Pattern Printing question asked in CGI Coding Round
Last Updated : 13 Mar, 2023
Write a program that receives a number as input and prints it in the following format as shown below.
Examples :
Input : n = 3 Output : 1*2*3*10*11*12 --4*5*8*9 ----6*7 Input : n = 4 Output : 1*2*3*4*17*18*19*20 --5*6*7*14*15*16 ----8*9*12*13 ------10*11
Asked in CGI coding round
Approach: The approach is to see the problem, not as a single task but three tasks which, on combining, complete the main task. The three tasks are printing the left-half of the pattern, printing dashes(-), and printing the right-half of the pattern. Combining all three tasks, we would be able to print the pattern.
left-half of pattern 1*2*3* --4*5* ----6* A function printdashes() to print the "-". right-half of pattern 10*11*12 *8*9 7
Below is the implementation.
C++ // C program to print the given pattern #include <stdio.h> // utility function to print "-" in every // row. This will take care of printing // "-" in the start of every row void printdashes(int k) { int i; for (i = 1; i <= k; i++) printf("-"); } // function to print the pattern void pattern(int n){ // variables for vertical left half /* 1*2*3* --4*5* ----6* */ int row, column, dashes = 0; int i, j, dash_counter = 0; int value = 1; // variables for vertical right half /* 10*11*12 *8*9 7 */ int k, l, decrementor = 0; int column_decrementor = 0; int support = n - 1; int temp = ((n * n) + 1); int temp1 = (n * 2) - 1; int z = temp; int tracker; for (i = 1; i <= n; i++) { printdashes(dash_counter); // This part will take care of the vertical // left half of the pattern for (j = 1; j <= (2 * n) - dash_counter; j++) { // Printing the "*" in even positions if (j % 2 == 0) printf("*"); else { printf("%d", value); value++; } } // This part will take care of the vertical // right half of the pattern for (k = 1; k <= (temp1 - decrementor); k++) { // Printing the "*" in even positions if (k % 2 == 0) printf("*"); else { if (k == 1) tracker = temp; printf("%d", temp); temp++; } } decrementor += 2; temp = tracker - support; support--; // In every row, the number of dash counts // is increased by 2 dash_counter += 2; printf("\n"); } } // driver program int main() { int n = 3; pattern(n); return 0; } // Java program to print the given pattern class GFG { // utility function to print "-" in every // row. This will take care of printing // "-" in the start of every row static void printdashes(int k) { int i; for (i = 1; i <= k; i++) System.out.print("-"); } // function to print the pattern static void pattern(int n) { // variables for vertical left half /* 1*2*3* --4*5* ----6* */ int row, column, dashes = 0; int i, j, dash_counter = 0; int value = 1; // variables for vertical right half /* 10*11*12 *8*9 7 */ int k, l, decrementor = 0; int column_decrementor = 0; int support = n - 1; int temp = ((n * n) + 1); int temp1 = (n * 2) - 1; int z = temp; int tracker = 0; for (i = 1; i <= n; i++) { printdashes(dash_counter); // This part will take care of the vertical // left half of the pattern for (j = 1; j <= (2 * n) - dash_counter; j++) { // Printing the "*" in even positions if (j % 2 == 0) System.out.print("*"); else { System.out.print(value); value++; } } // This part will take care of the vertical // right half of the pattern for (k = 1; k <= (temp1 - decrementor); k++) { // Printing the "*" in even positions if (k % 2 == 0) System.out.print("*"); else { if (k == 1) tracker = temp; System.out.print(temp); temp++; } } decrementor += 2; temp = tracker - support; support--; // In every row, the number of dash counts // is increased by 2 dash_counter += 2; System.out.print("\n"); } } // Driver code public static void main(String arg[]) { int n = 