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    Saturday, June 6

    Star Pattern Programs

    Updated: Algorithms No Comments3 Mins Readdevangini123By

    Pattern 1: Print nxn Star Square

    Print a square pattern of stars (*) of size n x n.

    Output

    * * * *
* * * *
* * * *
* * * *

    Approach:

    • ● Outer Loop (Rows): Run from i = 0 to i = n - 1
    • ● Inner Loop (Columns): For each row, loop from j = 0 to j = n - 1
    • ● Build Row String: Append *in each inner loop iteration.
    • ● Print Row:After the inner loop, print the complete row string.

    Time & Space Complexity:

    Time Complexity:O(n^2)

    Space Complexity:O(n)(temporary row string) 

    
let n = 4;
for (let i = 0; i < n; i++) {
  let row = "";
  for (let j = 0; j < n; j++) {
    row += "*";
  }
  console.log(row);
}

    
n = 4
for i in range(n):
    row = ""
    for j in range(n):
        row += "*"
    print(row)

    
public class StarSquare {
  public static void main(String[] args) {
    int n = 4;
    for (int i = 0; i < n; i++) {
      String row = "";
      for (int j = 0; j < n; j++) {
        row += "*";
      }
      System.out.println(row);
    }
  }
}

    
#include <iostream>
using namespace std;

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    string row = "";
    for (int j = 0; j < n; j++) {
      row += "*";
    }
    cout << row << endl;
  }
  return 0;
}

    
#include <stdio.h>
int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}

    
using System;
class Program {
  static void Main() {
    int n = 4;
    for (int i = 0; i < n; i++) {
      string row = "";
      for (int j = 0; j < n; j++) {
        row += "*";
      }
      Console.WriteLine(row);
    }
  }
}

    Pattern 2: Right-Angled Star Triangle

    Print a right-angled triangle of stars with n rows.

    Output

    *
* *
* * *
* * * *

    Approach:

    • ● Outer Loop (Rows): Run a loop from i = 0 to i = n - 1. Each iteration represents one row.
    • ● Inner Loop (Stars per Row): For each row irun another loop from j = 0 to j = iand append a *character to a string.
    • ● Print Row:Print the string after the inner loop, completes for each row.

    Time & Space Complexity:

    Time Complexity:O(n^2)because the total number of stars printed is1+2+3+.......+n = n(n+1)/2.

    Space Complexity:O(n)for the temporary string variable storing each row. 

    
let n = 4;
for (let i = 0; i < n; i++) {
  let row = "";
  for (let j = 0; j <= i; j++) {
    row += "*";
  }
  console.log(row);
}

    
n = 4
for i in range(n):
    row = ""
    for j in range(i + 1):
        row += "*"
    print(row)

    
public class StarTriangle {
  public static void main(String[] args) {
    int n = 4;
    for (int i = 0; i < n; i++) {
      String row = "";
      for (int j = 0; j <= i; j++) {
        row += "*";
      }
      System.out.println(row);
    }
  }
}

    
#include <iostream>
using namespace std;

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    string row = "";
    for (int j = 0; j <= i; j++) {
      row += "*";
    }
    cout << row << endl;
  }
  return 0;
}

    
#include <stdio.h>
int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}

    
using System;
class Program {
    static void Main() {
        int n = 4;
        for (int i = 0; i < n; i++) {
            string row = "";
            for (int j = 0; j <= i; j++) {
                row += "*";
            }
            Console.WriteLine(row);
        }
    }
}

    Pattern 3: Print a Right-Angled Number Triangle

    Write a program that prints a right-angled triangle of numbers of heightn.

    Output

    1
1 2
1 2 3
1 2 3 4

    Approach:

    • ● Outer loop (Rows): Run a loop from i = 0to i < n.Each iteration represents a new row.
    • ● Inner loop (Numbers):Run an inner loop from j = 0to j <= i, and append j+1to the row.
    • ● Build and Print: Construct a string for the row and print it after the inner loop ends.

    Time & Space Complexity:

    Time Complexity: O(n^2)Each row can have up to nnumbers.

