GFG POTD: 21/10/2023 | Sum of All Divisors from 1 to n | Problem of the Day

[@impalash_ag]

Sounds like a lecturer of a college rather than an engineer at a company

[@garvgoel1743]

true

[@Rajesh-fj5hm]

true, he himself says something else, write something else

[@savitore]

Harmonic lemma is not used

[@akhilesh_14_]

easy to understand and better solution thanks bhai

[@VivekSharma-sk3vp]

really nice explanation bhaiya!!

[@GauravKumar-qe7iu]

what if we only want the sum of divisor of MAX k element. how can we do that in that case?

[@ayushyadav1894]

Can we use i*i in for loop i^²

[@garvgoel1743]

above approach is better optimised than yours.

[@theDollarSign.]

WOW ,what a nice solution❤👍

[@SaurabhKumar-jv7wh]

aisa lg rha h bhaiya mst phook k bethe h

[@pratikbarye2527]

for(int i=1;i>1;i++)
{
cout

[@SaurabhKumar-jv7wh]

import java.util.ArrayList;
public class sumofalldivisorfrom1ton {

static long sumOfDivisors(int N) {
ArrayList arr = new ArrayList();
while (N>0) {
for (int i = 1; i * i

[@Manoj-hl2jp]

I think the time complexity of your solution is O(N2)
But they are expecting lesser time complexity link mentioned below : O(N)
static long sumOfDivisors(int n){
Long sumOfDivisor= 0L;
for(int i=1;i

[@rajdeepsharma8073]

kaise teacher rbaithe hai yaar gfg main ,,,worst explanation ek badhiya engineer zaruari nahi hai ki acha teacher ho

[@NayanMongia-pm3ho]

worst explanation understood nothing at all