3; pattern(n); } } // This code is contributed by Anant Agarwal. # Python program to print # the given pattern # utility function to # print "-" in every # row. This will take # care of printing # "-" in the start of every row def printdashes(k): for i in range(1,k+1): print("-",end="") # function to print the pattern def pattern(n): # variables for vertical left half ''' 1*2*3* --4*5* ----6* ''' dashes = 0 dash_counter = 0 value = 1 # variables for vertical right half ''' 10*11*12 *8*9 7 ''' decrementor = 0 column_decrementor = 0 support = n - 1 temp = ((n * n) + 1) temp1 = (n * 2) - 1 z = temp for i in range(1,n+1): printdashes(dash_counter) # This part will take # care of the vertical # left half of the pattern for j in range(1,(((2 * n) - dash_counter)+1)): # Printing the "*" in even positions if (j % 2 == 0): print("*",end="") else: print(value,end="") value=value+1 # This part will take # care of the vertical # right half of the pattern for k in range(1,((temp1 - decrementor)+1)): # Printing the "*" in even positions if (k % 2 == 0): print("*",end="") else: if (k == 1): tracker = temp print(temp,end="") temp=temp + 1 decrementor =decrementor + 2 temp = tracker - support support=support - 1 # In every row, the number of dash counts # is increased by 2 dash_counter =dash_counter + 2 print("") # driver program n = 3 pattern(n) # This code is contributed # by Anant Agarwal. // C# program to print the given pattern using System; class GFG { // utility function to print "-" in every // row. This will take care of printing // "-" in the start of every row static void printdashes(int k) { int i; for (i = 1; i <= k; i++) Console.Write("-"); } // function to print the pattern static void pattern(int n) { // variables for vertical left half /* 1*2*3* --4*5* ----6* */ int i, j, dash_counter = 0; int value = 1; // variables for vertical right half /* 10*11*12 *8*9 7 */ int k, decrementor = 0; int support = n - 1; int temp = ((n * n) + 1); int temp1 = (n * 2) - 1; int tracker = 0; for (i = 1; i <= n; i++) { printdashes(dash_counter); // This part will take care of the vertical // left half of the pattern for (j = 1; j <= (2 * n) - dash_counter; j++) { // Printing the "*" in even positions if (j % 2 == 0) Console.Write("*"); else { Console.Write(value); value++; } } // This part will take care of the vertical // right half of the pattern for (k = 1; k <= (temp1 - decrementor); k++) { // Printing the "*" in even positions if (k % 2 == 0) Console.Write("*"); else { if (k == 1) tracker = temp; Console.Write(temp); temp++; } } decrementor += 2; temp = tracker - support; support--; // In every row, the number of dash counts // is increased by 2 dash_counter += 2; Console.WriteLine(); } } // Driver code public static void Main() { int n = 3; pattern(n); } } // This code is contributed by vt_m. <?php // PHP program to print the given pattern // function to print the pattern function pattern($n) { // variables for vertical left half $dashes = 0; $dash_counter = 0; $value = 1; // variables for vertical right half $decrementor = 0; $column_decrementor = 0; $support = $n - 1; $temp = (($n * $n) + 1); $temp1 = ($n * 2) - 1; $z = $temp; for ($i = 1; $i <= $n; $i++) { // loop for printing dash for ($dd = 1; $dd <= $dash_counter; $dd++) printf("-"); // This part will take care of the // vertical left half of the pattern for ($j = 1; $j <= (2 * $n) - $dash_counter; $j++) { // Printing the "*" in even // positions if ($j % 2 == 0) printf("*"); else { printf($value); $value++; } } // This part will take care // of the vertical right half // of the pattern for ($k = 1; $k <= ($temp1 - $decrementor); $k++) { // Printing the "*" in even // positions if ($k % 2 == 0) printf("*"); else { if ($k == 1) $tracker = $temp; printf($temp); $temp++; } } $decrementor += 2; $temp = $tracker - $support; $support--; // In every row, the number of // dash counts is increased by 2 $dash_counter += 2; printf("\n"); } } // Driver code $n = 3; pattern($n); // This code is contributed by mits ?