Hard Number Patterns in C, Java, Python [Interview Questions]

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In interviews, once you are giving proper answers to beginner or intermediate questions, the interviewer might start asking the hardest questions. 

In this article, I have covered the hardest number patterns that are asked by interviewers to satisfy their ego or attitude. That is the fact, but nobody will tell you about it. Don’t worry — just go with better preparation and improve your technical skills.

1. Pattern: Custom Descending Number Pattern

Expected Output: 

1234567
12345
123
1

Solution in C Program:

#include <stdio.h> int main() { // Outer loop for the number of rows (starting from 7 to 1) for (int i = 7; i >= 1; i -= 2) { // Inner loop to print numbers from 1 to i for (int j = 1; j <= i; j++) { printf("%d", j); // Print the current number } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class NumberPattern { public static void main(String[] args) { // Outer loop for the number of rows (starting from 7 down to 1) for (int i = 7; i >= 1; i -= 2) { // Inner loop to print numbers from 1 to i for (int j = 1; j <= i; j++) { System.out.print(j); // Print the current number } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

# Outer loop for the number of rows (starting from 7 down to 1) for i in range(7, 0, -2): # Inner loop to print numbers from 1 to i for j in range(1, i + 1): print(j, end='') # Print the current number, stay on the same line print() # Move to the next line after each row

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2. Pattern: Mixed Ascending and Descending Number Pattern

Expected Output:

12345
4321
123
21
1

Solution in C Program:

#include <stdio.h> int main() { // Outer loop for the number of rows (starting from 5 to 1) for (int i = 5; i >= 1; i -= 2) { // Print ascending numbers for odd rows if ((5 - i) % 4 == 0) { for (int j = 1; j <= i; j++) { printf("%d", j); // Print ascending number } } // Print descending numbers for even rows else { for (int j = i; j >= 1; j--) { printf("%d", j); // Print descending number } } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class NumberPattern { public static void main(String[] args) { // Outer loop for the number of rows (starting from 5 to 1) for (int i = 5; i >= 1; i -= 2) { // Print ascending numbers for odd rows if ((5 - i) % 4 == 0) { for (int j = 1; j <= i; j++) { System.out.print(j); // Print ascending number } } // Print descending numbers for even rows else { for (int j = i; j >= 1; j--) { System.out.print(j); // Print descending number } } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

# Outer loop for the number of rows (starting from 5 down to 1) for i in range(5, 0, -2): # Print ascending numbers for odd rows if (5 - i) % 4 == 0: for j in range(1, i + 1): print(j, end='') # Print ascending number, stay on the same line # Print descending numbers for even rows else: for j in range(i, 0, -1): print(j, end='') # Print descending number, stay on the same line print() # Move to the next line after each row

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3. Pattern: Alternating Binary Pattern

Expected Output:

1
01
101
0101

Solution in C Program:

#include <stdio.h> int main() { // Outer loop for the number of rows (1 to 4) for (int i = 1; i <= 4; i++) { // Inner loop to print alternating 1 and 0 for (int j = 1; j <= i; j++) { if (j % 2 == 1) { printf("1"); // Print 1 for odd positions } else { printf("0"); // Print 0 for even positions } } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class BinaryPattern { public static void main(String[] args) { // Outer loop for the number of rows (1 to 4) for (int i = 1; i <= 4; i++) { // Inner loop to print alternating 1 and 0 for (int j = 1; j <= i; j++) { if (j % 2 == 1) { System.out.print("1"); // Print 1 for odd positions } else { System.out.print("0"); // Print 0 for even positions } } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

# Outer loop for the number of rows (1 to 4) for i in range(1, 5): # Inner loop to print alternating 1 and 0 for j in range(1, i + 1): if j % 2 == 1: print(1, end='') # Print 1 for odd positions else: print(0, end='') # Print 0 for even positions print() # Move to the next line after each row

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4. Pattern: Descending Odd Number Pattern

Expected Output:

13579
3579
579
79
9

Solution in C Program:

#include <stdio.h> int main() { // Outer loop for the number of rows (starting from 5 down to 1) for (int i = 1; i <= 5; i++) { // Inner loop to print odd numbers starting from 2*i-1 to 9 for (int j = 2 * i - 1; j <= 9; j += 2) { printf("%d", j); // Print odd numbers starting from 2*i - 1 } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class OddNumberPattern { public static void main(String[] args) { // Outer loop for the number of rows (starting from 5 down to 1) for (int i = 1; i <= 5; i++) { // Inner loop to print odd numbers starting from 2*i-1 to 9 for (int j = 2 * i - 1; j <= 9; j += 2) { System.out.print(j); // Print odd numbers starting from 2*i - 1 } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

# Outer loop for the number of rows (starting from 5 down to 1) for i in range(1, 6): # Inner loop to print odd numbers starting from 2*i-1 to 9 for j in range(2 * i - 1, 10, 2): print(j, end='') # Print odd numbers starting from 2*i - 1 print() # Move to the next line after each row

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5. Pattern: Alternating Odd and Even Number Pattern

Expected Output:

1
2 4
1 3 5
2 4 6 8
1 3 5 7 9

Solution in C Program:

#include <stdio.h> int main() { // Outer loop for the number of rows (1 to 5) for (int i = 1; i <= 5; i++) { // Inner loop to print numbers based on row parity (odd or even) for (int j = 1; j <= i; j++) { if (i % 2 == 1) { // Odd rows (1st, 3rd, 5th) printf("%d ", 2 * j - 1); // Print odd numbers (1, 3, 5, etc.) } else { // Even rows (2nd, 4th) printf("%d ", 2 * j); // Print even numbers (2, 4, 6, etc.) } } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class NumberPattern { public static void main(String[] args) { // Outer loop for the number of rows (1 to 5) for (int i = 1; i <= 5; i++) { // Inner loop to print numbers based on row parity (odd or even) for (int j = 1; j <= i; j++) { if (i % 2 == 1) { // Odd rows (1st, 3rd, 5th) System.out.print((2 * j - 1) + " "); // Print odd numbers (1, 3, 5, etc.) } else { // Even rows (2nd, 4th) System.out.print((2 * j) + " "); // Print even numbers (2, 4, 6, etc.) } } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

# Outer loop for the number of rows (1 to 5) for i in range(1, 6): # Inner loop to print numbers based on row parity (odd or even) for j in range(1, i + 1): if i % 2 == 1: # Odd rows (1st, 3rd, 5th) print(2 * j - 1, end=' ') # Print odd numbers (1, 3, 5, etc.) else: # Even rows (2nd, 4th) print(2 * j, end=' ') # Print even numbers (2, 4, 6, etc.) print() # Move to the next line after each row

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6. Pattern: Descending Number Pattern with 5s

Expected Output:

55555
45555
34555
23455
12345

Solution in C Program:

#include <stdio.h> int main() { // Outer loop for the number of rows (1 to 5) for (int i = 5; i >= 1; i--) { // Print descending numbers starting from i for (int j = i; j >= 1; j--) { printf("%d", j); // Print the descending number } // Print 5s for the remaining positions for (int j = i + 1; j <= 5; j++) { printf("5"); // Print 5 for the remaining positions } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class NumberPattern { public static void main(String[] args) { // Outer loop for the number of rows (5 down to 1) for (int i = 5; i >= 1; i--) { // Print descending numbers starting from i for (int j = i; j >= 1; j--) { System.out.print(j); // Print the descending number } // Print 5s for the remaining positions for (int j = i + 1; j <= 5; j++) { System.out.print("5"); // Print 5 for the remaining positions } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

# Outer loop for the number of rows (5 down to 1) for i in range(5, 0, -1): # Print descending numbers starting from i for j in range(i, 0, -1): print(j, end='') # Print the descending number # Print 5s for the remaining positions for j in range(i + 1, 6): print(5, end='') # Print 5 for the remaining positions print() # Move to the next line after each row

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7. Pattern: Binary Alternating Pattern

Expected Output:

1
10
101
1010
10101

Solution in C Program:

#include <stdio.h> int main() { // Outer loop for the number of rows (1 to 5) for (int i = 1; i <= 5; i++) { // Inner loop to print alternating 1 and 0 for (int j = 1; j <= i; j++) { if (j % 2 == 1) { printf("1"); // Print 1 for odd positions } else { printf("0"); // Print 0 for even positions } } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class BinaryAlternatingPattern { public static void main(String[] args) { // Outer loop for the number of rows (1 to 5) for (int i = 1; i <= 5; i++) { // Inner loop to print alternating 1 and 0 for (int j = 1; j <= i; j++) { if (j % 2 == 1) { System.out.print("1"); // Print 1 for odd positions } else { System.out.print("0"); // Print 0 for even positions } } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

