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Data Structures PYQs for GATE DA

GATE DA Data Structures: 3 chapters, 8 previous year questions (26% of Programming, Data Structures and Algorithms), 452 practice questions and one solved que

A question from this chapter

Question 1

2025 PYQ

Level 3: Exam Standard

Consider the following pseudocode. Create empty stack S Set x=0, flag=0, sum=0 Push x onto S while (S is not empty){ if (flag equals 0){ Set x = x+1 Push x onto S} if (x equals 8): Set flag=1 if (flag equals 1){ x = Pop(S) if (x is odd): Pop(S) Set sum = sum + x} } Output sum The value of sum output by a program executing the above pseudocode is (Answer in integer)

Question 2

2025 PYQ

Level 3: Exam Standard

Consider a hash table of size 10 with indices {0, 1, ..., 9}, with the hash function h (x) = 3 x (mod 10), where linear probing is used to handle collisions. The hash table is initially empty and then the following sequence of keys is inserted into the hash table: 1, 4, 5, 6, 14, 15 . The indices where the keys 14 and 15 are stored are, respectively

Question 3

2026 PYQ

Level 3: Exam Standard

You are given the following Pre-order and In-order traversals of a Binary Tree T with nodes E, F, G, P, Q, R, S. Pre-order: P Q S E R F G In-order: S Q E P F R G Which of the following statements is/are true about the Binary Tree T?

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Data Structures PYQs for GATE DA

GATE DA Data Structures: 3 chapters, 8 previous year questions (26% of Programming, Data Structures and Algorithms), 452 practice questions and one solved question from each chapter.

About Data Structures Previous Year Questions (PYQs)

8 previous year questions from Data Structures in GATE DA, grouped by chapter with the exam year, answer key and step-by-step solution for each.

Data Structures Weightage in GATE DA

Data Structures accounts for 8 of 31 Programming, Data Structures and Algorithms previous year questions in our bank (26%), about 2.7 per paper across 3 papers.

Data Structures Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Stacks, Queues and Deques	Stack Operations and Simulation, Queue and Deque Operations, ADT Properties: FIFO, LIFO and Lookup	3	38%	161
Hash Tables and Collision Resolution	Hash Tables with Open Addressing and Linear Probing, Uniform Hashing and Probe Complexity	2	25%	126
Binary Trees: Properties and Traversals	Binary Tree Traversals and Reconstruction, Binary Tree Structural Properties	3	38%	165

One Solved Question from Each Data Structures Chapter

Question 1 · Stacks, Queues and Deques · 2025 NAT

Consider the following pseudocode.

Create empty stack S

Set x=0, flag=0, sum=0

Push x onto S

while (S is not empty){

    if (flag equals 0){

        Set x = x+1

        Push x onto S}

    if (x equals 8):

    Set flag=1

    if (flag equals 1){

    x = Pop(S)

    if (x is odd):

    Pop(S)

    Set sum = sum + x}

}

Output sum

The value of sum output by a program executing the above pseudocode is
(Answer in integer)

Correct Answer:

Step-by-Step Solution

Key idea: This is a stack simulation question, recognizable because it provides pseudocode with stack operations (Push, Pop) and conditional logic, asking for the final state of a variable.

Step 1: Initialize variables. $x = 0$ , $f l a g = 0$ , $s u m = 0$ . Stack $S$ is initialized with $0$ , so $S = [0]$ .

Step 2: Trace the loop while $S$ is not empty.

Iterations 1 to 7: $f l a g$ is $0$ . $x$ increments by $1$ each time and is pushed. $S$ becomes $[0, 1, 2, 3, 4, 5, 6, 7]$ . $x$ reaches $7$ .
Iteration 8: $f l a g = 0$ , so $x$ becomes $8$ and is pushed. $S = [0, 1, 2, 3, 4, 5, 6, 7, 8]$ . Now $x = 8$ , so $f l a g$ becomes $1$ . Since $f l a g = 1$ , we execute the pop block: $x = P o p (S) = 8$ . $8$ is even, so the inner pop is skipped. $s u m = 0 + 8 = 8$ .
Iteration 9: $f l a g = 1$ . $x = P o p (S) = 7$ . $7$ is odd, so we $P o p (S)$ again, removing $6$ . $S$ is now $[0, 1, 2, 3, 4, 5]$ . $s u m = 8 + 7 = 15$ .
Iteration 10: $f l a g = 1$ . $x = P o p (S) = 5$ . $5$ is odd, so we $P o p (S)$ again, removing $4$ . $S$ is now $[0, 1, 2, 3]$ . $s u m = 15 + 5 = 20$ .
Iteration 11: $f l a g = 1$ . $x = P o p (S) = 3$ . $3$ is odd, so we $P o p (S)$ again, removing $2$ . $S$ is now $[0, 1]$ . $s u m = 20 + 3 = 23$ .
Iteration 12: $f l a g = 1$ . $x = P o p (S) = 1$ . $1$ is odd, so we $P o p (S)$ again, removing $0$ . $S$ is now $[]$ . $s u m = 23 + 1 = 24$ .

