Zoho Basic Programming Round 2025

C++

#include <iostream>
using namespace std;
void printNumberStaircase(int n) {
    for (int i = 1; i <= n; i++) {
        for (int j = i; j < i + n; j++) {
            cout << (j - 1) % n + 1 << " ";
        }
        cout << endl;
    }
}
int main() {
    printNumberStaircase(5);
    cout << endl;
    printNumberStaircase(3);
    return 0;
}

Java

public class NumberStaircase {
    public static void printNumberStaircase(int n) {
        for (int i = 1; i <= n; i++) {
            for (int j = i; j < i + n; j++) {
                System.out.print((j - 1) % n + 1 + " ");
            }
            System.out.println();
        }
    }
    public static void main(String[] args) {
        printNumberStaircase(5);
        System.out.println();
        printNumberStaircase(3);
    }
}

Python

def print_number_staircase(n):
    for i in range(1, n + 1):
        row = []
        for j in range(i, i + n):
            row.append((j - 1) % n + 1)
        print(" ".join(map(str, row)))
print_number_staircase(5)
print()
print_number_staircase(3)

Explanation of logic:

• Outer Loop (i): Iterates through each row from 1 to n.
• Inner Loop (j): Generates numbers for each row starting from i and wrapping around when the number exceeds n.
• (j - 1) % n + 1: Ensures numbers cycle back to 1 after n.
• Print Numbers: Each number is printed followed by a space, and a newline is printed after completing the row.
• Main Function: Demonstrates the function with test cases for n = 5 and n = 3.

Example Outputs:
For n = 5:

1 2 3 4 5
2 3 4 5 1
3 4 5 1 2
4 5 1 2 3
5 1 2 3 4

For n = 3:

1 2 3
2 3 1
3 1 2

C++

#include <iostream>
#include <string>
using namespace std;
string find_day(int date, int month, int year) {
    string days[] = {"sunday", "monday", "tuesday", "wednesday", "thursday", "friday", "saturday"};
    int month_days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    bool isLeap = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
    if (isLeap) month_days[1] = 29;
    int total_days = 0;
    total_days += (year - 1) * 365 + (year - 1) / 4 - (year - 1) / 100 + (year - 1) / 400;
    for (int i = 0; i < month - 1; i++) {
        total_days += month_days[i];
    }
    total_days += date;
    return days[total_days % 7];
}
int main() {
    cout << find_day(15, 8, 1947) << endl;
    cout << find_day(26, 1, 1950) << endl;
    return 0;
}

Java

public class FindDay {
    public static String find_day(int date, int month, int year) {
        String[] days = {"sunday", "monday", "tuesday", "wednesday", "thursday", "friday", "saturday"};
        int[] month_days = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        boolean isLeap = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
        if (isLeap) month_days[1] = 29;
        int total_days = 0;
        total_days += (year - 1) * 365 + (year - 1) / 4 - (year - 1) / 100 + (year - 1) / 400;
        for (int i = 0; i < month - 1; i++) {
            total_days += month_days[i];
        }
        total_days += date;
        return days[total_days % 7];
    }
    public static void main(String[] args) {
        System.out.println(find_day(15, 8, 1947));
        System.out.println(find_day(26, 1, 1950));
    }
}

Python

def find_day(date, month, year):
    days = ["sunday", "monday", "tuesday", "wednesday", "thursday", "friday", "saturday"]
    month_days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    is_leap = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
    if is_leap:
        month_days[1] = 29
    total_days = 0
    total_days += (year - 1) * 365 + (year - 1) // 4 - (year - 1) // 100 + (year - 1) // 400
    total_days += sum(month_days[:month - 1])
    total_days += date
    return days[total_days % 7]
print(find_day(15, 8, 1947))
print(find_day(26, 1, 1950))

Explanation of Logic:

• Odd Days Concept: The day of the week repeats in cycles of 7.
• Leap Year Handling: Leap years have an extra day in February.
• Steps:
  • Count days for complete years up to year - 1.
  • Add days for complete months in the given year.
  • Add the given date.
  • Use modulo 7 to find the day of the week.