> <script> // javascript program to print the given pattern // utility function to print "-" in every // row. This will take care of printing // "-" in the start of every row function printdashes(k) { var i; for (i = 1; i <= k; i++) document.write("-"); } // function to print the pattern function pattern(n) { // variables for vertical left half /* 1*2*3* --4*5* ----6* */ var row, column, dashes = 0; var i, j, dash_counter = 0; var value = 1; // variables for vertical right half /* 10*11*12 *8*9 7 */ var k, l, decrementor = 0; var column_decrementor = 0; var support = n - 1; var temp = ((n * n) + 1); var temp1 = (n * 2) - 1; var z = temp; var tracker = 0; for (i = 1; i <= n; i++) { printdashes(dash_counter); // This part will take care of the vertical // left half of the pattern for (j = 1; j <= (2 * n) - dash_counter; j++) { // Printing the "*" in even positions if (j % 2 == 0) document.write("*"); else { document.write(value); value++; } } // This part will take care of the vertical // right half of the pattern for (k = 1; k <= (temp1 - decrementor); k++) { // Printing the "*" in even positions if (k % 2 == 0) document.write("*"); else { if (k == 1) tracker = temp; document.write(temp); temp++; } } decrementor += 2; temp = tracker - support; support--; // In every row, the number of dash counts // is increased by 2 dash_counter += 2; document.write("<br>"); } } // Driver code var n = 3; pattern(n); // This code is contributed by 29AjayKumar </script> Output1*2*3*10*11*12 --4*5*8*9 ----6*7
Time Complexity: O(n2)
Auxiliary Space: O(1), As constant extra space is used.
Another approach :
C++ // C++ program to print the given pattern #include <iostream> using namespace std; // function to print the pattern void printPattern(int row) { int x = 1; int z = (row * row) + 1; int col = row == 1 ? 1 : (row * 4) - 1; for(int i = 1; i <= row; i++) { int t = z; for(int j = 1; j <= col - ((i - 1) * 2); j++) { if ((i * 2) - 2 >= j) { cout << "-"; } else { if(col == 1) { cout << x; } else if(j <= col/2 && j % 2 == 1) { cout << x; x++; } else if(j > col/2 && j % 2 == 1) { cout << t; t++; } else { cout << "*"; } } } z = (z - row) + i; cout << "\n"; } } // Driver code int main() { int row = 3; printPattern(row); return 0; } // This code is contributed by shubhamsingh10 // C program to print the given pattern #include<stdio.h> // function to print the pattern void printPattern(int row) { int x = 1; int z = (row * row) + 1; int col = row == 1 ? 1 : (row * 4) - 1; for(int i = 1; i <= row; i++) { int t = z; for(int j = 1; j <= col - ((i - 1) * 2); j++) { if ((i * 2) - 2 >= j) { printf("-"); } else { if(col == 1) { printf("%d", x); } else if(j <= col/2 && j % 2 == 1) { printf("%d", x); x++; } else if(j > col/2 && j % 2 == 1) { printf("%d", t);; t++; } else { printf("*"); } } } z = (z - row) + i; printf("\n"); } } // Driver code int main() { int row = 3; printPattern(row); return 0; } // This code is contributed by ankurmishra1794 // Java program to print the given pattern class GFG { // function to print the pattern static void printPattern(int row) { int x = 1; int z = (row * row) + 1; int col = row == 1 ? 