Oracle India Placement Papers 2026

Meta Description: Oracle India placement papers 2026 with latest exam pattern, coding questions, database interview tips, and preparation strategy. Crack Oracle with PapersAdda!

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Introduction

Oracle India is a global leader in database technology, cloud solutions, and enterprise software. With major development centers in Bangalore, Hyderabad, and Pune, Oracle offers excellent career opportunities for software engineers, database administrators, and cloud specialists. This comprehensive guide covers Oracle India's 2026 placement process, including exam patterns, coding questions, and preparation strategies tailored for Oracle's unique interview style.

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Oracle India Exam & Interview Pattern 2026

Round	Type	Duration	Topics Covered
Round 1	Online Assessment	90-120 mins	Aptitude, Coding, Verbal, Technical MCQs
Round 2	Technical Interview 1	45-60 mins	Coding, Data Structures, Algorithms
Round 3	Technical Interview 2	45-60 mins	Database, SQL, System Design
Round 4	Technical Interview 3	45-60 mins	Advanced topics, Project deep-dive
Round 5	HR Interview	30-45 mins	Behavioral, Compensation discussion

Key Highlights:

• Strong emphasis on Database and SQL concepts
• Coding questions focus on data structure implementation
• Questions on Oracle products and cloud services
• Puzzles and logical reasoning questions are common

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Practice Questions with Detailed Answers

Question 1: Department Highest Salary (SQL)

Problem: Write a SQL query to find employees who have the highest salary in each department.

Solution:

-- Method 1: Using JOIN
SELECT d.Name AS Department, e.Name AS Employee, e.Salary
FROM Employee e
JOIN Department d ON e.DepartmentId = d.Id
JOIN (
    SELECT DepartmentId, MAX(Salary) as MaxSalary
    FROM Employee
    GROUP BY DepartmentId
) max_sal ON e.DepartmentId = max_sal.DepartmentId 
    AND e.Salary = max_sal.MaxSalary;

-- Method 2: Using Window Function
SELECT Department, Employee, Salary
FROM (
    SELECT 
        d.Name AS Department,
        e.Name AS Employee,
        e.Salary,
        RANK() OVER (PARTITION BY e.DepartmentId ORDER BY e.Salary DESC) as rnk
    FROM Employee e
    JOIN Department d ON e.DepartmentId = d.Id
) ranked
WHERE rnk = 1;

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Question 2: Nth Highest Salary (SQL)

Problem: Write a SQL query to get the nth highest salary from the Employee table.

Solution:

-- Using LIMIT/OFFSET
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
    SET N = N - 1;
    RETURN (
        SELECT DISTINCT Salary
        FROM Employee
        ORDER BY Salary DESC
        LIMIT 1 OFFSET N
    );
END;

-- Using DENSE_RANK
SELECT DISTINCT Salary
FROM (
    SELECT Salary, DENSE_RANK() OVER (ORDER BY Salary DESC) as rnk
    FROM Employee
) ranked
WHERE rnk = N;

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Question 3: Consecutive Numbers (SQL)

Problem: Write a SQL query to find all numbers that appear at least three times consecutively.

Solution:

SELECT DISTINCT l1.Num AS ConsecutiveNums
FROM Logs l1, Logs l2, Logs l3
WHERE l1.Id = l2.Id - 1 
  AND l2.Id = l3.Id - 1
  AND l1.Num = l2.Num 
  AND l2.Num = l3.Num;

-- Alternative using LEAD/LAG
SELECT DISTINCT Num AS ConsecutiveNums
FROM (
    SELECT Num,
        LAG(Num, 1) OVER (ORDER BY Id) as prev1,
        LAG(Num, 2) OVER (ORDER BY Id) as prev2
    FROM Logs
) t
WHERE Num = prev1 AND Num = prev2;

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Question 4: Binary Tree Level Order Traversal

Problem: Given the root of a binary tree, return the level order traversal of its nodes' values.

Solution:

from collections import deque

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def levelOrder(root):
    if not root:
        return []
    
    result = []
    queue = deque([root])
    
    while queue:
        level_size = len(queue)
        level = []
        
        for _ in range(level_size):
            node = queue.popleft()
            level.append(node.val)
            
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        
        result.append(level)
    
    return result

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Question 5: Validate Binary Search Tree

Problem: Given the root of a binary tree, determine if it is a valid binary search tree.