Example Outputs:
For find_day(15, 8, 1947) the output is "friday".
For find_day(26, 1, 1950) the output is "thursday".

C++

#include <iostream>
#include <cmath>
using namespace std;
bool is_valid_move(int sourceRow, int sourceCol, int destRow, int destCol, int boardSize) {
    if (destRow < 0 || destRow >= boardSize || destCol < 0 || destCol >= boardSize) {
        return false;
    }
    if (sourceRow == destRow || sourceCol == destCol || abs(sourceRow - destRow) == abs(sourceCol - destCol)) {
        return true;
    }
    return false;
}
int main() {
    cout << is_valid_move(2, 1, 1, 0, 5) << endl;
    cout << is_valid_move(2, 1, 4, 2, 5) << endl;
    cout << is_valid_move(2, 1, 3, 3, 5) << endl;
    return 0;
}

Java

public class ChessQueen {
    public static boolean is_valid_move(int sourceRow, int sourceCol, int destRow, int destCol, int boardSize) {
        if (destRow < 0 || destRow >= boardSize || destCol < 0 || destCol >= boardSize) {
            return false;
        }
        if (sourceRow == destRow || sourceCol == destCol || Math.abs(sourceRow - destRow) == Math.abs(sourceCol - destCol)) {
            return true;
        }
        return false;
    }
    public static void main(String[] args) {
        System.out.println(is_valid_move(2, 1, 1, 0, 5));
        System.out.println(is_valid_move(2, 1, 4, 2, 5));
        System.out.println(is_valid_move(2, 1, 3, 3, 5));
    }
}

Python

def is_valid_move(sourceRow, sourceCol, destRow, destCol, boardSize):
    if destRow < 0 or destRow >= boardSize or destCol < 0 or destCol >= boardSize:
        return False
    if sourceRow == destRow or sourceCol == destCol or abs(sourceRow - destRow) == abs(sourceCol - destCol):
        return True
    return False
print(is_valid_move(2, 1, 1, 0, 5))
print(is_valid_move(2, 1, 4, 2, 5))
print(is_valid_move(2, 1, 3, 3, 5))

Explanation of Logic:

• Check Board Bounds: Ensure the destination is within the board dimensions.
• Queen's Movement Rules:
  • Horizontal Move: sourceRow == destRow
  • Vertical Move: sourceCol == destCol
  • Diagonal Move: abs(sourceRow - destRow) == abs(sourceCol - destCol)
• Return Results: If any condition is met, return true; otherwise, return false.

Example Outputs:
For is_valid_move(2, 1, 1, 0, 5) the output is true.
For is_valid_move(2, 1, 4, 2, 5) the output is true.
For is_valid_move(2, 1, 3, 3, 5) the output is false.

C++

#include <iostream>
#include <iomanip>
using namespace std;
void numeric_diamond(int n) {
    int current = 1;
    // Upper Triangle
    for (int i = 1; i <= n; ++i) {
        cout << string(n - i, ' ');
        int temp = current + i - 1;
        for (int j = 0; j < i; ++j) {
            cout << temp--;
            if (j < i - 1) cout << " ";
        }
        current += i;
        cout << endl;
    }
    current -= n;
    // Lower Triangle
    for (int i = n; i >= 1; --i) {
        cout << string(n - i, ' ');
        int temp = current + i - 1;
        for (int j = 0; j < i; ++j) {
            cout << temp--;
            if (j < i - 1) cout << " ";
        }
        current -= i;
        cout << endl;
    }
}
int main() {
    int n = 4;
    numeric_diamond(n);
    return 0;
}

