Sample Oracle Coding Round Questions and Answers for 2025

Published: 23 Aug 2025 | Reading Time: 12 min read

Overview

Oracle was established in 1977 and is among the top industry leaders in database management and cloud computing with multiple careers in diverse fields. Oracle recruitment could be challenging, and job seekers must show technical and analytical skills. This comprehensive guide provides an overview of different Oracle job roles, Oracle recruitment process stages, fundamental coding concepts to learn for the Oracle coding round interview questions, and suggestions on cracking the recruitment process.

Table of Contents

• Stages in the Oracle Recruitment Process
• Concepts to Look for Oracle Coding Questions
• Top Coding Questions Asked in Oracle
• Tips to Crack the Oracle Recruitment
• Conclusion
• Frequently Asked Questions

Stages in the Oracle Recruitment Process

The Oracle recruitment process consists of five distinct stages:

1. Resume Submission

You submit your CV through Oracle's careers portal.

2. Aptitude Test

You have logical reasoning, quantitative aptitude, and verbal ability.

3. Technical Interview

You deal with programming, algorithms, data structures, and solving problems. Sometimes there might be live coding.

4. HR Interview

Assessing fit within the culture, language skills, and disposition to work in Oracle.

5. Offer Letter

If successful, you will receive an offer letter detailing the job role and pay package.

Concepts to Look for Oracle Coding Questions

The Oracle coding interview question basics are embedded in the following core areas:

Data Structures

• Arrays
• Linked Lists
• Stacks
• Queues
• Trees (Binary Trees, Binary Search Trees, Heaps)
• Graphs
• Hashmaps

Algorithms

• Sorting
• Search (Binary Search)
• Dynamic Programming
• Divide and Conquer
• Greedy Algorithms
• Backtracking

Core Database

• SQL queries
• Normalization
• Indexing
• Joins
• Transactions

Object-Oriented Programming

• Classes
• Objects
• Inheritance
• Polymorphism
• Encapsulation
• Abstraction

Concurrency and Threads

• Simple multi-threading fundamentals
• Synchronization
• Race conditions

Top Coding Questions Asked in Oracle

Here are 12 Oracle coding test questions asked in the technical assessment or interview:

Problem Statement 1: Sort Colors (Dutch National Flag Problem)

Problem Description:

You are tasked with organising a group of people based on a simple classification system. Each person is assigned a number that can only be 0, 1, or 2, representing different types of roles in a team. Your task is to arrange the group so that all 0s are at the front, followed by all 1s, and then 2s. This way, the group can be easily segmented for different activities. The arrangement must be done efficiently since the group can consist of many people.

Input:

5 2 0 1 2 0

Output:

0 0 1 2 2

Solution in C++

#include <iostream>
using namespace std;

void sortColors(int arr[], int n) {
    int count[3] = {0};

    for (int i = 0; i < n; i++) {
        count[arr[i]]++;
    }

    for (int i = 0; i < count[0]; i++) arr[i] = 0;
    for (int i = 0; i < count[1]; i++) arr[count[0] + i] = 1;
    for (int i = 0; i < count[2]; i++) arr[count[0] + count[1] + i] = 2;
}

int main() {
    int n;
    cin >> n;
    int arr[n];

    for (int i = 0; i < n; i++) {
        cin >> arr[i];
    }

    sortColors(arr, n);

    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
    cout << endl;

    return 0;
}

Solution in Python

def sort_colors(arr):
    count = [0, 0, 0]  # To count 0s, 1s, and 2s
    for num in arr:
        count[num] += 1

    index = 0
    for i in range(3):
        for _ in range(count[i]):
            arr[index] = i
            index += 1

# Input and Output
n = int(input())
arr = list(map(int, input().split()))
sort_colors(arr)
print(" ".join(map(str, arr)))

