Oracle Coding Questions With Solutions

Here are the key topics you need to focus on for the Oracle coding assessment in 2024:

• Arrays
• Linked Lists
• Stacks
• Queues
• Hash Tables
• Heaps
• Searching Algorithms (e.g., Binary Search)
• Sorting Algorithms (e.g., Quick Sort, Merge Sort)
• Complexity Theory (Time and Space Complexity)
• Procedures, Functions, and Scope

Here are the sample Oracle coding round questions with solutions:

1. Ratan is a wealthy investor who wants to maximize his profits by buying and selling stocks. Given the stock prices for several days, calculate the maximum profit he could earn.

Input Format

• The first line contains an integer n, the number of days.
• The next n lines contain the stock prices for each day.

Output Format

• Output a single integer representing the maximum profit Ratan could make.

Example 

Input
7
1
9
2
11
1
9
2
Output
10

Code Solutions

C

#include <stdio.h>

int main() {
    int n, profit = 0, min_price = 1e9;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        int price;
        scanf("%d", &price);
        if (price < min_price) min_price = price;
        profit = (price - min_price > profit) ? price - min_price : profit;
    }
    printf("%d\n", profit);
    return 0;
}

Java

import java.util.Scanner;

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

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

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)

2. You have received loan offers from two banks with varying interest rates over different tenures. Calculate the total interest paid for each bank and decide which offer is cheaper.

Input Format

• First line: Principal amount P.
• Second line: Tenure T in years.
• Third line: Number of slabs N1 for Bank A, followed by N1 lines of interest rates and their corresponding periods.
• Next, the same for Bank B with N2 slabs.

Output Format

• Output either "Bank A" or "Bank B" based on which bank offers the lesser total interest.

Example

Input
10000
20
3
5 9.5
10 9.6
5 8.5
3
10 6.9
5 8.5
5 7.9

Output
Bank B

Code Solutions

C

#include <stdio.h>
#include <math.h>

double calculate_total_emi(double p, int years, int slabs) {
    double total = 0;
    for (int i = 0; i < slabs; i++) {
        int period;
        double rate;
        scanf("%d %lf", &period, &rate);
        double monthlyRate = rate / 1200; // converting to monthly
        double emi = (p * monthlyRate) / (1 - pow(1 + monthlyRate, -period * 12));
        total += emi * period * 12; // total interest paid
    }
    return total;
    }

int main() {
    double P;
    int T, N1, N2;
    scanf("%lf %d", &P, &T);

    scanf("%d", &N1);
    double bankA_total = calculate_total_emi(P, T, N1);

    scanf("%d", &N2);
    double bankB_total = calculate_total_emi(P, T, N2);

    if (bankA_total < bankB_total) printf("Bank A\n");
    else printf("Bank B\n");
    
    return 0;
}

Java

import java.util.Scanner;

public class BestBank {
    static double calculateTotalEMI(double principal, int slabs, Scanner scanner) {
        double total = 0;
        for (int i = 0; i < slabs; i++) {
            int period;
            double rate;
            period = scanner.nextInt();
            rate = scanner.nextDouble();
            double monthlyRate = rate / 1200; // converting to monthly
            double emi = (principal * monthlyRate) / (1 - Math.pow(1 + monthlyRate, -period * 12));
            total += emi * period * 12; // total interest paid
        }
        return total;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        double P = scanner.nextDouble();
        int T = scanner.nextInt();
        int N1 = scanner.nextInt();
        double bankA_total = calculateTotalEMI(P, N1, scanner);
        int N2 = scanner.nextInt();
        double bankB_total = calculateTotalEMI(P, N2, scanner);

        System.out.println(bankA_total < bankB_total ? "Bank A" : "Bank B");
    }
}

3. Determine how many borrow operations are needed to subtract two numbers, or return "Not possible" if the subtraction can't be performed.

Input Format

• Two lines containing the two numbers as strings.

Output Format

• Number of borrows or "Not possible".

