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.

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

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

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

Amazing 😀😀😀

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

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)

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.

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 🙏🙏

You are simply superb sir Awesome explanation

great explanation....thanks a lot sir

nice

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))

wonderful lecture.

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

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

classes are in not the correct order please correct it!

You are explaining code directly It is hard to understand

low sound.

Low sound boring ,speak loudly sir

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

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

not running in python mrcf

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

sound quality is not satisfactory