Solution in Java

import java.util.Scanner;

public class SortColors {
    public static void sortColors(int[] arr) {
        int[] count = new int[3];  // To count 0s, 1s, and 2s
        for (int num : arr) {
            count[num]++;
        }

        int index = 0;
        for (int i = 0; i < count[0]; i++) arr[index++] = 0;
        for (int i = 0; i < count[1]; i++) arr[index++] = 1;
        for (int i = 0; i < count[2]; i++) arr[index++] = 2;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] arr = new int[n];

        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }

        sortColors(arr);

        for (int num : arr) {
            System.out.print(num + " ");
        }
        System.out.println();
    }
}

Explanation

• Using the count array, we count how many 0s, 1s, and 2s are in the array.
• We then use the count values to place the respective numbers (0s, 1s, 2s) back into the original array in the correct order.
• This solution uses an efficient counting technique with time complexity O(n).

Problem Statement 2: Best Time to Buy and Sell Stock

Problem Description:

Imagine you're an investor who buys and sells stocks. You are given a list of stock prices for each day. You can buy a stock on one day and sell it on a later day. You aim to maximize your profit by buying at the lowest price and selling at the highest possible price. However, you can only perform one buy and one sell operation. Determine your maximum profit using this strategy while ensuring the solution runs efficiently with larger data sets.

Input:

7
1 9 2 11 1 9 2

Output:

10

Solution in C++

#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    int n, profit = 0, min_price = INT_MAX;
    cin >> n;

    for (int i = 0; i < n; i++) {
        int price;
        cin >> price;
        min_price = min(min_price, price);
        profit = max(profit, price - min_price);
    }

    cout << profit << endl;
    return 0;
}

Solution in Python

n = int(input())
profit = 0
min_price = float('inf')

for _ in range(n):
    price = int(input())
    min_price = min(min_price, price)
    profit = max(profit, price - min_price)

print(profit)

Solution in Java

import java.util.Scanner;

public class MaxProfit {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int profit = 0, minPrice = Integer.MAX_VALUE;

        for (int i = 0; i < n; i++) {
            int price = sc.nextInt();
            minPrice = Math.min(minPrice, price);
            profit = Math.max(profit, price - minPrice);
        }

        System.out.println(profit);
    }
}

Explanation

• We initialize min_price to infinity and profit to 0.
• We update min_price to the minimum of the current price and min_price for each price.
• We calculate the potential profit for each day by subtracting min_price from the current price, and we update profit with the maximum value.
• This approach runs in O(n) time, making it efficient for significant inputs.

Problem Statement 3: Count Borrow Operations in Subtraction

Problem Description:

You are a software developer working on a financial application that requires you to implement subtraction of two large numbers. However, these numbers may involve borrow operations, which are common when subtracting digits that are smaller than the corresponding digits in the other number. Your task is to count the number of borrow operations required to perform the subtraction, or return "Not possible" if the subtraction cannot be performed (e.g., if the first number is smaller than the second). The solution must be optimal for large numbers with varying lengths.

Constraints:

• Both numbers are non-negative integers represented as strings.
• Length of the numbers is at most 10^5.

Solution in C++

#include <iostream>
#include <cstring>
using namespace std;

int countBorrows(char* num1, char* num2) {
    int len1 = strlen(num1), len2 = strlen(num2);
    if (len1 < len2 || (len1 == len2 && strcmp(num1, num2) < 0)) {
        return -1;
    }

    int borrow = 0, count = 0;
    for (int i = 0; i < len1; i++) {
        int digit1 = num1[len1 - 1 - i] - '0';
        int digit2 = (i < len2) ? num2[len2 - 1 - i] - '0' : 0;

        if (digit1 < digit2 + borrow) {
            borrow = 1;
            count++;
        } else {
            borrow = 0;
        }
    }
    return count;
}

int main() {
    char num1[100], num2[100];
    cin >> num1 >> num2;
    int result = countBorrows(num1, num2);
    if (result == -1) {
        cout << "Not possible" << endl;
    } else {
        cout << result << endl;
    }
    return 0;
}