1 : (row * 4) - 1; for(int i = 1; i <= row; i++) { int t = z; for(int j = 1; j <= col -((i - 1) * 2); j++) { if ((i * 2) - 2 >= j) { System.out.print("-"); } else { if(col == 1) { System.out.print(x); } else if(j <= col/2 && j % 2 == 1) { System.out.print(x); x++; } else if(j > col/2 && j % 2 == 1) { System.out.print(t); t++; } else { System.out.print("*"); } } } z = (z - row) + i; System.out.print("\n"); } } // Driver code public static void main(String[] args) { int row = 3; printPattern(row); } } /* This code is contributed by PrinciRaj1992 */ # Python3 program to print the given pattern # Function to print the pattern def printPattern(row): x = 1 z = (row * row) + 1 if row == 1: col = 1 else: col = (row * 4) - 1 for i in range(1, row + 1): t = z for j in range(1, col - ((i - 1) * 2) + 1): if ((i * 2) - 2 >= j): print("", end = "-") else: if (col == 1): print(x, end = "") elif (j <= col / 2 and j % 2 == 1): print(x, end = "") x += 1 elif (j > col / 2 and j % 2 == 1): print(t, end = "") t += 1 else: print("*", end = "") z = (z - row) + i print() # Driver code row = 3 printPattern(row) # This code is contributed by shivani // C# program to print the given pattern using System; class GFG { // function to print the pattern static void printPattern(int row) { int x = 1; int z = (row * row) + 1; int col = row == 1 ? 1 : (row * 4) - 1; for(int i = 1; i <= row; i++) { int t = z; for(int j = 1; j <= col -((i - 1) * 2); j++) { if ((i * 2) - 2 >= j) { Console.Write("-"); } else { if(col == 1) { Console.Write(x); } else if(j <= col/2 && j % 2 == 1) { Console.Write(x); x++; } else if(j > col/2 && j % 2 == 1) { Console.Write(t); t++; } else { Console.Write("*"); } } } z = (z - row) + i; Console.Write("\n"); } } // Driver code public static void Main(String[] args) { int row = 3; printPattern(row); } } // This code is contributed by 29AjayKumar <script> // javascript program to print the given pattern // function to print the pattern function printPattern(row) { var x = 1; var z = (row * row) + 1; var col = row == 1 ? 1 : (row * 4) - 1; for(var i = 1; i <= row; i++) { var t = z; for(var j = 1; j <= col -((i - 1) * 2); j++) { if ((i * 2) - 2 >= j) { document.write("-"); } else { if(col == 1) { document.write(x); } else if(j <= col/2 && j % 2 == 1) { document.write(x); x++; } else if(j > col/2 && j % 2 == 1) { document.write(t); t++; } else { document.write("*"); } } } z = (z - row) + i; document.write("<br>"); } } // Driver code var row = 3; printPattern(row); // This code is contributed by 29AjayKumar </script> Output1*2*3*10*11*12 --4*5*8*9 ----6*7
Time Complexity: O(n2)
Auxiliary Space: O(1), As constant extra space is used.
Another approach :
C++ #include <bits/stdc++.h> using namespace std; void pattern(int n) { int size = n * (n + 1); // prev1 will be used to keep track of last number // printed in left half of pattern int prev1 = 0; // prev2 will be used to keep track of last number // printed in right half of pattern int prev2 = size; for (int i = 0; i < n; i++) { // print the '-' for (int j = 0; j < 2 * i; j++) cout << "-"; // l1 to store numbers of left half to be printed vector<int> l1; for (int j = prev1 + 1; j <= prev1 + n - i; j++) l1.push_back(j); // l2 to store numbers of right half to be printed vector<int> l2; for (int j = prev2 - (n - i) + 1; j <= prev2; j++) l2.push_back(j); // combine l1 and l2 and print the list separated by * for (int j = 0; j < l1.size(); j++) cout << l1[j] << "*"; for (int j = 0; j < l2.size(); j++) cout << l2[j] << "*"; cout << endl; // decrease prev2 and increase prev1 prev2 -= (n - i); prev1 += (n - i); } } // driver program int main() { int n = 3; pattern(n); return 0; } // This code is contributed by poojaagarwal2. import java.util.ArrayList; import java.util.List; public class Main { public static void pattern(int n) { int size = n * (n + 1); // prev1 will be used to keep track of last number // printed in left half of pattern int prev1 = 0; // prev2 will be used to keep track of last number // printed