POTD- 01/11/2023 | Frequencies of Limited Range Array Elements | Problem of the Day GeeksforGeeks

[@kumarqwerty4452]

Karan Mashru is the best teacher GfG has ever produced.Thanks man for wonderful explanation.

[@jyoti3427]

great explanation

[@himansuiii]

for(int i=0;iN)
arr[i]=0;
int i=0;
while(i

[@prateekbhaisora]

A cleaner code using modular arithematic:
void frequencyCount(vector& arr,int n, int p)
{
p++;
for(int i=0; i= n)
continue;
int cnt = arr[ind] / p;
int ele = arr[ind] % p;
cnt ++;
arr[ind] = cnt * p + ele;
}
for (int i=0; i

[@piyushphaske3271]

Thanks

[@daaamn6414]

Solution giving Run time error.

[@miracleguy1330]

Easiest Solution -> int[] freq = new int[N + 1];
// Counting the frequency of each element
for (int i = 0; i < N; i++) {
if (arr[i]

[@antimr2808]

Your answer is right and easy but there is a little point I want to make
They mentioned time complexity as O(1) that means we can not use extra space (int[] freq = new int[N + 1];) becasue this will make the time complexity as O(n).

[@miracleguy1330]

@@antimr2808 ya I didn't take that in account. As It was accepted on gfg. 🥲

[@zeeshanali4502]

wow you give new meaning to Inplace 😆

[@srinijabhogoju9033]

Thank you. Very well explained.

[@chandruram7553]

Tq💖