Solution in Python

def count_borrows(num1, num2):
    if len(num1) < len(num2) or (len(num1) == len(num2) and num1 < num2):
        return -1

    borrow = 0
    count = 0
    for i in range(len(num1)):
        digit1 = int(num1[-1 - i])
        digit2 = int(num2[-1 - i]) if i < len(num2) else 0

        if digit1 < digit2 + borrow:
            borrow = 1
            count += 1
        else:
            borrow = 0
    return count

num1 = input().strip()
num2 = input().strip()
result = count_borrows(num1, num2)
if result == -1:
    print("Not possible")
else:
    print(result)

Solution in Java

import java.util.Scanner;

public class BorrowNumber {
    public static int countBorrows(String num1, String num2) {
        if (num1.length() < num2.length() || (num1.length() == num2.length() && num1.compareTo(num2) < 0)) {
            return -1;
        }

        int borrow = 0, count = 0;
        for (int i = 0; i < num1.length(); i++) {
            int digit1 = num1.charAt(num1.length() - 1 - i) - '0';
            int digit2 = (i < num2.length()) ? num2.charAt(num2.length() - 1 - i) - '0' : 0;

            if (digit1 < digit2 + borrow) {
                borrow = 1;
                count++;
            } else {
                borrow = 0;
            }
        }
        return count;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String num1 = sc.next();
        String num2 = sc.next();
        int result = countBorrows(num1, num2);
        if (result == -1) {
            System.out.println("Not possible");
        } else {
            System.out.println(result);
        }
    }
}

Explanation

• First, check if num1 is smaller than num2. If so, return "Not possible".
• Start from the least significant digit (rightmost), subtract the digits of num2 from num1 considering any borrow carried over from the previous step.
• Count how many times a borrow is needed, and return that count.

Problem Statement 4: Count Distinct Elements in Range

Problem Description:

You have an array of some unique transactions of a company. You want to know how many unique types of transactions were happening within two days of the timeline of the company. A transaction is defined by an integer, so you have some range of days. Your goal is to find out how many unique types of transactions were happening in that range. This needs to be achieved in a time-effective manner as the array could be enormous and you must give a quick output for any given range of days.

Constraints:

• n is between 1 and 10^5.
• The elements in the array are integers.

Input:

5 1 2 2 3 4 1 3

Output:

3

Solution in C++

#include <iostream>
#include <unordered_set>
using namespace std;

int distinctCount(int arr[], int l, int r) {
    unordered_set<int> distinct;
    for (int i = l; i <= r; i++) {
        distinct.insert(arr[i]);
    }
    return distinct.size();
}

int main() {
    int n, l, r;
    cin >> n;
    int arr[n];
    for (int i = 0; i < n; i++) {
        cin >> arr[i];
    }
    cin >> l >> r;
    cout << distinctCount(arr, l, r) << endl;
    return 0;
}

Solution in Python

def distinct_count(arr, l, r):
    return len(set(arr[l:r+1]))

n = int(input())
arr = list(map(int, input().split()))
l, r = map(int, input().split())
print(distinct_count(arr, l, r))

Solution in Java

import java.util.HashSet;
import java.util.Scanner;

public class DistinctCount {
    public static int distinctCount(int[] arr, int l, int r) {
        HashSet<Integer> distinctSet = new HashSet<>();
        for (int i = l; i <= r; i++) {
            distinctSet.add(arr[i]);
        }
        return distinctSet.size();
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }
        int l = sc.nextInt();
        int r = sc.nextInt();
        System.out.println(distinctCount(arr, l, r));
    }
}

Explanation

• We use an unordered_set to store unique transaction types in the given range.
• The size of the set gives the number of distinct transaction types in the specified range.
• The time complexity is O(n) for inserting elements into the set.

