BS-10. Finding Sqrt of a number using Binary Search

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[takeUforward]

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[varun]

The check "mid * mid <= x" may lead to integer overflow. instead, by dividing mid on both sides, we can write "mid <= x / mid"

[piyushraj]

Had been desperately waiting for the series to restart. Thanks a lot.

[KunalSingh-mf]

i never thought this could be the solution for finding sqrt ....mind blowing

[suyashmishra]

Understood!!
Here are the time stamps:
00:00 - 00:15 Intro
00:16 - 1:15 -Problem description
1:16 - 4:00 - Brute(linear)
4:01-5:20 - Binary search intuition
5:20 - 16:16 - Optimal solution
16:17 - 16:45- code
16:45 - 17:10 - outro

[Dipanshutripathi]

Hats Off to the content and it's creator ! Understood each and every part of the video.

[mayankshakya]

Correction at time 12:45 the time complexity of the brute force solution is not O(n) as suggested by striver but it is O(sqrt n) according to me as the loop will break after that condition always

And plz striver bhaiya vidios jaldi jaldi daal do bhot wait kar liya

[jasrajsehmbey]

You can also take high as n/2 rather than n intially as it is certain that ans will lie in a range of 1 to n/2

[adarshkumarrao]

Excellent explanation ❣.. Waiting for linked list series💛💛

[jfzodgb]

Well explained!! I finally understand why the answer converges to "right" especially from 12:00

[pushankarmakar]

dada, cant we optimize it further by reducing the search space. we can keep high as n//2 initially cause a square root of a number can never be greater than half of the number.

[Manishgupta]

Good to see from brute force approach to optimal approach.

[sohamchatterjee]

I aam doing dsa from tha last 6 months and i found u yesterday and now i am a fannnn...

[iWontFakeIt]

first naive approach had TC of O(sqrt(n)) but binary search has TC of O(log(n)) which is eventually slightly better than former one for very large values.

[mihirkumar]

int mySqrt(int n) {
if(n==0)return 0;if(n==1)return 1;
int ans=-1;
int start=1;int end=n/2;
while(start<=end){
long mid=start+(end-start)/2;
if(mid*mid<=n){
ans=mid;
start=mid+1;
}else{
end=mid-1;
}
}
return ans;

}

[kiransoorya]

Bhaiya why can't we use high as n/2 instead of n at start in the function(instead of 28 we can use 14 ) 13:14

[impalash_ag]

Hi Raj,

I think the solution can be bit more efficient with the TC of O(log n/2) instead of O(log n). Since, we know square root of any number n can't be bigger than half of that number i.e, n/2 therefore we would take high as n/2 instead of n. Here's the code:

class Solution {
public int mySqrt(int x) {

if(x == 1)
return 1;

int low = 1, high = x/2;
int result = 0;

while(low <= high) {
int mid = low + (high-low)/2;

if((long)mid*mid <= x) {
result = mid;
low = mid+1;
} else
high = mid-1;
}
return result;
}
}

[Anushka]

we can further optimise this by taking range of only x//2 because the square root number will be smaler than that except for x = 1 that case can be covered separately

[SuvradipDasPhotographyOfficial]

amazing explanation striver, understood.

[subhankarkanrar]

Respect to Striver ❤ for such a easy explanation