Revature Previous Coding Questions

July 24, 2025 NextGenKodingHub

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🔥 Revature Previous Coding Questions with Java & Python Solutions

Welcome to this specially curated post featuring Revature Previous Coding Questions.
Each problem includes:

A clear problem statement
Java and Python solutions
Example input/output
Concise and clear explanation

Let’s get started! 🚀

✅ Problem 1: Sum of Arithmetic Progression

📝 Problem Statement

You are given an arithmetic progression (AP) with N terms.
Your task is to calculate the sum of the series.
You may derive the first term A and common difference D from the input.

📐 Formula Used

Sum = (N / 2) × (2A + (N – 1) × D)

🔢 Input

Integer N — number of terms
N space-separated integers forming the AP

🧾 Output

A single integer — the sum of the AP

✅ Example

Input:
5
2 4 6 8 10

Output:
30

import java.util.*;

public class APSum {
    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 &lt; N; i++)
            arr[i] = sc.nextInt();
        
        int A = arr[0];
        int D = arr[1] - arr[0];
        int sum = (N * (2 * A + (N - 1) * D)) / 2;
        
        System.out.println(sum);
    }
}

import java.util.*;

public class APSum {

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 A = arr[0];

int D = arr[1] - arr[0];

int sum = (N * (2 * A + (N - 1) * D)) / 2;

System.out.println(sum);

}

N = int(input())
arr = list(map(int, input().split()))

A = arr[0]
D = arr[1] - arr[0]
sum_ap = (N * (2 * A + (N - 1) * D)) // 2

print(sum_ap)

N = int(input())

arr = list(map(int, input().split()))

A = arr[0]

D = arr[1] - arr[0]

sum_ap = (N * (2 * A + (N - 1) * D)) // 2

print(sum_ap)

🔍 Explanation

We extract the first element A and compute the common difference D.
Using the arithmetic progression sum formula, we calculate and print the total.

✅ Problem 2: Pointer Conversion in Code

📝 Problem Statement

Given C-style code, convert expressions like a*b into a->b,
except on lines that start with comments (//).

🔢 Input

Up to 100 lines of C-style code
Each line ≤ 50 characters

🧾 Output

Modified code with * replaced by -> where appropriate

✅ Example

Input:
// ignore this line
a*b = 5;
c*d = a*b;

Output:
// ignore this line
a->b = 5;
c->d = a->b;

🔣Code

Java
Python

import java.util.*;

public class PointerConversion {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        List&lt;String> lines = new ArrayList&lt;>();

        while (sc.hasNextLine()) {
            String line = sc.nextLine();
            if (line.trim().startsWith("//")) {
                lines.add(line);
            } else {
                lines.add(line.replaceAll("(\\w+)\\*(\\w+)", "$1->$2"));
            }
        }

        for (String line : lines)
            System.out.println(line);
    }
}

import java.util.*;

public class PointerConversion {

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

List<String> lines = new ArrayList<>();

while (sc.hasNextLine()) {

String line = sc.nextLine();

if (line.trim().startsWith("//")) {

lines.add(line);

} else {

lines.add(line.replaceAll("(\\w+)\\*(\\w+)", "$1->$2"));

}

for (String line : lines)

System.out.println(line);

}

import re
import sys

lines = sys.stdin.read().splitlines()

for line in lines:
    if line.strip().startswith("//"):
        print(line)
    else:
        print(re.sub(r'(\w+)\*(\w+)', r'\1->\2', line))

import re

import sys

lines = sys.stdin.read().splitlines()

for line in lines:

if line.strip().startswith("//"):

print(line)

else:

print(re.sub(r'(\w+)\*(\w+)', r'\1->\2', line))

🔍 Explanation

Using regex, we safely convert pointer-like expressions to the -> format
without affecting comment lines that start with //.

✅ Problem 3: Previous 5 Characters

📝 Problem Statement

Given a capital letter C, print the 5 letters that precede it alphabetically.