Problem Statement 5: Count Vowels in String

Problem Description:

You are working on a text processing application where you need to analyze the frequency of vowels in various sentences. Your task is to count how many vowels (a, e, i, o, u) are in a given text string. The string could be a long paragraph, and your solution should be efficient enough to handle large text sizes. The input string can contain uppercase and lowercase letters, and your program must accurately account for vowels regardless of their case. This operation is critical for text-based analytics, including readability assessments.

Constraints:

• The string contains only English letters (both lowercase and uppercase).
• Length of the string is at most 10^5.

Input:

Hello World

Output:

3

Solution in C++

#include <iostream>
using namespace std;

int countVowels(string s) {
    int count = 0;
    for (char c : s) {
        if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
            c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U') {
            count++;
        }
    }
    return count;
}

int main() {
    string s;
    getline(cin, s);
    cout << countVowels(s) << endl;
    return 0;
}

Solution in Python

def count_vowels(s):
    vowels = "aeiouAEIOU"
    return sum(1 for char in s if char in vowels)

s = input().strip()
print(count_vowels(s))

Solution in Java

import java.util.Scanner;

public class VowelCount {
    public static int countVowels(String s) {
        int count = 0;
        for (char c : s.toCharArray()) {
            if ("aeiouAEIOU".indexOf(c) != -1) {
                count++;
            }
        }
        return count;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        System.out.println(countVowels(s));
    }
}

Explanation

• Loop through each character of the string.
• Check if the character is a vowel (either lowercase or uppercase).
• Count and return the number of vowels.

Problem Statement 6: Compare Loan Interest Rates

Problem Description:

You are a financial analyst evaluating loan offers from two banks. Both banks offer different interest rates and loan tenures with varying periods. Your task is calculating the total interest paid over the full tenure for each loan offer, based on the principal amount. You must compare both offers and determine which bank has the cheaper total interest cost.

Input:

1000 5 3 6 2

Output:

Bank 1 Interest: 150.00
Bank 2 Interest: 120.00
Bank 2 offers the cheaper interest.

Solution in C++

#include <iostream>
using namespace std;

int main() {
    double p, r1, t1, r2, t2;
    cin >> p >> r1 >> t1 >> r2 >> t2;
    double interest1 = p * r1 * t1 / 100;
    double interest2 = p * r2 * t2 / 100;
    cout << "Bank 1 Interest: " << interest1 << "\nBank 2 Interest: " << interest2 << endl;
    cout << (interest1 < interest2 ? "Bank 1" : "Bank 2") << " offers the cheaper interest." << endl;
    return 0;
}

Solution in Python

p, r1, t1, r2, t2 = map(float, input().split())
interest1 = p * r1 * t1 / 100
interest2 = p * r2 * t2 / 100
print(f"Bank 1 Interest: {interest1}\nBank 2 Interest: {interest2}")
print("Bank 1" if interest1 < interest2 else "Bank 2", "offers the cheaper interest.")

Solution in Java

import java.util.Scanner;

public class LoanInterest {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        double p = scanner.nextDouble(), r1 = scanner.nextDouble(), t1 = scanner.nextDouble(), r2 = scanner.nextDouble(), t2 = scanner.nextDouble();
        double interest1 = p * r1 * t1 / 100;
        double interest2 = p * r2 * t2 / 100;
        System.out.printf("Bank 1 Interest: %.2f\nBank 2 Interest: %.2f\n", interest1, interest2);
        System.out.println((interest1 < interest2 ? "Bank 1" : "Bank 2") + " offers the cheaper interest.");
    }
}

Explanation

• The formula for interest calculation is: Interest = Principal * Rate * Time / 100.
• We calculate the interest for both banks and compare them to determine which bank offers the cheaper total interest.

Problem Statement 7: Subset Sum Problem

Problem Description:

You have a list of donation values (integers), and your function is to determine whether it's possible to pick a subset which sums up to a given target value. This is helpful in planning charity donations.

