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    Sunday, July 13

    Star Pattern Programs in JavaScript, Java, C++, C, and Python

    C/C++ No Comments2 Mins ReadakshayBy

    Pattern 1 – Print n x n Star Square

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

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

    Pattern 2 - Right-Angled Star Triangle Pattern

    Write a program to print a right-angled triangle of stars (*) with n rows.

    Example Output

    
*
**
***
****

    Approach to Print Right-Angled Star Triangle

    • 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 i, run another loop from j = 0 to j = i and 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 is 1 + 2 + ... + 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);
}
    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;
}
    n = 4
for i in range(n):
    row = ""
    for j in range(i + 1):
        row += "*"
    print(row)

    Pattern 3 - Print a Right-Angled Number Triangle

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

    Example Output

    
1
12
123
1234

    Step-by-Step Approach

    • Outer Loop (Rows): Run a loop from i = 0 to i < n. Each iteration represents a new row.
    • Inner Loop (Numbers): Run an inner loop from j = 0 to j <= i, and append j + 1 to 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(n²) — Each row can have up to n numbers.
    • 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);
}
    
    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;
}
    
    n = 4
for i in range(n):
    row = ""
    for j in range(i + 1):
        row += str(j + 1)
    print(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.

    Example Output

    
1
22
333
4444

    Step-by-Step Approach

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

    Time & Space Complexity

    • Time Complexity: O(n²)
    • 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);
}
    
    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;
}
    
    n = 4
for i in range(n):
    row = ""
    for j in range(i + 1):
        row += str(i + 1)
    print(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.

    Example Output (n = 4)

    
1234
123
12
1

    Approach

    • Outer Loop (Rows): Loop i from 0 to n - 1. Each iteration represents a row.
    • Inner Loop (Print Numbers): For each row, loop j from 0 to n - i - 1 and append j + 1 to a row string.
    • Print Row: After the inner loop, print the row string.

    Time and Space Complexity

    • Time Complexity: O(n²)
    • 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);
}
    
    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;
}
    
    n = 4
for i in range(n):
    row = ""
    for j in range(n - i):
        row += str(j + 1)
    print(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.

    Example Output (n = 4)

    
   *
  **
 ***
****

    Approach

    1. Outer Loop (Rows): Loop i from 0 to n - 1. Each iteration is a new row.
    2. Inner Loop 1 (Spaces): For each row, add n - i - 1 spaces before the stars to right-align the triangle.
    3. Inner Loop 2 (Stars): Add i + 1 stars after the spaces.
    4. Print Row: Combine the spaces and stars, then print the row.

    Time and Space Complexity

    • Time Complexity: O(n²)
    • 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);
}
    
    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;
}
    
    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)

    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.

    Example Output (n = 4)

    
1
10
101
1010

    Approach

    1. Outer Loop (Rows): Loop i from 0 to n - 1.
    2. Initialize toggle = 1: Start each row with toggle = 1.
    3. Inner Loop (Columns): For each row, loop j from 0 to i. On each iteration:
      • Append toggle to the row string.
      • Flip toggle between 1 and 0.
    4. Print Row: After inner loop, print the row string.

    Time and Space Complexity

    • Time Complexity: O(n²)
    • 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);
}
    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;
}
    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)

    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 (n = 4)

    
1
01
010
1010

    Approach

    1. Global Toggle Variable: Declare toggle = 1 before the outer loop.
    2. Outer Loop: Loop i from 0 to n - 1.
    3. Inner Loop: Loop j from 0 to i. On each iteration:
      • Append toggle to the row string.
      • Flip toggle: 1 → 0 and 0 → 1.
    4. Print the row after the inner loop.

    Key Difference

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

    Time & Space Complexity

    • Time Complexity: O(n²)
    • 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);
}
    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;
}
    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)

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