others were explaining the same approach but your explanation stuck in my mind now i will never forget this . Thanks (bring a dsa sheet solution one -two question a day) it will be great
Bhaiya ek chota point mention kijiye ki in i*(N/i) , the value of N/i is calculated first and is rounded to the integer eg if N/i is 6/4 = 1.5, so it becomes 1. Ye point thoda miss hojata hain. In learning DSA and doing problems we forget basic c/c++ programming roots. Explanation is Very good.
others were explaining the same approach but your explanation stuck in my mind now i will never forget this . Thanks (bring a dsa sheet solution one -two question a day) it will be great
thank sir, you discussed all the possible approaches
Bhaiya ek chota point mention kijiye ki in i*(N/i) , the value of N/i is calculated first and is rounded to the integer eg if N/i is 6/4 = 1.5, so it becomes 1. Ye point thoda miss hojata hain. In learning DSA and doing problems we forget basic c/c++ programming roots. Explanation is Very good.
Hi bro hope doing well, could u plz continue❤❤❤
Yes bro will continue
Nice please keep this up it really helps if we get stuck in our POTD
and it also saves us from breaking our long streak
class Solution
{
public:
long long sumOfDivisors(int N)
{
long long sum=0;
for(int i=1;i
kahan ho bhai...aaj kal...we need you...more Graph questions please...!!!
Sure amaan will continue.
nice explanation
Op
hi
Hi bro