# Outer loop for the number of rows (1 to 5) for i in range(1, 6): # Inner loop to print alternating 1 and 0 for j in range(1, i + 1): if j % 2 == 1: print(1, end='') # Print 1 for odd positions else: print(0, end='') # Print 0 for even positions print() # Move to the next line after each row

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8. Pattern: Sequential Number Pattern

Expected Output:

1
23
456
78910

Solution in C Program:

#include <stdio.h> int main() { int num = 1; // Start with 1 // Outer loop for the number of rows (1 to 4) for (int i = 1; i <= 4; i++) { // Inner loop to print numbers in each row for (int j = 1; j <= i; j++) { printf("%d", num); // Print current number num++; // Increment the number for the next position } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class SequentialNumberPattern { public static void main(String[] args) { int num = 1; // Start with 1 // Outer loop for the number of rows (1 to 4) for (int i = 1; i <= 4; i++) { // Inner loop to print numbers in each row for (int j = 1; j <= i; j++) { System.out.print(num); // Print current number num++; // Increment the number for the next position } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

num = 1 # Start with 1 # Outer loop for the number of rows (1 to 4) for i in range(1, 5): # Inner loop to print numbers in each row for j in range(1, i + 1): print(num, end='') # Print current number num += 1 # Increment the number for the next position print() # Move to the next line after each row

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9. Pattern: Palindromic Number Pattern with Stars

Expected Output:

12344321
123**321
12****21
1******1

Solution in C Program:

#include <stdio.h> int main() { // Outer loop for the number of rows (1 to 4) for (int i = 1; i <= 4; i++) { // Print increasing numbers from 1 to (5 - i) for (int j = 1; j <= 5 - i; j++) { printf("%d", j); } // Print stars in the middle for (int j = 1; j <= 2 * i - 2; j++) { printf("*"); } // Print decreasing numbers from (5 - i) to 1 for (int j = 5 - i; j >= 1; j--) { printf("%d", j); } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class PalindromicNumberPattern { public static void main(String[] args) { // Outer loop for the number of rows (1 to 4) for (int i = 1; i <= 4; i++) { // Print increasing numbers from 1 to (5 - i) for (int j = 1; j <= 5 - i; j++) { System.out.print(j); } // Print stars in the middle for (int j = 1; j <= 2 * i - 2; j++) { System.out.print("*"); } // Print decreasing numbers from (5 - i) to 1 for (int j = 5 - i; j >= 1; j--) { System.out.print(j); } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

# Outer loop for the number of rows (1 to 4) for i in range(1, 5): # Print increasing numbers from 1 to (5 - i) for j in range(1, 5 - i + 1): print(j, end='') # Print stars in the middle for j in range(1, 2 * i): print("*", end='') # Print decreasing numbers from (5 - i) to 1 for j in range(5 - i, 0, -1): print(j, end='') print() # Move to the next line after each row

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10. Pattern: Right-Aligned Number Pattern

Expected Output:

1 2 3 4 5 6 7 8 9

Solution in C Program:

#include <stdio.h> int main() { int num = 1; // Start with 1 // Outer loop for the number of rows (3 rows) for (int i = 1; i <= 3; i++) { // Print leading spaces for right alignment for (int j = 1; j <= 3 - i; j++) { printf(" "); // Print spaces to right-align the numbers } // Print the numbers in the row for (int j = 1; j <= 2 * i - 1; j++) { printf("%d ", num); // Print the current number num++; // Increment the number } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class RightAlignedNumberPattern { public static void main(String[] args) { int num = 1; // Start with 1 // Outer loop for the number of rows (3 rows) for (int i = 1; i <= 3; i++) { // Print leading spaces for right alignment for (int j = 1; j <= 3 - i; j++) { System.out.print(" "); // Print spaces to right-align the numbers } // Print the numbers in the row for (int j = 1; j <= 2 * i - 1; j++) { System.out.print(num + " "); // Print the current number num++; // Increment the number } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

num = 1 # Start with 1 # Outer loop for the number of rows (3 rows) for i in range(1, 4): # Print leading spaces for right alignment for j in range(1, 4 - i): print(" ", end="") # Print spaces to right-align the numbers # Print the numbers in the row for j in range(1, 2 * i): print(num, end=" ") # Print the current number num += 1 # Increment the number print() # Move to the next line after each row