in right half of pattern int prev2 = size; for (int i = 0; i < n; i++) { // print the '-' for (int j = 0; j < 2 * i; j++) System.out.print("-"); // l1 to store numbers of left half to be // printed List<Integer> l1 = new ArrayList<>(); for (int j = prev1 + 1; j <= prev1 + n - i; j++) l1.add(j); // l2 to store numbers of right half to be // printed List<Integer> l2 = new ArrayList<>(); for (int j = prev2 - (n - i) + 1; j <= prev2; j++) l2.add(j); // combine l1 and l2 and print the list // separated by * for (int j = 0; j < l1.size(); j++) System.out.print(l1.get(j) + "*"); for (int j = 0; j < l2.size(); j++) System.out.print(l2.get(j) + "*"); System.out.println(); // decrease prev2 and increase prev1 prev2 -= (n - i); prev1 += (n - i); } } public static void main(String[] args) { int n = 3; pattern(n); } } def pattern(n): size = n*(n+1) # prev1 will be used to keep track of last number # printed in left half of pattern prev1 = 0 # prev2 will be used to keep track of last number # printed in right half of pattern prev2 = size for i in range(n): # print the '-' print('-'*(2*i), end = '') # l1 to store numbers of left half to be printed l1 = [j for j in range(prev1+1, prev1+n-i+1)] # l2 to store numbers of right half to be printed l2 = [j for j in range(prev2-(n-i)+1,prev2+1)] # combine l1 and l2 and print the list separated by * print(*l1+l2, sep = '*') # decrease prev2 and increase prev1 prev2 -= (n-i) prev1 += (n-i) # driver program n = 3 pattern(n) # This code is contributed # by Akash Jain (ultrainstinct). //C# code for the above approach using System; using System.Collections.Generic; class GFG { public static void pattern(int n) { int size = n * (n + 1); // prev1 will be used to keep track of last number // printed in left half of pattern int prev1 = 0; // prev2 will be used to keep track of last number // printed in right half of pattern int prev2 = size; for (int i = 0; i < n; i++) { // print the '-' for (int j = 0; j < 2 * i; j++) Console.Write("-"); // l1 to store numbers of left half to be // printed List<int> l1 = new List<int>(); for (int j = prev1 + 1; j <= prev1 + n - i; j++) l1.Add(j); // l2 to store numbers of right half to be // printed List<int> l2 = new List<int>(); for (int j = prev2 - (n - i) + 1; j <= prev2; j++) l2.Add(j); // combine l1 and l2 and print the list // separated by * for (int j = 0; j < l1.Count; j++) Console.Write(l1[j] + "*"); for (int j = 0; j < l2.Count; j++) Console.Write(l2[j] + "*"); Console.WriteLine(); // decrease prev2 and increase prev1 prev2 -= (n - i); prev1 += (n - i); } } public static void Main(string[] args) { int n = 3; pattern(n); } } function pattern( n) { let size = n * (n + 1); // prev1 will be used to keep track of last number // printed in left half of pattern let prev1 = 0; // prev2 will be used to keep track of last number // printed in right half of pattern let prev2 = size; for (let i = 0; i < n; i++) { // print the '-' for (let j = 0; j < 2 * i; j++) console.log( "-"); // l1 to store numbers of left half to be printed let l1=[]; for (let j = prev1 + 1; j <= prev1 + n - i; j++) l1.push(j); // l2 to store numbers of right half to be printed let l2=[]; for (let j = prev2 - (n - i) + 1; j <= prev2; j++) l2.push(j); // combine l1 and l2 and print the list separated by * for (let j = 0; j < l1.length; j++) console.log( l1[j] + "*"); for (let j = 0; j < l2.length; j++) console.log( l2[j] + "*"); console.log("<br>"); // decrease prev2 and increase prev1 prev2 -= (n - i); prev1 += (n - i); } } // driver program let n = 3; pattern(n); // This code is contributed by ratiagrawal. Output1*2*3*10*11*12* --4*5*8*9* ----6*7*
Time Complexity: O(n2)
Auxiliary Space: O(1)
As constant extra space is used.