🔢 Input

A single uppercase letter C (C > ‘E’)

🧾 Output

5 space-separated characters before C

✅ Example

Input:
K

Output:
F G H I J

🔣Code

Java
Python

import java.util.Scanner;

public class PrevFive {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        char ch = sc.next().charAt(0);

        for (int i = 5; i >= 1; i--) {
            System.out.print((char)(ch - i) + " ");
        }
    }
}

import java.util.Scanner;

public class PrevFive {

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

char ch = sc.next().charAt(0);

for (int i = 5; i >= 1; i--) {

System.out.print((char)(ch - i) + " ");

}

ch = input().strip()

for i in range(5, 0, -1):
    print(chr(ord(ch) - i), end=' ')

ch = input().strip()

for i in range(5, 0, -1):

print(chr(ord(ch) - i), end=' ')

🔍 Explanation

We use ASCII math to get previous characters using ord() and chr(),
then loop from -5 to -1 to print the correct sequence.

✅ Problem 4: Average of Middle Digits

📝 Problem Statement

Given an integer:

If it has even digits → take 2 middle digits
If odd → take 3 middle digits

Return the floor average of these digits.

🔢 Input

A single integer with at least 2 digits

🧾 Output

Integer — floor of the average

✅ Example

Input:
123456

Output:
4
(Middle digits: 3 and 4 → average = (3+4)//2 = 3)

🔣Code

Java
Python

import java.util.Scanner;

public class MiddleDigitAverage {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String num = sc.next();

        int len = num.length();
        int sum;

        if (len % 2 == 0) {
            int mid1 = num.charAt(len/2 - 1) - '0';
            int mid2 = num.charAt(len/2) - '0';
            sum = (mid1 + mid2) / 2;
        } else {
            int mid1 = num.charAt(len/2 - 1) - '0';
            int mid2 = num.charAt(len/2) - '0';
            int mid3 = num.charAt(len/2 + 1) - '0';
            sum = (mid1 + mid2 + mid3) / 3;
        }

        System.out.println(sum);
    }
}

import java.util.Scanner;

public class MiddleDigitAverage {

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

String num = sc.next();

int len = num.length();

int sum;

if (len % 2 == 0) {

int mid1 = num.charAt(len/2 - 1) - '0';

int mid2 = num.charAt(len/2) - '0';

sum = (mid1 + mid2) / 2;

} else {

int mid1 = num.charAt(len/2 - 1) - '0';

int mid2 = num.charAt(len/2) - '0';

int mid3 = num.charAt(len/2 + 1) - '0';

sum = (mid1 + mid2 + mid3) / 3;

}

System.out.println(sum);

}

num = input().strip()
length = len(num)

if length % 2 == 0:
    mid = length // 2
    digits = [int(num[mid-1]), int(num[mid])]
else:
    mid = length // 2
    digits = [int(num[mid-1]), int(num[mid]), int(num[mid+1])]

print(sum(digits) // len(digits))

num = input().strip()

length = len(num)

if length % 2 == 0:

mid = length // 2

digits = [int(num[mid-1]), int(num[mid])]

else:

mid = length // 2

digits = [int(num[mid-1]), int(num[mid]), int(num[mid+1])]

print(sum(digits) // len(digits))

🔍 Explanation

We get the middle digits based on string length and
calculate the integer (floor) average using integer division.

✅ Problem 5: First 4-Letter Word Position

📝 Problem Statement

Given a sentence, find the 1-based position of the first word with exactly 4 characters.

🔢 Input

A sentence

🧾 Output

Index of the first 4-letter word or -1

✅ Example

Input:
We love open code!