Input:

5 1 2 3 4 5 10

Output:

Subset found

Solution in C++

#include <iostream>
#include <vector>
using namespace std;

bool subsetSum(vector<int>& arr, int target) {
    vector<bool> dp(target + 1, false);
    dp[0] = true;
    for (int num : arr)
        for (int i = target; i >= num; i--) dp[i] |= dp[i - num];
    return dp[target];
}

int main() {
    int n, target;
    cin >> n;
    vector<int> arr(n);
    for (int i = 0; i < n; i++) cin >> arr[i];
    cin >> target;
    cout << (subsetSum(arr, target) ? "Subset found" : "Subset not found") << endl;
    return 0;
}

Solution in Python

def subset_sum(arr, target):
    dp = [False] * (target + 1)
    dp[0] = True
    for num in arr:
        for i in range(target, num - 1, -1):
            dp[i] |= dp[i - num]
    return dp[target]

n = int(input())
arr = list(map(int, input().split()))
target = int(input())
print("Subset found" if subset_sum(arr, target) else "Subset not found")

Solution in Java

import java.util.Scanner;

public class SubsetSum {
    public static boolean subsetSum(int[] arr, int target) {
        boolean[] dp = new boolean[target + 1];
        dp[0] = true;
        for (int num : arr)
            for (int i = target; i >= num; i--) dp[i] |= dp[i - num];
        return dp[target];
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) arr[i] = scanner.nextInt();
        int target = scanner.nextInt();
        System.out.println(subsetSum(arr, target) ? "Subset found" : "Subset not found");
    }
}

Explanation

• We use dynamic programming (DP) to solve this problem.
• dp[i] is true if a sum of i can be formed from the array.
• We update the possible sums in the DP array by iterating through each element.
• Finally, check if dp[target] is true, indicating that the target sum is achievable.

Problem Statement 8: Reverse a Linked List

Problem Description:

You are working on a system that stores customer orders in a linked list. You need to reverse the list to process them in the reverse order.

Input:

5 1 2 3 4 5

Output:

5 4 3 2 1

Solution in C++

#include <iostream>
using namespace std;

struct Node { int data; Node* next; };

void reverse(Node*& head) {
    Node *prev = nullptr, *cur = head, *next = nullptr;
    while (cur) { next = cur->next; cur->next = prev; prev = cur; cur = next; }
    head = prev;
}

int main() {
    int n, val;
    Node *head = nullptr;
    cin >> n;
    while (n--) { cin >> val; Node* newNode = new Node{val, head}; head = newNode; }
    reverse(head);
    for (Node* temp = head; temp; temp = temp->next) cout << temp->data << " ";
    cout << endl;
    return 0;
}

Solution in Python

class Node:
    def __init__(self, data):
        self.data, self.next = data, None

def reverse(head):
    prev, current = None, head
    while current:
        current.next, prev, current = prev, current, current.next
    return prev

n = int(input())
head = None
for _ in range(n):
    val = int(input())
    new_node = Node(val)
    new_node.next = head
    head = new_node

head = reverse(head)
while head:
    print(head.data, end=" ")
    head = head.next
print()

Solution in Java

import java.util.Scanner;

class Node {
    int data;
    Node next;
    Node(int data) { this.data = data; }
}

public class ReverseLinkedList {
    public static Node reverse(Node head) {
        Node prev = null, current = head, next = null;
        while (current != null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        return prev;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        Node head = null;
        for (int i = 0; i < n; i++) {
            Node newNode = new Node(scanner.nextInt());
            newNode.next = head;
            head = newNode;
        }

        head = reverse(head);
        while (head != null) {
            System.out.print(head.data + " ");
            head = head.next;
        }
        System.out.println();
    }
}

Explanation

• A Node stores an integer (data) and a reference/pointer to the next node (next).
• The reverse function iterates through the list, adjusting following pointers to reverse the list.
• Reads input values to construct the linked list, calls the reverse function, and prints the reversed list.

Problem Statement 9: Maximum Product of Two Numbers

Problem Description:

You are provided with an array of integers. Your objective is to determine the maximum product of two different numbers from the array. This will maximize the choice of two numbers so that their product is optimized in real life applications like stock market prediction or finance modeling.

