SDE - 1

Tip 1: Don't Forget fundamentals !
Tip 2: It's about dynamic allocation and static allocation.

Tip 1: Dynamic Allocation -> Heaps , Static Allocation -> Stacks

Step 1: the first Solution is obviously brute force! increment the counter from N till we get the digit which has all 0 except one non zero.
But this question was not asked for this solution.
Step 2: the more efficient solution would be :
1. Get the no digits in the Number. [Lets say 'x']
2. then find use this formule to find the smallest no in that x digit : 10^(x-1) [Lets say 'y']
3. Then find the diffrence between y and number : lets say 'z'
4. then result would be = Number - z + y
Step 3: But the best solution would be very very simple;
1. Increment the first digit to + 1 [Lets say number is 930 then make it 1030]
2. Make all the rest numbers 0.

I knew the code since i practised it 100s of times before !
Soln: [Most efficent one]
reverseLL(N* curr, N* prev, N** head) {
if (!curr->next) {
*head = curr;
curr->next = prev;
return;
}
N* next = curr->next;
curr->next = prev;
reverseLL(next, curr, head);
}

Tip 1: First make simple Api Endpoints
Tip 2: Add expected response
Tip 3: Learn About Redis [Cache management server]
Tip 4: At last go into jest [unit testing]

Tip 1: I said I need atleast 20% more than what i am currently expected to get at my current job.