First negative integer in every window of size k | Love Babbar DSA sheet

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Problems:
1. First negative integer in every window of size k:
practice.geeks...
solution:
practice.geeks...
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[@royalsingh9400]

There is no need of j variable in this problem. Instead of j we can use (i-k)

[@manthanpatil2268]

Great explanation buddy!

[@adityarajiv6346]

Glad you liked it!

[@saniashaikh23b82]

is this in O SPACE ? If not how can we optimise

[@adityarajiv6346]

im not sure about further optimization. but yeah this solution works!

[@ankitdas7532]

This isn't O(1) space since we are using a queue to store the negative elements.
For O(1) we can maintain a pointer which keeps track of next negative element . Once our window crosses that pointer , we move the pointer to next negative element.
Code( accepted in gfg) :
vector sol ;
long long n =0 ; // pointer to negative elements
while(n < N && A[n] >=0)n++;
for(long long i=0 ; i=s && n=0)n++; // move pointer to next negative element
}
}
return sol;

[@rajaryansharma5001]

Thanks!

[@adityarajiv6346]

Your welcome 😀.
Please share the channel with all your friends and subscribe 😀🙏.

[@mdazazali7332]

Can't we do this with vector..?

[@adityarajiv6346]

try it maybe its not possible

[@avinashyadav8598]

Bro use list

[@Rajat_maurya]

@@adityarajiv6346 through vector
vector v;
int neg=0;
for(int i=0;i

[@ishanikasingh9321]

Hi Aditya, queue works fine:
vector printFirstNegativeInteger(long long int A[],
long long int n, long long int k) {
vector v;
queue q;
int i=0;
int j=0;
while(j

[@avinashyadav8598]

Bhai explanation is not up to mark.

[@adityarajiv6346]

okay is shall c if i can explain better next time

[@mohammedwaseem8599]

Bro is competitive programming necessary??

[@adityarajiv6346]

It's good way to build problem solving skills

[@Priyanshukumar-yl5nh]

vector printFirstNegativeInteger(long long int a[],
long long int n, long long int k) {
vectorans;
queueq;
for(int i=0;i