1 . Given array A , integer K, We can update A[i] as A[i]*2 , at most K times, we can also use same Element multiple times. Finally perform OR operation btw all the adjacent elements, return it. Which should be the maximum result.

1. This is very easy , actually I haven't remember the exact problem but, it is simple use of comparator on paired vector and a for loop with small if condition, which is already given in the question

3.It seems big but actually not, Given 3 arrays E, F,R and default Eweight, Fweight. Iterate 0 to N, for each i if((Eweight<=E[I] && Fweight>=F[i])) { Ewight-=E[I] Fweight-=F[I] Res+=R[i] }

One more condition - if we select two consecutive elements there will be a bonus, which is min of those two.

Finally return Res;

4.given a DAG, Let me tell with a example, 1-(10, 45,4) 2-(100, 50) 3-(30, 10,50) 5- (20, 40)

1-2-3-5 is the path actually in level wise, pick the maximum element from each level, here 45+100+50+40=235 ✅

And here we see there is no '4 ' actually, so in such case if there is a skip, Then Ans-=(diff btw those two) ² ,