improving naive gcd

[@ANAND-zw1on]

We can further simplify this by starting the loop from min(m, n) /2 to 1 and seperate sentence for testing the min(m, n) . Eg: if we take 100 and want to know the factors, first 100 is a factor, then there is no factor from 99 to 51. So we can skip that part completely. This is true for every other numbers.

[@tavilefty]

exactly man. this comment needs to be pinned.

[@girish8579]

What about odd numbers?

[@mrtalksick6178]

@@girish8579 instead of dividing by 2, take the quotient of the division i.e. min(m, n)//2.

[@abhishek-wt5xm]

i dont know why people are getting difficulties in understanding but i would must say i there cant be an easier explanation. really a great resource for understanding the algorithm syntax and know the best solution you also explained the way to decrease the time and ans space complexity of the algorithm thankyou

[@coldcoke9254]

probably because it s the first lecture and we are already using concepts of functions and loops and lists...I'd recommend watching a one shot lecture on python by codewithharry and then coming to this that's what I did and now all this seems really simple...don't even need to watch all of it only till chapter 8.

[@ruturajjadhav8905]

def gcd(num1,num2): return max(list(set([i for i in range(1,num1+1) if num1 % i == 0]).intersection(set([i for i in range(1,num1+1) if num1 % i == 0])))) ## intersection : combines the common factors of given two sets ##just two lines

[@SlickShitt]

after intersection you used the first parameter instead of num2

[@adityamehra7238]

My brain literally exploded after realising there was 1 another way to write code for this gcd...... Meanwhile professor : you said 1? Wait hold my mike...... 💥💥... A beautiful lecture

[@sameers8510]

You are simply superb sir Awesome explanation

[@coldcoke9254]

probably because it s the first lecture and we are already using concepts of functions and loops and lists...I'd recommend watching a one shot lecture on python by codewithharry and then coming to this that's what I did and now all this seems really simple...don't even need to watch all of it only till chapter 8.

[@ashishkumarsharma1323]

def gcd(m, n): i= min(m, n) while(i>0): if (m%i == 0 and n%i== 0): return(i) else: i=i- 1 x=gcd(21, 9) print(x)

[@mandarchincholkar7250]

Marvelous logic... Amazing... 👏👏👏👏🎉🎉🎊🎊🎊 only experienced programmers will get it..🔥🔥🔥

[@panthrohit]

Not an experienced programmer, still I get it... :)

[@mandarchincholkar7250]

@@panthrohit then you nailed it hardly.. 😎😎😎😎

[@panthrohit]

@@mandarchincholkar7250 This reminds me of the saying 'working hard or working hardly'. Whats my takeaway here....🤔😛😬

[@mandarchincholkar7250]

@@panthrohit bhai baat ko mat ghumaya kar.. 😂😂😂😂😂

[@panthrohit]

@@mandarchincholkar7250 😉

[@Jaimin_Bariya]

Jp here again [comment number 51] Thank you, sir, :)

[@bhargavdobariya5859]

great explanation....thanks a lot sir

[@mritunjay3723]

A one line code -- print(next(a for a in range(63,0,-1) if 8888%a==0 and 63%a==0))

[@anshulojha5699]

The algo that u explain if I will learn that methods to minimize the all codes with time and memory then I will be the rank 1 in Google contests 🙏🙏

[@sravanchilumula5213]

classes are in not the correct order please correct it!

[@arpitaingermany]

The shorter python program was confusing min(m, n) this part

[@vatsalgupta00]

Divisor of a number cannot be larger than the number. Since we are trying to find common divisors, only divisors of smaller number are suffcient.

[@DineshJunjariyaBEE]

min(m,n) gives least of the two numbers

[@anshulojha5699]

Like min(6,8) so dono me vo common no hoga vo 6 se thoe hamesha less hi hoga like 2 here . That's why they used this concept ✅

[@vatsal_gamit]

def gcd3(m,n): return max([i for i in range(1,min(m,n)+1) if ((m%i)==0 and (n%i)==0)]) print(gcd3(3575,978))

[@BiswajitDas-lk7pp]

Amazing 😀😀😀

[@tavilefty]

wonderful lecture.

[@bhupiagra1]

you said actually any common factor must be less than min(m, n) but how this statement is justifying gcd of (2, 4) ?

[@tabassumhebbalkar]

including the minimum of both.

[@panthrohit]

Must be less than or equal to min(m,n)

[@DineshJunjariyaBEE]

he meant min or less than min

[@sanjayguptacg486]

low sound.

[@prashnatdivase8523]

def gcd(m,n): fm=[] # list down factor of m in fm fn=[] # list down factor of m in fn for i in range(1,m+1): if (m%i ==0): fm.append(i) print("fm is",fm) for j in range(1,n+1): if (n%j ==0): fn.append(j) print("fn is",fn) cf=[] # for returning largest value for k in fm: if k in fm: cf.append(k) print("The largest GCD is: %d "%cf[-1]) gcd(10,8) above programs give correct output but i would like to understand why below program is not giving same output: def gcd(m,n): i=min(m,n) while i>=0: if m%i==0 and n%i==0: return i else: i=i-1 v=gcd(10,8) print(v) Sorry if i am missing anything and would be great if someone explain last optimize example in details

[@coldcoke9254]

the code works just fine tho, just chane

[@venkatashasidharreddychali7485]

give the brackets correctly, it worked for me, I alos got the wrong output when i used this (m%i) ==0 & (n%i) ==0 after tweaking it to this ((m%i) ==0) & ((n%i) ==0), i got the correct output

[@prashnatdivase8523]

@@venkatashasidharreddychali7485 thank you

[@exploringlife2904]

not running in python mrcf

[@izharkhankhattak]

nice

[@gauravjoshi7589]

i am not getting any error and also not getting any output

[@pratapkhandekar8322]

You are explaining code directly It is hard to understand

[@sagarsagar-xr7bo]

Low sound boring ,speak loudly sir

[@sumitbhattacharyya2058]

sound quality is not satisfactory

[@pk15692]

You can use a volume booster for chrome to enhance the sound volume

[@sagarsagar-xr7bo]

@@pk15692 try chesaa ,no use

[@LAVETI-28]

Use ear phones