Input:

5 1 -10 3 7 2

Output:

Maximum Product: 70

Solution in C++

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> arr(n);
    for (int i = 0; i < n; i++) cin >> arr[i];
    sort(arr.begin(), arr.end());
    int product = max(arr[0] * arr[1], arr[n-1] * arr[n-2]);
    cout << "Maximum Product: " << product << endl;
    return 0;
}

Solution in Python

n = int(input())
arr = list(map(int, input().split()))
arr.sort()
product = max(arr[0] * arr[1], arr[-1] * arr[-2])
print(f"Maximum Product: {product}")

Solution in Java

import java.util.Scanner;
import java.util.Arrays;

public class MaxProduct {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) arr[i] = scanner.nextInt();
        Arrays.sort(arr);
        int product = Math.max(arr[0] * arr[1], arr[n - 1] * arr[n - 2]);
        System.out.println("Maximum Product: " + product);
    }
}

Explanation

• We first read the array and sort it.
• The maximum product is either from the two most minor numbers (if they are negative) or the two most significant positive numbers.
• We use max to select the higher product between the two possibilities.

Problem Statement 10: First Non-Repeating Character

Problem Description:

You are given a string provided to you. Your job is to identify the first non-repeating character in the string. An operation of this sort is useful for programs like text parsing, in which individual characters can be used to identify errors or key words from user input.

Input:

swiss

Output:

First non-repeating character: w

Solution in C++

#include <iostream>
#include <unordered_map>
using namespace std;

int main() {
    string s;
    cin >> s;
    unordered_map<char, int> freq;
    for (char c : s) freq[c]++;
    for (char c : s) {
        if (freq[c] == 1) {
            cout << "First non-repeating character: " << c << endl;
            return 0;
        }
    }
    cout << "No non-repeating character found." << endl;
    return 0;
}

Solution in Python

s = input()
freq = {}
for c in s:
    freq[c] = freq.get(c, 0) + 1
for c in s:
    if freq[c] == 1:
        print(f"First non-repeating character: {c}")
        break
else:
    print("No non-repeating character found.")

Solution in Java

import java.util.HashMap;
import java.util.Scanner;

public class FirstNonRepeating {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s = scanner.next();
        HashMap<Character, Integer> freq = new HashMap<>();
        for (char c : s.toCharArray()) {
            freq.put(c, freq.getOrDefault(c, 0) + 1);
        }
        for (char c : s.toCharArray()) {
            if (freq.get(c) == 1) {
                System.out.println("First non-repeating character: " + c);
                return;
            }
        }
        System.out.println("No non-repeating character found.");
    }
}

Explanation

• We use a hash map to store the frequency of each character.
• We then iterate through the string to find the first character that occurs exactly once.
• If no such character is found, we return a message indicating so.

Problem Statement 11: Find Missing Number

Problem Description:

You are given an array of numbers from 1 to n with one missing number. Your task is to find the missing number. This problem helps analyze data gaps, such as finding missing entries in a sequence or identifying faulty data in data processing systems.

Input:

5
1 2 4 5

Output:

Missing Number: 3

Solution in C++

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    int sum = n * (n + 1) / 2;
    int arr_sum = 0;
    for (int i = 0; i < n - 1; i++) {
        int num;
        cin >> num;
        arr_sum += num;
    }
    cout << "Missing Number: " << sum - arr_sum << endl;
    return 0;
}

Solution in Python

n = int(input())
arr = list(map(int, input().split()))
total_sum = n * (n + 1) // 2
arr_sum = sum(arr)
print(f"Missing Number: {total_sum - arr_sum}")

Solution in Java

import java.util.Scanner;

public class MissingNumber {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int totalSum = n * (n + 1) / 2;
        int arrSum = 0;
        for (int i = 0; i < n - 1; i++) {
            arrSum += scanner.nextInt();
        }
        System.out.println("Missing Number: " + (totalSum - arrSum));
    }
}

Explanation

• We calculate the sum of numbers from 1 to n using the formula n * (n + 1) / 2.
• We then calculate the sum of the numbers in the array.
• The missing number is the difference between the expected sum and the sum of the array.