Step 3: The stack is now empty, so the loop terminates. The final value of $s u m$ is $24$ .

Answer: 24

Question 2 · Hash Tables and Collision Resolution · 2025 MCQ

Consider a hash table of size

10

with indices

{0, 1, ..., 9}

, with the hash function

h (x) = 3 x (mod 10),

where linear probing is used to handle collisions. The hash table is initially empty and then the following sequence of keys is inserted into the hash table:

1, 4, 5, 6, 14, 15

. The indices where the keys

14

and

15

are stored are, respectively

A.
$2$ and $5$
B.
$2$ and $6$
C.
$4$ and $5$
D.
$4$ and $6$

Correct Answer:

Step-by-Step Solution

Key idea: This is a hash table tracing question, recognizable because it provides a specific hash function, collision resolution strategy, and a sequence of keys to insert.

Step 1: Understand the hash function and table size. The table has size 10 (indices 0 to 9). The hash function is $h (x) = 3 x (mod 10)$ .

Step 2: Trace the insertion of each key sequentially using linear probing.

Insert 1: $h (1) = 3 (1) (mod 10) = 3$ . Index 3 is empty. Place 1 at index 3.
Insert 4: $h (4) = 3 (4) (mod 10) = 12 (mod 10) = 2$ . Index 2 is empty. Place 4 at index 2.
Insert 5: $h (5) = 3 (5) (mod 10) = 15 (mod 10) = 5$ . Index 5 is empty. Place 5 at index 5.
Insert 6: $h (6) = 3 (6) (mod 10) = 18 (mod 10) = 8$ . Index 8 is empty. Place 6 at index 8.
Insert 14: $h (14) = 3 (14) (mod 10) = 42 (mod 10) = 2$ . Index 2 is occupied (by 4).
Probe 1: $(2 + 1) (mod 10) = 3$ . Index 3 is occupied (by 1).
Probe 2: $(3 + 1) (mod 10) = 4$ . Index 4 is empty. Place 14 at index 4.
Insert 15: $h (15) = 3 (15) (mod 10) = 45 (mod 10) = 5$ . Index 5 is occupied (by 5).
Probe 1: $(5 + 1) (mod 10) = 6$ . Index 6 is empty. Place 15 at index 6.

Step 3: Identify the final indices. Key 14 is at index 4, and key 15 is at index 6.

Answer: D

Question 3 · Binary Trees: Properties and Traversals · 2026 MSQ

You are given the following Pre-order and In-order traversals of a Binary Tree T with nodes E, F, G, P, Q, R, S.
Pre-order: P Q S E R F G
In-order: S Q E P F R G
Which of the following statements is/are true about the Binary Tree T?

A.
Node P is the root of T
B. The Post-order traversal of T is:
S E Q F G R P
C.
Node Q has only one child
D.
The left subtree of node R contains the node G

Correct Answer:

["A","B"]

Step-by-Step Solution

Insight: Reconstruct the tree using the standard Pre-order and In-order split method.

Exam route: Root is P. Left In-order has 3 elements, Right has 3. Split Pre-order accordingly. Build left and right subtrees. Check options.

Learning route:

Step 1: Root is the first element in Pre-order: P.

Step 2: Find P in In-order. Left of P is S, Q, E (size 3). Right of P is F, R, G (size 3).

Step 3: Split remaining Pre-order (Q, S, E, R, F, G) into Left Pre-order (Q, S, E) and Right Pre-order (R, F, G).

Step 4: Left subtree: Pre (Q, S, E), In (S, Q, E). Root is Q. Left In is S, Right In is E. So Q has left child S, right child E.

Step 5: Right subtree: Pre (R, F, G), In (F, R, G). Root is R. Left In is F, Right In is G. So R has left child F, right child G.

Step 6: Evaluate options.

A) P is root. (True)

B) Post-order is Left Post + Right Post + Root. Left Post: S, E, Q. Right Post: F, G, R. Total: S, E, Q, F, G, R, P. (True)

C) Q has two children (S and E). (False)

D) G is in the right subtree of R. (False)

A question from this chapter

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Honest progress

Unlock the whole course

Data Structures PYQs for GATE DA

About Data Structures Previous Year Questions (PYQs)

Data Structures Weightage in GATE DA

Data Structures Chapter Matrix

More from Programming, Data Structures and Algorithms

One Solved Question from Each Data Structures Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution