBaraich]

Given a list of sorted integers and a ‘key’, find and return the index of the first integer that is not less than the ‘key’. If no such integer exists, return -1. The complexity shall be O(logn). (Marks 10) Example 1: {1, 2, 4, 5, 5, 6}, key = 2, returned index 2; Example 2: {1, 2, 4, 5, 5, 6}, key = 3, returned index 3; Example 3: {1, 2, 4, 5, 5, 6}, key = 5, returned index 4; Example 4: {1, 2, 4, 5, 5, 6}, key = 7, returned index -1; (Note: In the example, the indexing starts with 1 and not 0.) Only PSEUDO-CODE is acceptable. Sir kindly Solve this question

[@lerinkuriakose]

Super🤩

[@pranalinaik2447]

Thank you so much😇

[@TAPACADEMY]

You're welcome 😊

[@srinivasareddymandem9204]

anybody clear me what if encounter if the first occurence at first index how should we approach we cant compare mid-1 because 0 is the last

[@avinashnoubade7272]

String problem please

[@krishnavar7219]

sir,paste code in chat or give patreon link

[@fariyabkhan8095]

class Solution { public int[] searchRange(int[] nums, int target) { int[] res={-1,-1}; int beg=0,end=nums.length-1; int mid=0; while(begtarget) end=mid-1; else beg=mid+1; } beg=0; end=nums.length-1; mid=0; while(begtarget) end=mid-1; else beg=mid+1; } return res; } }

[@aryanverma1233]

THIS CODE IS NOT WORKING