Java

public class NumericDiamond {
    public static void numericDiamond(int n) {
        int current = 1;
        for (int i = 1; i <= n; i++) {
            for (int space = 1; space <= n - i; space++) {
                System.out.print(" ");
            }
            int temp = current + i - 1;
            for (int j = 0; j < i; j++) {
                System.out.print(temp--);
                if (j < i - 1) System.out.print(" ");
            }
            current += i;
            System.out.println();
        }
        current -= n;
        for (int i = n; i >= 1; i--) {
            for (int space = 1; space <= n - i; space++) {
                System.out.print(" ");
            }
            int temp = current + i - 1;
            for (int j = 0; j < i; j++) {
                System.out.print(temp--);
                if (j < i - 1) System.out.print(" ");
            }
            current -= i;
            System.out.println();
        }
    }
    public static void main(String[] args) {
        int n = 4;
        numericDiamond(n);
    }
}

Python

def numeric_diamond(n):
    current = 1
    for i in range(1, n + 1):
        print(" " * (n - i), end="")
        temp = current + i - 1
        for j in range(i):
            print(temp, end="")
            temp -= 1
            if j < i - 1:
                print(" ", end="")
        current += i
        print()
    current -= n
    for i in range(n, 0, -1):
        print(" " * (n - i), end="")
        temp = current + i - 1
        for j in range(i):
            print(temp, end="")
            temp -= 1
            if j < i - 1:
                print(" ", end="")
        current -= i
        print()

numeric_diamond(4)

Explanation of Logic:

• Upper Triangle:
  • Numbers are printed in descending order, starting from current.
  • Each row increases in size, and the number of spaces decreases.
  • current is updated after each row.
• Lower Triangle:
  • Mirrors the upper triangle with numbers decreasing.
  • current is decremented after printing each row.

Example Output for n = 4:

   1
  3 2
 6 5 4
10 9 8 7
10 9 8 7
 6 5 4
  3 2
   1

C++

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

void snake_pattern(int n) {
    int number = 1;
    for (int i = 0; i < n; i++) {
        if (i % 2 == 0) {
            // Print row left to right
            for (int j = 0; j < n; j++) {
                cout << setw(4) << number++;
            }
        } else {
            // Print row right to left
            int start = number + n - 1;
            for (int j = 0; j < n; j++) {
                cout << setw(4) << start--;
            }
            number += n;
        }
        cout << endl;
    }
}

int main() {
    int n = 5;
    snake_pattern(n);
    return 0;
}

Java

public class SnakePattern {
    public static void snakePattern(int n) {
        int number = 1;
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0) {
                // Print row left to right
                for (int j = 0; j < n; j++) {
                    System.out.printf("%4d", number++);
                }
            } else {
                // Print row right to left
                int start = number + n - 1;
                for (int j = 0; j < n; j++) {
                    System.out.printf("%4d", start--);
                }
                number += n;
            }
            System.out.println();
        }
    }
    public static void main(String[] args) {
        int n = 4;
        snakePattern(n);
    }
}

Python

def snake_pattern(n):
    number = 1
    for i in range(n):
        if i % 2 == 0:
            # Print row left to right
            for j in range(n):
                print(f"{number:4}", end="")
                number += 1
        else:
            # Print row right to left
            start = number + n - 1
            for j in range(n):
                print(f"{start:4}", end="")
                start -= 1
            number += n
        print()

# Test the function
snake_pattern(5)

Explanation:

• Odd Rows (Left-to-Right):
  • The numbers are printed sequentially from number.
  • The value of number is incremented with each step.
• Even Rows (Right-to-Left):
  • Calculate the starting number for the row as number + n - 1.
  • Print numbers in reverse order, decrementing from the starting value.
  • Increment number by n after processing the row.

Example Output for n = 4:

 1   2   3   4  
 8   7   6   5  
 9  10  11  12  
16  15  14  13  

Example Output for n = 5:

 1   2   3   4   5  
10   9   8   7   6  
11  12  13  14  15  
20  19  18  17  16  
21  22  23  24  25