    Space Complexity:O(n)Temporary string to build each row.

    
let n = 4;
for (let i = 0; i < n; i++) {
  let row = "";
  for (let j = 0; j <= i; j++) {
    row += (j + 1);
  }
  console.log(row);
}

    
n = 4
for i in range(n):
    row = ""
    for j in range(i + 1):
        row += str(j + 1)
    print(row)

    
public class NumberTriangle {
  public static void main(String[] args) {
    int n = 4;
    for (int i = 0; i < n; i++) {
      String row = "";
      for (int j = 0; j <= i; j++) {
        row += (j + 1);
      }
      System.out.println(row);
    }
  }
}

    
#include <iostream>
using namespace std;

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    string row = "";
    for (int j = 0; j <= i; j++) {
      row += to_string(j + 1);
    }
    cout << row << endl;
  }
  return 0;
}

    
#include <stdio.h>

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      printf("%d", j + 1);
    }
    printf("\n");
  }
  return 0;
}

    
using System;
class Program {
    static void Main() {
        int n = 4;
        for (int i = 0; i < n; i++) {
            string row = "";
            for (int j = 0; j <= i; j++) {
                row += (j + 1);
            }
            Console.WriteLine(row);
        }
    }
}

    Pattern 4: Print a Right-Angled Triangle of Repeated Numbers

    Write a program that prints a right-angled triangle where each row contains the same number repeated.

    Output

    1
2 2
3 3 3
4 4 4 4

    Approach:

    • ● Outer loop (Rows): Loop from i = 0to i < n.
    • ● Inner loop (Numbers):Loop fromj = 0to j <= i, appending i+1as a string.
    • ● Build and Print: Build the row string and print it.

    Time & Space Complexity:

    Time Complexity: O(n^2)

    Space Complexity:O(n)for the temporary row string.

    
let n = 4;
for (let i = 0; i < n; i++) {
  let row = "";
  for (let j = 0; j <= i; j++) {
    row += (i + 1);
  }
  console.log(row);
}

    
n = 4
for i in range(n):
    row = ""
    for j in range(i + 1):
        row += str(i + 1)
    print(row)

    
public class NumberRepeatedTriangle {
  public static void main(String[] args) {
    int n = 4;
    for (int i = 0; i < n; i++) {
      String row = "";
      for (int j = 0; j <= i; j++) {
        row += (i + 1);
      }
      System.out.println(row);
    }
  }
}

    
#include <iostream>
using namespace std;

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    string row = "";
    for (int j = 0; j <= i; j++) {
      row += to_string(i + 1);
    }
    cout << row << endl;
  }
  return 0;
}

    
#include <stdio.h>

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      printf("%d", i + 1);
    }
    printf("\n");
  }
  return 0;
}

    
using System;
class Program {
    static void Main() {
        int n = 4;
        for (int i = 0; i < n; i++) {
            string row = "";
            for (int j = 0; j <= i; j++) {
                row += (i + 1).ToString();
            }
            Console.WriteLine(row);
        }
    }
}

    Pattern 5: Print a Reverse Right-Angled Triangle of Increasing Numbers

    Write a program that prints a reverse right-angled triangle where each row starts from 1 and the number of elements decreases with each row.

    Output

    1 2 3 4
1 2 3
1 2
1

    Approach:

    • ● Outer loop (Rows): Loopifrom0 to n - 1.Each iteration represents a row.
    • ● Inner loop (Print Numbers):For each row, loopjfrom0to n-i-1 and append j+1to a row string.
    • ● Print Row: After the inner loop, print the row string.