Output:
2

🔣Code

Java
Python

import java.util.*;

public class First4LetterWord {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String[] words = sc.nextLine().split("\\s+");

        for (int i = 0; i &lt; words.length; i++) {
            if (words[i].length() == 4) {
                System.out.println(i + 1);
                return;
            }
        }

        System.out.println(-1);
    }
}

import java.util.*;

public class First4LetterWord {

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

String[] words = sc.nextLine().split("\\s+");

for (int i = 0; i < words.length; i++) {

if (words[i].length() == 4) {

System.out.println(i + 1);

return;

}

System.out.println(-1);

}

words = input().split()

idx = 1
found = False

for word in words:
    if len(word) == 4:
        print(idx)
        found = True
        break
    idx += 1

if not found:
    print(-1)

words = input().split()

idx = 1

found = False

for word in words:

if len(word) == 4:

print(idx)

found = True

break

idx += 1

if not found:

print(-1)

🔍 Explanation

The Java and Python programs take a line of text, break it into words, and find the first word that has exactly 4 letters. They then print its position (starting from 1), or -1 if no such word exists.

✅ Problem 6: Third Smallest Element Position

📝 Problem Statement

Given a list of integers, find the 1-based position of the third smallest unique element.
If there are fewer than 3 unique elements, return -1.

🔢 Input

First line: integer n
Second line: n space-separated integers

🧾 Output

Index of third smallest unique number or -1

✅ Example

Input:
6
4 1 2 1 2 3

Output:
6

🔣Code

Java
Python

import java.util.*;

public class ThirdSmallestPosition {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        List&lt;Integer> list = new ArrayList&lt;>();
        for (int i = 0; i &lt; n; i++)
            list.add(sc.nextInt());

        Set&lt;Integer> set = new TreeSet&lt;>(list);

        if (set.size() &lt; 3) {
            System.out.println(-1);
        } else {
            int count = 0, third = 0;
            for (int num : set) {
                count++;
                if (count == 3) {
                    third = num;
                    break;
                }
            }

            for (int i = 0; i &lt; list.size(); i++) {
                if (list.get(i) == third) {
                    System.out.println(i + 1);
                    break;
                }
            }
        }
    }
}

import java.util.*;

public class ThirdSmallestPosition {

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

int n = sc.nextInt();

List<Integer> list = new ArrayList<>();

for (int i = 0; i < n; i++)

list.add(sc.nextInt());

Set<Integer> set = new TreeSet<>(list);

if (set.size() < 3) {

System.out.println(-1);

} else {

int count = 0, third = 0;

for (int num : set) {

count++;

if (count == 3) {

third = num;

break;

}

for (int i = 0; i < list.size(); i++) {

if (list.get(i) == third) {

System.out.println(i + 1);

break;

}

n = int(input())
arr = list(map(int, input().split()))

unique = sorted(set(arr))

if len(unique) &lt; 3:
    print(-1)
else:
    third = unique[2]
    for i, val in enumerate(arr):
        if val == third:
            print(i + 1)
            break

n = int(input())

arr = list(map(int, input().split()))

unique = sorted(set(arr))

if len(unique) < 3:

print(-1)

else:

third = unique[2]

for i, val in enumerate(arr):

if val == third:

print(i + 1)

break

🔍 Explanation

We create a sorted set to get unique values.
The third element is selected, and we find its first occurrence index (1-based).

🚨 Revature Overview!

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🔥 Revature Previous Coding Questions with Java & Python Solutions

✅ Problem 1: Sum of Arithmetic Progression

📝 Problem Statement

📐 Formula Used

🔢 Input

🧾 Output

✅ Example

🔣Code

🔍 Explanation

✅ Problem 2: Pointer Conversion in Code

📝 Problem Statement

🔢 Input

🧾 Output

✅ Example

🔣Code

🔍 Explanation

✅ Problem 3: Previous 5 Characters

📝 Problem Statement

🔢 Input

🧾 Output

✅ Example

🔣Code

🔍 Explanation

✅ Problem 4: Average of Middle Digits

📝 Problem Statement

🔢 Input

🧾 Output

✅ Example

🔣Code

🔍 Explanation

✅ Problem 5: First 4-Letter Word Position

📝 Problem Statement

🔢 Input

🧾 Output

✅ Example

🔣Code

🔍 Explanation

✅ Problem 6: Third Smallest Element Position

📝 Problem Statement

🔢 Input

🧾 Output

✅ Example

🔣Code

🔍 Explanation

🚨 Revature Overview!

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