Lecture 23: Binary Search Different Problem || Search Insert Position || Sqrt(x) || Count occurrence

Binary Search
What is Binary Search
Binary search in C++.
1: Find First and Last Position of Element in Sorted Array

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[sanjanad]

I have tried a lot of But everyone is telling a concept verbally and directly write the Neither they dry run nor they tell how the loop is iterating.... 🥺❤ ..

[KaranKumar-bhij]

00 q1, 21:14 its code, 26:17 q2 40:47 its code, 42:00 sqrt question 1:00:26 its code 1:07:00 Homework questions

[mananbatra]

At 1:03:00, we can initialize "mid" as long long rather than int. This will resolve all errors automatically.

[rxi-geek]

Thankyou Rohit Bhaiya, you explain every topic very patiently and step by step. Amazing Lecture 🥰

[procrastinator]

sabkuch chamak ra bhaiya. Keep continuing the good work.

[hi]

36:46 we can directly print "first"
Because first will hold the possible index value of the not found element 👍

[sumitvishwakarma]

I was stuck on the sqrt(n) problem when you gave that in hw so when I checked the solution people were solving it using binary search but was not able to understand the code then but now
I think it will be made crystal clear by you bhaiya .
Thankyou bhaiya ❤❤

[atifmalik]

27:33
class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
int start =0, end =nums.size()-1;
int mid;
while(start<=end){
mid = start + (end-start)/2;
if(nums[mid] == target){
return mid;
}
else if(nums[mid]<target)
start = mid+1;
else
end = mid-1;
}
return start;
}
};

[thriveforsuccess]

bhaiya logic banana aur implement karna, vo chotey chotey points ko dimaag mein rakhna ab kaafi default sa lagney laga hai aapki videos dekhkar .... thank you for the efforts bhaiya

[tusharagarwal]

2nd question(search insert position) me normal binary search bhi use karr sakte hai bas end me -1 ki jagah start return karna hoga.

[narutodihargo]

00:00= Q1) First & Last position of an element in an Array, 04:25=When to use Linear search & Binary Search, 21:16= Leetcode platform,

[joydeep-halder]

Solution 5: Count zeroes ✅✅
int start = 0, end = n-1, mid, firstIndex = -1;

while(start<=end){
int mid = start + (end-start)/2;
if(arr[mid]==0){
firstIndex = mid;
if(mid==0 || arr[mid-1]!=0) break;
end = mid - 1;
}else{
start = mid + 1;
}
}

return firstIndex == -1 ? 0 : n-firstIndex;

[chinmayOP]

Bhaiya i watch your videos daily with ongoing college and it will definitely give me an edge in Future so thank you and keep this going❤️

[Jv_Singh]

00:01 Find First and Last
26:08 Insert Position
43:22 Sqrt(x)

[ManishKumar-dbj]

Ram Ram Rohit bhai series bahut baddiya chal rahi hai aur sab kuch chamak raha hai apki wajah se best DSA corse on the internet thanks for the GOAT content man love&&respect++;

[RonakAnand-gqur]

1:03:57 we can also do long mid = start+(end-start)/2

[SunnyKadian]

In Every video there is something new to learn. ❤❤❤❤

[joydeep-halder]

26:03 Optimization to Find First and Last Position of Element in Sorted Array 🚀🚀

We can check the adjacent left or right element matching or not, and this way we can avoid unnecessary search in left and right subarray.

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int firstIndex = -1, lastIndex = -1;
int start=0, end = nums.size() - 1, mid;

while(start<=end){
mid = start + (end-start)/2;
if(nums[mid]<target){
start = mid + 1;
}else if(nums[mid]>target){
end = mid - 1;
}else{
firstIndex = mid;
if(mid==0 || nums[mid-1]!=target) break;
end = mid - 1;
}
}

start = 0, end = nums.size() - 1;

while(start<=end){
mid = start + (end-start)/2;
if(nums[mid]<target){
start = mid + 1;
}else if(nums[mid]>target){
end = mid - 1;
}else{
lastIndex = mid;
if(mid==nums.size()-1 || nums[mid+1]!=target) break;
start = mid + 1;
}
}

return {firstIndex, lastIndex};
}
};

[abhaydeep]

Bhaiya i am in class 9th and your teaching methods are so good that i got to know every topic crystal clear . I am watching your videos from day 1 and i will definitely complete the 180 challenge
By Abhay Deep

[sheetalsharma]

Aapne youtube pr sbse best pdya ..ithe sare questions bhot ache smajae...