Example

Input
754
658

Output
2

Code Solutions

C

#include <stdio.h>
#include <string.h>

int countBorrows(char* num1, char* num2) {
    int len1 = strlen(num1);
    int 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];
    scanf("%s %s", num1, num2);
    int result = countBorrows(num1, num2);
    if (result == -1) {
        printf("Not possible\n");
    } else {
        printf("%d\n", result);
    }
    return 0;
}

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);
        }
    }
}

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)

4. Given an array of integers, count the number of distinct elements in a specified range.

Input Format

• First line: integer n (size of array).
• Second line: n integers.
• Third line: two integers representing the range [l, r].

Output Format

• Number of distinct elements in the range.

Example 

Input
5
1 2 2 3 4
1 3

Output
3

Code Solutions

C

#include <stdio.h>
#include <stdlib.h>

int distinctCount(int* arr, int l, int r) {
    int hash[1000] = {0}; // Adjust size as necessary
    for (int i = l; i <= r; i++) {
        hash[arr[i]] = 1;
    }
    int count = 0;
    for (int i = 0; i < 1000; i++) {
        if (hash[i] == 1) count++;
    }
    return count;
}

int main() {
    int n, l, r;
    scanf("%d", &n);
    int arr[n];
    for (int i = 0; i < n; i++) {
        scanf("%d", &arr[i]);
    }
    scanf("%d %d", &l, &r);
    printf("%d\n", distinctCount(arr, l, r));
    return 0;
}

Java

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

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

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))

5. Given an array consisting of colors represented as integers, sort the array so that all occurrences of color 0 come first, followed by 1s, and then 2s.

Input Format

• First line: integer n (size of array).
• Second line: n integers (0, 1, or 2).

Output Format

• Sorted array.

Example

Input
5
2 0 1 2 0

Output
0 0 1 2 2

Code Solutions

C

#include <stdio.h>

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;
    scanf("%d", &n);
    int arr[n];
    for (int i = 0; i < n; i++) {
        scanf("%d", &arr[i]);
    }
    sortColors(arr, n);
    for (int i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
    printf("\n");
    return 0;
}

Java

import java.util.Scanner;

public class SortColors {
    public static void sortColors(int[] arr) {
        int[] count = new int[3];
        for (int num : arr) {
            count[num]++;
        }
        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;
    }

    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();
    }
}

Python

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

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

6. Count the number of vowels in a given string.

Input Format

• A single line containing the string.

Output Format

• Number of vowels in the string.

Example

Input
Hello World

Output
3

Code Solutions

C

#include <stdio.h>
#include <string.h>

int countVowels(char* str) {
    int count = 0;
    for (int i = 0; str[i]; i++) {
        char c = str[i];
        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() {
    char str[100];
    fgets(str, sizeof(str), stdin);
    printf("%d\n", countVowels(str));
    return 0;
}

Java

import java.util.Scanner;

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

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

Python

def count_vowels(s):
    return sum(1 for char in s if char.lower() in 'aeiou')

s = input()
print(count_vowels(s))

Here are some essential tips for cracking Oracle coding questions:

• Focus on data structures, algorithms, SQL, and PL/SQL fundamentals.
• Use platforms like LeetCode and HackerRank for coding problems. The more you code the better you’ll get.
• Simulate real interview conditions to improve your communication and problem-solving skills.
• After solving discuss potential optimizations and edge cases.
• Review incorrect answers to reinforce concepts and improve.
• Maintain confidence and clarity under pressure during interviews.

Getting ready for Oracle coding questions takes time and effort, but with the right preparation, you can succeed. By understanding important concepts, practicing common questions, and using our tips, you'll be well-equipped for your Oracle coding interview. 

1. How can I improve my skills for Oracle Java coding interview questions?

Practice Java programming and learn about how Java works with Oracle databases using JDBC.

2. What resources can help me prepare for Oracle online coding test questions?

Look for online coding platforms, Oracle documentation, and coding challenge websites to practice.