Program to Print the Trapezium Pattern Last Updated : 13 Mar, 2023 Comments Improve Suggest changes Like Article Like Report Given 'num' which indicates number of lines.The task is to print a trapezium pattern in num lines.Examples: Input : 4 Output : 1*2*3*4*17*18*19*20 5*6*7*14*15*16 8*9*12*13 10*11 Input : 2 Output : 1*2*5*6 3*4 Algorithm : step 1. To read num which indicates the number of lines. step 2.We are diving the pattern into 2 halves that is LHS part and the RHS part. Ex : When num = 2 LHS - 1*2* 3*RHS - 5*6 4 step 3.Combining LHS and RHS we get the complete pattern. C++ // CPP program to print Trapezium Pattern #include <iostream> using namespace std; int main() { int num = 3; int space; int i, j, lterm, rterm; // The terms on the LHS of the pattern lterm = 1; // The terms on the RHS of the pattern rterm = num * num + 1; for (i = num; i > 0; i--) { // To print number of spaces for (space = num; space > i; space--) cout << " "; for (j = 1; j <= i; j++) { cout << lterm; cout << "*"; lterm++; } for (j = 1; j <= i; j++) { cout << rterm; if (j < i) printf("*"); rterm++; } // To get the next term on RHS of the Pattern rterm = rterm - (i - 1) * 2 - 1; cout << endl; } } Java // Java program to print Trapezium Pattern public class HelloWorld { public static void trapeziumPattern(int num) { int firsthalf = 1; int secondhalf = (num * num) + 1; int numOfSpaces = 0; // numOfLines is the line number for (int numOfLines = num; numOfLines >= 1; numOfLines--) { // Prints the spaces for each line for (int numOfSpacesCounter = numOfSpaces; numOfSpacesCounter >= 1; numOfSpacesCounter--) { System.out.print(" "); } // Prints the first half of the trapezium for (int firstHalfCounter = 1; firstHalfCounter <= numOfLines; firstHalfCounter++) { // If it is the last number for a line then // we don't print '*' if (firstHalfCounter == numOfLines) System.out.print((firsthalf++)); else System.out.print((firsthalf++) + "*"); } // Prints the second half of the trapezium for (int secondHalfCounter = 1; secondHalfCounter <= numOfLines; secondHalfCounter++) { System.out.print("*" + (secondhalf++)); } System.out.println(); // Calculates the number of Spaces for the next // line numOfSpaces += 2; // Calculates the first number of the // second half for the next iteration/line secondhalf = (secondhalf - 1) - ((numOfLines - 1) * 2); } } public static void main(String[] args) { trapeziumPattern( 4); // Passing the integer as the argument to // print trapezium pattern } } Python 3 # Python 3 program to print # Trapezium Pattern if __name__ == "__main__": num = 3 # The terms on the LHS # of the pattern lterm = 1 # The terms on the RHS # of the pattern rterm = num * num + 1 for i in range(num, -1, -1): # To print number of spaces for space in range(num, i-1, -1): print(" ", end ="") for j in range(1, i + 1): print(str(lterm)+"*", end ="") lterm += 1 for j in range(1, i + 1): print(rterm, end ="") if j < i: print("*", end ="") rterm += 1 # To get the next term on RHS of the Pattern rterm = rterm - (i - 1) * 2 - 1 print() # This code is contributed by ChitraNayal C# // C# program to print Trapezium Pattern using System; public class HelloWorld { public static void Main(String[] args) { // Scanner scn = new Scanner(System.in); int num = 3; int space; // System.out.println("Enter number of lines : "); // num = scn.nextInt(); int i, j, lterm, rterm; lterm = 1; // The terms on the LHS of the pattern // The terms on the RHS of the pattern rterm = num * num + 1; for (i = num; i > 0; i--) { // To print number of spaces for (space = num; space > i; space--) Console.Write(" "); for (j = 1; j <= i; j++) { Console.Write(lterm); Console.Write("*"); lterm++; } for (j = 1; j <= i; j++) { Console.Write(rterm); if (j < i) Console.Write("*"); rterm++; } // To get the next term on RHS of the Pattern rterm = rterm - (i - 1) * 2 - 1; Console.WriteLine(); } } } // This code is contributed by ankita_saini PHP <?php // PHP program to print // Trapezium Pattern $num = 3; $space; $i; $j; $lterm; $rterm; // The terms on the LHS // of the pattern $lterm = 1; // The terms on the // RHS of the pattern $rterm = $num * $num + 1; for ($i = $num; $i > 0; $i--) { // To print number of spaces for ($space = $num; $space > $i; $space--) echo " "; for ($j = 1; $j <= $i; $j++) { echo $lterm; echo "*"; $lterm++; } for ($j = 1; $j <= $i; $j++) { echo $rterm; if ($j < $i) echo "*"; $rterm++; } // To get the next term // on RHS of the Pattern $rterm = $rterm - ($i - 1) * 2 - 1; echo "\n"; } // This code is contributed // by Akanksha Rai(Abby_akku) ?> JavaScript <script> // JavaScript program to print Trapezium Pattern var num = 3; var space; var i, j, lterm, rterm; // The terms on the LHS of the pattern lterm = 1; // The terms on the RHS of the pattern rterm = num * num + 1; for (i = num; i > 0; i--) { // To print number of spaces for (space = num; space > i; space--) document.write(" "); for (j = 1; j <= i; j++) { document.write(lterm); document.write("*"); lterm++; } for (j = 1; j <= i; j++) { document.write(rterm); if (j < i) document.write("*"); rterm++; } // To get the next term on RHS of the Pattern rterm = rterm - (i - 1) * 2 - 1; document.write("<br>"); } </script> Output: Enter number of lines : 3 1*2*3*10*11*12 4*5*8*9 6*7 Time complexity: O(n2) space complexity: O(1) Comment More infoAdvertise with us Next Article Program to Print the Trapezium Pattern S sheikhmannansohail Follow Improve Article Tags : DSA pattern-printing Practice Tags : pattern-printing Similar Reads Program to print the pattern "GFG" In this article, given the value of n(length of the alphabet) and k(width of the alphabet) we will learn how to print the pattern "GFG" using stars and white-spaces. 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