Revature Previous Coding Questions
🔥 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 Nspace-separated integers forming the AP
🧾 Output
- A single integer — the sum of the AP
✅ Example
Input:
5
2 4 6 8 10
Output:
30
🔣Code
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)
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);
}
}
🔍 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
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 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);
}
}
🔍 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
ch = input().strip()
for i in range(5, 0, -1):
print(chr(ord(ch) - i), end=' ')
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) + " ");
}
}
}
🔍 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
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))
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);
}
}
🔍 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
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)
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);
}
}
🔍 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
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
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;
}
}
}
}
}
🔍 Explanation
We create a sorted set to get unique values.
The third element is selected, and we find its first occurrence index (1-based).
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