Another approach :
C++ #include <algorithm> #include <iostream> #include <string> #include <vector> using namespace std; int main() { int n = 4; int temp_number = (n * n) + n; int counter = 1; // loop through each row of the pattern for (int i = 0; i < n; i++) { vector<int> temp_list; // loop through each column in the current row for (int j = 0; j < n - i; j++) { // generate two numbers and add them to the // temp_list temp_list.push_back(counter); temp_list.push_back(temp_number - counter + 1); counter += 1; } // sort the numbers in the current row in ascending // order sort(temp_list.begin(), temp_list.end()); // print the appropriate number of dashes before the // row for (int k = 0; k < i; k++) { cout << "--"; } // print the numbers in the current row, separated // by asterisks for (int num : temp_list) { cout << num; if (num != temp_list.back()) { cout << "*"; } } cout << endl; } return 0; } // This code is contributed by divyansh2212 import java.util.ArrayList; import java.util.Collections; public class Main { public static void main(String[] args) { int n = 4; int temp_number = (n * n) + n; int counter = 1; // loop through each row of the pattern for (int i = 0; i < n; i++) { ArrayList<Integer> temp_list = new ArrayList<Integer>(); // loop through each column in the current row for (int j = 0; j < n - i; j++) { // generate two numbers and add them to the // temp_list temp_list.add(counter); temp_list.add(temp_number - counter + 1); counter += 1; } // sort the numbers in the current row in // ascending order Collections.sort(temp_list); // print the appropriate number of dashes before // the row for (int k = 0; k < i; k++) { System.out.print("--"); } // print the numbers in the current row, // separated by asterisks for (int num : temp_list) { System.out.print(num); if (num != temp_list.get(temp_list.size() - 1)) { System.out.print("*"); } } System.out.println(); } } } n = 4 temp_number = (n*n)+n counter = 1 for i in range(n): temp_list = [] for j in range(n-i): temp_list.append(counter) temp_list.append(temp_number-counter+1) counter += 1 temp_list.sort() temp_list = [str(each) for each in temp_list] [print("--", end="") for k in range(i)] print("*".join(temp_list)) let n = 4; let temp_number = n * n + n; let counter = 1; // loop through each row of the pattern for (let i = 0; i < n; i++) { let temp_list = []; // loop through each column in the current row for (let j = 0; j < n - i; j++) { // generate two numbers and add them to the temp_list temp_list.push(counter); temp_list.push(temp_number - counter + 1); counter += 1; } // sort the numbers in the current row in ascending order temp_list.sort(function (a, b) { return a - b; }); // print the appropriate number of dashes before the row let temp = ""; for (let k = 0; k < i; k++) { temp = temp+ "--"; } // print the numbers in the current row, separated by asterisks for (let num of temp_list) { temp= temp+ num.toString(); if (num != temp_list[temp_list.length - 1]) { temp = temp +"*"; } } console.log(temp); } // C# code implementation for the above approach using System; using System.Collections.Generic; using System.Linq; public class GFG { static public void Main() { // Code int n = 4; int temp_number = (n * n) + n; int counter = 1; // loop through each row of the pattern for (int i = 0; i < n; i++) { List<int> temp_list = new List<int>(); // loop through each column in the current row for (int j = 0; j < n - i; j++) { // generate two numbers and add them to the // temp_list temp_list.Add(counter); temp_list.Add(temp_number - counter + 1); counter += 1; } // sort the numbers in the current row in // ascending order temp_list.Sort(); // print the appropriate number of dashes before // the row for (int k = 0; k < i; k++) { Console.Write("--"); } // print the numbers in the current row, // separated by asterisks for (int idx = 0; idx < temp_list.Count; idx++) { Console.Write(temp_list[idx]); if (temp_list[idx] != temp_list.Last()) { Console.Write("*"); } } Console.WriteLine(); } } } // This code is contributed by karthik. Output1*2*3*4*17*18*19*20 --5*6*7*14*15*16 ----8*9*12*13 ------10*11
Time Complexity: O(n2)
Auxiliary Space: O(1)
As constant extra space is used.
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