Cognizant GenC Next Data Structure Questions 2025

Explanation : v-e+f=2 is used to calculate the number of bounded faces in any embedding of graphs on the plane.Here, in the given question we have v=4, e=14

So, f=2+(14-4) => f=12

Given array contains elements {2,1,4,3,7,9,0}.

First insertion will done in inserting 1 to its correct position , so array will become {1,2,4,3,7,9,0}

Second insertion will done in inserting 3 to its correct position , so array will become {1,2,3,4,7,9,0}

And third insertion will done in inserting 0  to its correct position , so array will become {0,1,2,3,4,7,9}

As Fun_POP() will delete the element so the index of top will get decremented by 1.

Explanation : Vertex cut is a subset of vertices of a graph which, if removed (or "cut")--together with any incident edges--disconnects the graph (i.e., forms a disconnected graph). A complete graph G, of order x will contain x-2 maximum cut vertices.

Let max_size=0 represents the required size and curr_size is the current size of the stack.

Initially ( will push into stack now curr_size=1, as curr_size > max_size so max_size =1

now,( will push into stack now curr_size=2, as curr_size > max_size so max_size =2

Now, ) comes then pop() operation will perform so curr_size become 1 ,

Now ( will push into stack now curr_size=2, as curr_size > max_size so max_size =2

Now again  ( will push into stack now curr_size=3, as curr_size > max_size so max_size =3

Now, ) comes then pop() operation will perform so curr_size become 2

Again ) comes then pop() operation will perform so curr_size become 1

Now ( will push into stack now curr_size=2, as curr_size > max_size so max_size =2

Now again  ( will push into stack now curr_size=3, as curr_size > max_size so max_size =3

Now, ) comes then pop() operation will perform so curr_size become 2

Again ) comes then pop() operation will perform so curr_size become 1

And at last ) comes then pop() operation will perform so curr_size become 1

So the max_size appears to be 3.

Explanation : The following code will return the total number of nodes in the tree, in the given code we pass the root of the tree which has 5 odes, ence the function will give output as 5.

In the given scenario: 

The status of the stack , when 7 elements are inserted into it is : S R Q P Z Y X.

As stack follows LIFO then  status of queue will be : X Y Z P Q.

Now the queue follows FIFO principle then the first deleted element will be X and second will be Y.

Initially the queue is empty,

Queue follows FIFO mechanism

Operation            Status of queue

ADD 23                23

DELETE               -

ADD 20                20

ADD 32                20  32

ADD 90                20  32  90

DELETE               32  90

DELETE               90

ADD 8                   90 8

ADD 27                 90 8 27

Hence, 90 is present at front end and 27 at rear end.

Initially the stack will contain A B C D E when stack is popped four times and element from stack will be inserted in queue then Queue contain E D C B and stack contain A only ,.

Now two elements will deleted from queue and inserted into stack then stack contain A E D and queue contain C B.

Now the popped element from stack is D.

 The time complexity will be O(n) , n is the length of the main string.