Problem Statement 12: Rotate Array

Problem Description:

You are given an array of integers and a number k. Your task is to rotate the array by k positions to the right. This problem is commonly used in real-time systems where items are rotated for processing, such as in circular buffers or data streams.

Input:

5
1 2 3 4 5
2

Output:

Rotated Array: 4 5 1 2 3

Solution in C++

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n, k;
    cin >> n;
    vector<int> arr(n);
    for (int i = 0; i < n; i++) cin >> arr[i];
    cin >> k;
    k = k % n; // handle cases where k > n
    reverse(arr.begin(), arr.end());
    reverse(arr.begin(), arr.begin() + k);
    reverse(arr.begin() + k, arr.end());
    for (int i : arr) cout << i << " ";
    cout << endl;
    return 0;
}

Solution in Python

n = int(input())
arr = list(map(int, input().split()))
k = int(input())
k = k % n  # handle cases where k > n
arr = arr[-k:] + arr[:-k]
print(" ".join(map(str, arr)))

Solution in Java

import java.util.Scanner;
import java.util.Arrays;

public class RotateArray {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) arr[i] = scanner.nextInt();
        int k = scanner.nextInt();
        k = k % n; // handle cases where k > n
        reverse(arr, 0, n - 1);
        reverse(arr, 0, k - 1);
        reverse(arr, k, n - 1);
        for (int i : arr) System.out.print(i + " ");
        System.out.println();
    }

    static void reverse(int[] arr, int start, int end) {
        while (start < end) {
            int temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }
}

Explanation

• The array is rotated by reversing parts of the array.
• We reverse the entire array, then reverse the first k and the remaining n-k elements to achieve the rotated order.

Tips to Crack the Oracle Recruitment

Here are the tips to crack Oracle coding round:

1. Prepare the Aptitude Tests

Practice quantitative reasoning and verbal ability questions.

2. Brush Up on Coding Skills

Concentrate on data structures and algorithms. Can practice with platforms such as LeetCode, HackerRank, and CodeSignal.

3. Be Informed of Oracle Products

Get accustomed to Oracle databases, cloud services, and technical solutions.

4. Mock Interview

Practice with technical and HR mock interviews to gain confidence.

5. Brush Up on SQL

As Oracle has a firm table on database-related ensuing queries, practice writing SQL queries and understanding database design.

6. Maintain Calm in the Interviews

Effectively showcase your problem-solving abilities.

Conclusion

In conclusion, coding questions in Oracle test a candidate's capability to solve problems and knowledge of data structures and algorithms. The questions can be simple and advanced, emphasising arrays, linked lists, trees, graphs, and algorithms such as sorting, searching, and dynamic programming. The Oracle coding round questions also focus on SQL and database concepts. A candidate must regularly practice coding problems, become a master at core algorithms, and write very efficient, optimised code. The way in which problems can be approached must be clear and well-defined.

Frequently Asked Questions

1. How many rounds are there in Oracle?

The typical recruitment process for Oracle comprises four rounds: an online aptitude test, technical interviews (a coding round and problem-solving), HR interviews, and sometimes another round of interviews, which is for specific roles or more senior positions.

2. Is Oracle easy to crack?

Cracking Oracle interviews could be tough to crack, as they usually want to check your problem-solving, algorithmic thinking, and coding aptitude. With steady ground work, rigorous coding practices, and a clear understanding of concepts, you can increase your odds against all others.

3. What does Oracle use for programming?

Oracle coding interviews consist of coding in different programming scripts such as Java, C++, Python, and occasionally SQL. The views check data structures, algorithms, and problem solving, with SQL being essential for database roles.

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Source: NxtWave - CCBP Blog

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