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11. Pattern: Column-wise Number Pattern

Expected Output:

1
2 6
3 7 10
4 8 11 13
5 9 12 14 15

Solution in C Program:

#include <stdio.h> int main() { int num = 1; // Start with 1 int matrix[5][5]; // To store the pattern numbers // Fill the matrix with the correct pattern for (int i = 0; i < 5; i++) { for (int j = 0; j <= i; j++) { matrix[j][i] = num; // Fill column-wise num++; // Increment the number } } // Print the matrix to show the pattern for (int i = 0; i < 5; i++) { for (int j = 0; j <= i; j++) { printf("%d ", matrix[j][i]); // Print the numbers in the correct order } printf("n"); // Move to the next line after each row } return 0; }

Solution in Java:

public class ColumnWiseNumberPattern { public static void main(String[] args) { int num = 1; // Start with 1 int[][] matrix = new int[5][5]; // To store the pattern numbers // Fill the matrix with the correct pattern for (int i = 0; i < 5; i++) { for (int j = 0; j <= i; j++) { matrix[j][i] = num; // Fill column-wise num++; // Increment the number } } // Print the matrix to show the pattern for (int i = 0; i < 5; i++) { for (int j = 0; j <= i; j++) { System.out.print(matrix[j][i] + " "); // Print the numbers in the correct order } System.out.println(); // Move to the next line after each row } } }

Solution in Python:

num = 1 # Start with 1 matrix = [[0 for _ in range(5)] for _ in range(5)] # To store the pattern numbers # Fill the matrix with the correct pattern for i in range(5): for j in range(i + 1): matrix[j][i] = num # Fill column-wise num += 1 # Increment the number # Print the matrix to show the pattern for i in range(5): for j in range(i + 1): print(matrix[j][i], end=" ") # Print the numbers in the correct order print() # Move to the next line after each row

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Oops… I forgot to add a 12-number pattern in the image. 

12. Pattern: Diamond Number Pattern

Expected Output:

1 123 12345 1234567 123456789 1234567 12345 123 1

Solution in C Program:

#include <stdio.h> int main() { int n = 5; // The maximum row length (central row length) // Upper half including the middle row for (int i = 1; i <= n; i++) { // Print leading spaces for (int j = 1; j <= n - i; j++) { printf(" "); } // Print increasing numbers for (int j = 1; j <= 2 * i - 1; j++) { printf("%d", j); } printf("n"); } // Lower half (excluding the middle row) for (int i = n - 1; i >= 1; i--) { // Print leading spaces for (int j = 1; j <= n - i; j++) { printf(" "); } // Print increasing numbers for (int j = 1; j <= 2 * i - 1; j++) { printf("%d", j); } printf("n"); } return 0; }

Solution in Java:

public class DiamondNumberPattern { public static void main(String[] args) { int n = 5; // The maximum row length (central row length) // Upper half including the middle row for (int i = 1; i <= n; i++) { // Print leading spaces for (int j = 1; j <= n - i; j++) { System.out.print(" "); } // Print increasing numbers for (int j = 1; j <= 2 * i - 1; j++) { System.out.print(j); } System.out.println(); } // Lower half (excluding the middle row) for (int i = n - 1; i >= 1; i--) { // Print leading spaces for (int j = 1; j <= n - i; j++) { System.out.print(" "); } // Print increasing numbers for (int j = 1; j <= 2 * i - 1; j++) { System.out.print(j); } System.out.println(); } } }

Solution in Python:

n = 5 # The maximum row length (central row length) # Upper half including the middle row for i in range(1, n + 1): # Print leading spaces print(" " * (n - i), end="") # Print increasing numbers for j in range(1, 2 * i): print(j, end="") print() # Lower half (excluding the middle row) for i in range(n - 1, 0, -1): # Print leading spaces print(" " * (n - i), end="") # Print increasing numbers for j in range(1, 2 * i): print(j, end="") print()

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