    Time & Space Complexity:

    Time Complexity: O(n^2)

    Space Complexity:O(n)for the temporary row string.

    
let n = 4;
for (let i = 0; i < n; i++) {
  let row = "";
  for (let j = 0; j < n - i; j++) {
    row += (j + 1);
  }
  console.log(row);
}

    
n = 4
for i in range(n):
    row = ""
    for j in range(n - i):
        row += str(j + 1)
    print(row)

    
public class ReverseNumberTriangle {
  public static void main(String[] args) {
    int n = 4;
    for (int i = 0; i < n; i++) {
      String row = "";
      for (int j = 0; j < n - i; j++) {
        row += (j + 1);
      }
      System.out.println(row);
    }
  }
}

    
#include <iostream>
using namespace std;

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    string row = "";
    for (int j = 0; j < n - i; j++) {
      row += to_string(j + 1);
    }
    cout  << row  << endl;
  }
  return 0;
}

    
#include <stdio.h>

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n - i; j++) {
      printf("%d", j + 1);
    }
    printf("\n");
  }
  return 0;
}

    
using System;
class Program {
    static void Main() {
        int n = 4;
        for (int i = 0; i < n; i++) {
            string row = "";
            for (int j = 0; j < n - i; j++) {
                row += (j + 1).ToString();
            }
            Console.WriteLine(row);
        }
    }
}

    Pattern 6: Print a Right-Aligned Right-Angled Triangle of Stars

    Write a program that prints a right-aligned triangle of stars increasing row by row, with leading spaces for alignment.

    Output

          *
    * *
  * * *
* * * *

    Approach:

    • ● Outer loop (Rows): Loopi = 0from0ton-1.Each iteration is a new row.
    • ● Inner loop 1(Spaces):For each row, addn - i - 1spaces before the stars to right-align the triangle.
    • ● Inner loop 2(Stars):Addi+1stars after the spaces.
    • ● Print Row: Combine the spaces and stars, then print the row.

    Time & Space Complexity:

    Time Complexity: O(n^2)

    Space Complexity:O(n)for the row string.

    
let n = 4;
for (let i = 0; i < n; i++) {
  let row = "";
  for (let j = 0; j < n - (i + 1); j++) {
    row += " ";
  }
  for (let k = 0; k < i + 1; k++) {
    row += "*";
  }
  console.log(row);
}

    
n = 4
for i in range(n):
    row = ""
    for j in range(n - (i + 1)):
        row += " "
    for k in range(i + 1):
        row += "*"
    print(row)

    
public class RightAlignedTriangle {
  public static void main(String[] args) {
    int n = 4;
    for (int i = 0; i < n; i++) {
      String row = "";
      for (int j = 0; j < n - (i + 1); j++) {
        row += " ";
      }
      for (int k = 0; k < i + 1; k++) {
        row += "*";
      }
      System.out.println(row);
    }
  }
}

    
#include <iostream>
using namespace std;

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    string row = "";
    for (int j = 0; j < n - (i + 1); j++) {
      row += " ";
    }
    for (int k = 0; k < i + 1; k++) {
      row += "*";
    }
    cout << row << endl;
  }
  return 0;
}

    
#include <stdio.h>

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n - (i + 1); j++) {
      printf(" ");
    }
    for (int k = 0; k < i + 1; k++) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}

    
using System;
class Program {
    static void Main(){
        int n = 4;
        for (int i = 0; i < n; i++) {
            string row = "";
            for (int j = 0; j < n - (i + 1); j++) {
                row += " ";
            }
            for (int k = 0; k < i + 1; k++) {
                row += "*";
            }
            Console.WriteLine(row);
        }
    }
}

    Pattern 7: Print a Right-Angled Triangle of Alternating 1s and 0s

    Write a program that prints a triangle of alternating 1s and 0s starting with 1 on each row.

    Output

    1
1 0
1 0 1
1 0 1 0

    Approach:

    • ● Outer loop (Rows): Loopifrom0ton-1.
    • ● Initialize toggle = 1:Start each row withtoggle = 1..
    • ● Inner loop (Columns):For each row, loopjfrom 0toi. On each iteration:
    • Appendtoggleto the row string.
    • Fliptogglebetween 1 and 0.
    • ● Print Row: After inner loop, print the row string.