Solution:

def isValidBST(root):
    def validate(node, low, high):
        if not node:
            return True
        
        if node.val <= low or node.val >= high:
            return False
        
        return (validate(node.left, low, node.val) and
                validate(node.right, node.val, high))
    
    return validate(root, float('-inf'), float('inf'))

# Iterative approach
def isValidBSTIterative(root):
    if not root:
        return True
    
    stack = [(root, float('-inf'), float('inf'))]
    
    while stack:
        node, low, high = stack.pop()
        
        if not node:
            continue
        
        if node.val <= low or node.val >= high:
            return False
        
        stack.append((node.right, node.val, high))
        stack.append((node.left, low, node.val))
    
    return True

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Question 6: Clone Graph

Problem: Given a reference of a node in a connected undirected graph, return a deep copy of the graph.

Solution:

from collections import deque

class Node:
    def __init__(self, val=0, neighbors=None):
        self.val = val
        self.neighbors = neighbors if neighbors else []

def cloneGraph(node):
    if not node:
        return None
    
    # BFS approach
    old_to_new = {}
    queue = deque([node])
    old_to_new[node] = Node(node.val)
    
    while queue:
        current = queue.popleft()
        
        for neighbor in current.neighbors:
            if neighbor not in old_to_new:
                old_to_new[neighbor] = Node(neighbor.val)
                queue.append(neighbor)
            old_to_new[current].neighbors.append(old_to_new[neighbor])
    
    return old_to_new[node]

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Question 7: Min Stack

Problem: Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Solution:

class MinStack:
    def __init__(self):
        self.stack = []
        self.min_stack = []
    
    def push(self, val):
        self.stack.append(val)
        if not self.min_stack or val <= self.min_stack[-1]:
            self.min_stack.append(val)
    
    def pop(self):
        if self.stack:
            val = self.stack.pop()
            if val == self.min_stack[-1]:
                self.min_stack.pop()
    
    def top(self):
        return self.stack[-1] if self.stack else None
    
    def getMin(self):
        return self.min_stack[-1] if self.min_stack else None

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Question 8: Evaluate Reverse Polish Notation

Problem: Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Solution:

def evalRPN(tokens):
    stack = []
    operators = {'+', '-', '*', '/'}
    
    for token in tokens:
        if token not in operators:
            stack.append(int(token))
        else:
            b = stack.pop()
            a = stack.pop()
            
            if token == '+':
                stack.append(a + b)
            elif token == '-':
                stack.append(a - b)
            elif token == '*':
                stack.append(a * b)
            else:
                stack.append(int(a / b))
    
    return stack[0]

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Question 9: Daily Temperatures

Problem: Given an array of daily temperatures, return an array where each element is the number of days until a warmer temperature.

Solution:

def dailyTemperatures(temperatures):
    n = len(temperatures)
    result = [0] * n
    stack = []
    
    for i in range(n):
        while stack and temperatures[i] > temperatures[stack[-1]]:
            prev_idx = stack.pop()
            result[prev_idx] = i - prev_idx
        stack.append(i)
    
    return result

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Question 10: Find Median from Data Stream

Problem: Implement a data structure that supports adding integers and finding the median.

Solution:

import heapq

class MedianFinder:
    def __init__(self):
        # max heap for lower half (store negatives)
        self.small = []
        # min heap for upper half
        self.large = []
    
    def addNum(self, num):
        heapq.heappush(self.small, -num)
        
        # Balance: max of small <= min of large
        if self.small and self.large and (-self.small[0]) > self.large[0]:
            val = -heapq.heappop(self.small)
            heapq.heappush(self.large, val)
        
        # Size balancing
        if len(self.small) > len(self.large) + 1:
            val = -heapq.heappop(self.small)
            heapq.heappush(self.large, val)
        
        if len(self.large) > len(self.small):
            val = heapq.heappop(self.large)
            heapq.heappush(self.small, -val)
    
    def findMedian(self):
        if len(self.small) > len(self.large):
            return -self.small[0]
        return (-self.small[0] + self.large[0]) / 2

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Question 11: Trie Implementation

Problem: Implement a trie with insert, search, and startsWith methods.

