CareerCup

your while loop (i+j>k ) . think both array >k so i=j=k and than analyse

your order will be O(K) not log(k)

keep on hashing every number before hashing any number check if it already exist . the duplicate will be found in O(n).

create an array c and fill it from right to left c[i]= max { b[j] i<j<n}
then 2nd pass create an array d left to right d[i]= min{b[j} 0<j<i}

3rd pass find the ci , di where diff is max corresponding points will give you buy and sell time.

create an array c and fill it from right to left c[i]= max { b[j] i<j<n}
then 2nd pass create an array d left to right d[i]= min{b[j} 0<j<i}

3rd pass find the ci , di where diff is max corresponding points will give you buy and sell time.

if they are binary trees than add at any leaves. In case of BST than the complexity would be mlogn or nlogm .

(1)take 2pointers initialize to head of the list
(2)increment both pointer till they reach end of the list and calculate list length l1,l2
(3) calculate absolute difference between l1-l2 ie n=|l1-l2|
(4) calculate nth element from end of the list.

your ordering will become wrong:
5,15,3,10

your answer will be 3,15 but the i<j is invalid

will not work for : 3 5 1 2

Simply find smallest and largest element in the array. ie answere

(1) take 2 points i, j both point head of list.
(2) Increment j to j->next for K times.
(3)After that increment both pointer by single step .
(4)When the j pointer reach j->next ==null then your i pointer points to kth element from the last

depends on what position you are applying also regional ,similar college politics do exist in big coorporates

I told the similar solution but then he said at the end the sum of all probabilities will become > 1 . So i suggested that as the next move is different experiment so according to theory of probability it should be multiple of p(n,x,y)x p(n-1 , x' ,y').