    Time & Space Complexity:

    Time Complexity: O(n^2)

    Space Complexity:O(n)per row

    
let n = 4;
for (let i = 0; i < n; i++) {
  let row = "";
  let toggle = 1;
  for (let j = 0; j < i + 1; j++) {
    row += toggle;
    toggle = toggle === 1 ? 0 : 1;
  }
  console.log(row);
}

    
n = 4
for i in range(n):
    row = ""
    toggle = 1
    for j in range(i + 1):
        row += str(toggle)
        toggle = 0 if toggle == 1 else 1
    print(row)

    
public class BinaryTriangle {
  public static void main(String[] args) {
    int n = 4;
    for (int i = 0; i < n; i++) {
      String row = "";
      int toggle = 1;
      for (int j = 0; j < i + 1; j++) {
        row += toggle;
        toggle = (toggle == 1) ? 0 : 1;
      }
      System.out.println(row);
    }
  }
}

    
#include <iostream>
using namespace std;

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    string row = "";
    int toggle = 1;
    for (int j = 0; j < i + 1; j++) {
      row += to_string(toggle);
      toggle = (toggle == 1) ? 0 : 1;
    }
    cout << row << endl;
  }
  return 0;
}

    
#include <stdio.h>

int main() {
  int n = 4;
  for (int i = 0; i < n; i++) {
    int toggle = 1;
    for (int j = 0; j < i + 1; j++) {
      printf("%d", toggle);
      toggle = (toggle == 1) ? 0 : 1;
    }
    printf("\n");
  }
  return 0;
}

    
using System;
class Program {
    static void Main() {
        int n = 4;
        for (int i = 0; i < n; i++) {
            int toggle = 1;
            for (int j = 0; j < i + 1; j++) {
                Console.Write(toggle);
                toggle = (toggle == 1) ? 0 : 1;
            } Console.WriteLine();
        }
    }
}

    Pattern 8: Right-Angled Triangle of Alternating 1s and 0s (Global Toggle)

    Write a program to print a triangle of alternating 1s and 0s, but the toggle continues globally across rows.

    Output

    1
0 1
0 1 0
1 0 1 0

    Approach:

    • ● Global Toggle Variable: Declaretoggle = 1before the outer loop.
    • ● Outer Loop:Loopifrom0ton-1.
    • ● Inner loop:Loopjfrom0toi.On each iteration:
    • ● Append:toggleto the row string.
    • ● Flip:toggle1 -> 0and0 -> 1.
    • ● Print the Row: after the inner loop.

    Key Difference:

    In the previous pattern, toggle = 1was reset each row. Here, the toggle continues globally across the entire pattern.

    Time & Space Complexity:

    Time Complexity: O(n^2)

    Space Complexity:O(n)per row.

    
let n = 4;
let toggle = 1;
for (let i = 0; i < n; i++) {
  let row = "";
  for (let j = 0; j < i + 1; j++) {
    row += toggle;
    toggle = toggle === 1 ? 0 : 1;
  }
  console.log(row);
}

    
n = 4
toggle = 1
for i in range(n):
    row = ""
    for j in range(i + 1):
        row += str(toggle)
        toggle = 0 if toggle == 1 else 1
    print(row)

    
public class GlobalToggleTriangle {
  public static void main(String[] args) {
    int n = 4;
    int toggle = 1;
    for (int i = 0; i < n; i++) {
      String row = "";
      for (int j = 0; j < i + 1; j++) {
        row += toggle;
        toggle = (toggle == 1) ? 0 : 1;
      }
      System.out.println(row);
    }
  }
}

    
#include <iostream>
using namespace std;

int main() {
  int n = 4;
  int toggle = 1;
  for (int i = 0; i < n; i++) {
    string row = "";
    for (int j = 0; j < i + 1; j++) {
      row += to_string(toggle);
      toggle = (toggle == 1) ? 0 : 1;
    }
    cout << row << endl;
  }
  return 0;
}

    
#include <stdio.h>

int main() {
  int n = 4;
  int toggle = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < i + 1; j++) {
      printf("%d", toggle);
      toggle = (toggle == 1) ? 0 : 1;
    }
    printf("\n");
  }
  return 0;
}

    
using System;
class Program {
    static void Main() {
        int n = 4;
        int toggle = 1;
        for (int i = 0; i < n; i++) {
            string row = "";
            for (int j = 0; j < i + 1; j++) {
                row += toggle.ToString();
                toggle = (toggle == 1) ? 0 : 1;
            } Console.WriteLine(row);
        }
    }
}

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