Solution:

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end = False

class Trie:
    def __init__(self):
        self.root = TrieNode()
    
    def insert(self, word):
        node = self.root
        for char in word:
            if char not in node.children:
                node.children[char] = TrieNode()
            node = node.children[char]
        node.is_end = True
    
    def search(self, word):
        node = self._find_node(word)
        return node is not None and node.is_end
    
    def startsWith(self, prefix):
        return self._find_node(prefix) is not None
    
    def _find_node(self, word):
        node = self.root
        for char in word:
            if char not in node.children:
                return None
            node = node.children[char]
        return node

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Question 12: Design HashMap

Problem: Design a HashMap without using any built-in hash table libraries.

Solution:

class ListNode:
    def __init__(self, key=-1, val=-1, next=None):
        self.key = key
        self.val = val
        self.next = next

class MyHashMap:
    def __init__(self):
        self.size = 1000
        self.buckets = [None] * self.size
    
    def _hash(self, key):
        return key % self.size
    
    def put(self, key, value):
        idx = self._hash(key)
        if not self.buckets[idx]:
            self.buckets[idx] = ListNode()
        
        curr = self.buckets[idx]
        while curr.next:
            if curr.next.key == key:
                curr.next.val = value
                return
            curr = curr.next
        curr.next = ListNode(key, value)
    
    def get(self, key):
        idx = self._hash(key)
        curr = self.buckets[idx]
        
        while curr and curr.next:
            if curr.next.key == key:
                return curr.next.val
            curr = curr.next
        return -1
    
    def remove(self, key):
        idx = self._hash(key)
        curr = self.buckets[idx]
        
        while curr and curr.next:
            if curr.next.key == key:
                curr.next = curr.next.next
                return
            curr = curr.next

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Preparation Strategy

Step 1: Database & SQL Mastery (Weeks 1-3)

• Master SQL queries: JOINs, Subqueries, Window Functions
• Practice complex queries on LeetCode Database section
• Study database design and normalization
• Learn indexing and query optimization

Step 2: Core Programming (Weeks 4-6)

• Strengthen Java/C++ fundamentals
• Practice data structure implementation from scratch
• Focus on Trees, Graphs, and Hash Tables

Step 3: Algorithms (Weeks 7-9)

• Dynamic Programming patterns
• Graph algorithms (BFS, DFS, Dijkstra)
• String algorithms and Trie

Step 4: Oracle-Specific Topics (Week 10)

• Study Oracle Database architecture
• Learn about Oracle Cloud Infrastructure (OCI)
• Understand PL/SQL basics
• Review Oracle's product portfolio

Step 5: Mock Tests & Revision (Weeks 11-12)

• Solve previous year Oracle papers
• Practice timed coding tests
• Prepare for puzzles and logical reasoning

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Resources & Books to Follow

SQL & Database

• LeetCode Database - Practice SQL problems
• Mode Analytics SQL Tutorial - Free SQL learning
• "SQL Cookbook" by Anthony Molinaro
• "Database Management Systems" by Raghu Ramakrishnan

Programming

• LeetCode - Primary coding platform
• GeeksforGeeks - Oracle-specific questions
• "Cracking the Coding Interview" by Gayle Laakmann McDowell

Oracle-Specific

• Oracle Documentation (docs.oracle.com)
• Oracle Cloud Learning
• PL/SQL Programming by Steven Feuerstein

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Oracle India Salary & Package Details 2026

Role	Level	Experience	CTC (LPA)
Associate Software Engineer	IC1	0-2 years	₹12-18 LPA
Software Engineer	IC2	2-4 years	₹20-30 LPA
Senior Software Engineer	IC3	4-7 years	₹35-50 LPA
Principal Software Engineer	IC4	7+ years	₹55-80 LPA

Benefits:

• Health and life insurance
• Employee stock purchase plan
• Flexible work arrangements
• Learning and development programs
• Retirement benefits

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Conclusion

Oracle India offers excellent opportunities for database enthusiasts and cloud professionals. The interview process emphasizes SQL proficiency alongside strong coding skills. Focus on mastering database concepts, practice complex queries, and demonstrate your understanding of scalable systems.

For more placement papers and interview resources, visit PapersAdda.com - Your ultimate job preparation companion!

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Keywords: Oracle placement papers 2026, Oracle interview questions, Oracle SQL questions, Oracle recruitment process, Oracle India careers