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Mirror image of triangle

Easy

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Average time to solve is 10m

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5 upvotes

Asked in companies

Problem statement

Ninja’s younger sister got a project to print the pattern for the given number of rows, say ‘N’. She finds it difficult to write such a pattern for every ‘N’.

Ninja decided to help his sister and decided to write a code for this but wasn't able to write the code for the same. Ninja wants help to write a code and asks you for help. Help Ninja.

Example:

Pattern for N = 2
    0
   101
  21012

Detailed explanation ( Input/output format, Notes, Images )

Input format :

The first line of input contains an integer ‘T’ denoting the number of test cases. And each test case follows:

Each test case contains an integer ‘N’ representing the number of rows.

Output format :

Each test case print 'N' strings denoting the required pattern of integers for a given integer ‘N’.

The output of each test case will be printed on a separate line

Constraints :

1 <= T <= 5
1 <= N <= 100

Time Limit: 1 sec.

Sample Input 1 :

2
1
2

Sample Output 1 :

Explanation for Sample Input 1 :

Test Case 1:
 For ‘N’  = 1 , we need to print 2 rows - in first row 0, and second row 1 0 1.
Test Case 2:
The Number of rows for ‘N’ = 2 will be 3.

Sample Input 2 :

2
3
4

Sample Output 2 :

    0
   101
  21012
 3210123
432101234
     0
    101
   21012
  3210123
 432101234
54321012345

Hint 1

Observe the pattern and try using nested iterations.

Approaches (1)

Approach 1

Nested Iterations

The idea is to use nested iterations for generating spaces and the digits inside the pattern.

Approach :

First, initialize two integer variables, say ‘number1’ and ‘number2’ to print the digits in the pattern and initialize both the variables with value = 1;
Make an iteration with iterator pointer ‘i’ to print the number of rows which will iterate from 0 to ‘N’.
- Make a nested iteration to generate the spaces in the pattern with an iterator pointer ‘j’ which will iterate from ‘N’ - 1 to ‘i’ and print the spaces.
- Make an iteration to print the digits inside the pattern with an iterator pointer ’k’ from 1 to ‘number1’and print the value = (‘k’ - ‘number2’).
- Increment ‘number1’ with value 2.
- Increment ‘number2’ with value 1.
- Leave a line after that.

Time Complexity

O(N ^ 2), where ‘N’ is the given integer.

As, we are making an iteration ‘N’ number of times to print different rows in the pattern and at max (2 * (N + 1)) columns for the pattern with nested iterations. Therefore, overall time complexity will be O(N ^ 2).

Space Complexity

O(1)

As we are using constant space. Therefore, space complexity will be O(1).

(100% EXP penalty)

Mirror image of triangle

Jenil Panchal

Interview problems

Easy python solution

def pattern(n):

s = ""

for i in range(1, n+2):

count = i-1

for k in range(n-i+1):

s += " "

for j in range(2*i):

if count > 0:

s += str(count)

count -= 1

print(s + "".join(map(str, list(range(0, i)))))

s = ""

6 views

0 replies

0 upvotes

Jahanvi Parmar

Interview problems

python

def pattern(n):

for i in range(n+1):

for j in range(i,n):

print(" ",end="")

for j in range(i,-1,-1):

print(j,end="")

for j in range(1,i+1):

print(j,end="")

print()

20 views

0 replies

0 upvotes

ARJUN BALAJI

Interview problems

Why i cant see any of my output being returned

def pattern(n):

triangle = []

row = [0]

triangle.append(row)

for i in range(1, n + 1):

row2 = [0] * ((2 * i) + 1)

temp=i

increment=True

for j in range(0, len(row2)):

if temp>=0:

# print(temp)

row2[j]=temp

if temp==0:

increment=False

if increment:

temp-=1

else:

temp+=1

triangle.append(row2)

return triangle

python

21 views

0 replies

1 upvote

kartik garg

Interview problems

Mirror image of triangle

#include <bits/stdc++.h>

void pattern(int n)

{

int space=n;

string a="0";

int i=0;

while(n-->=0)

{

int k=space;

while(k--)

cout<<" ";

if(i!=0)

{

a=to_string(i)+a+to_string(i);

}

cout<<a<<" ";

cout<<endl;

i++;

space--;

}

88 views

0 replies

0 upvotes

Busireddy Gari Aishvarya

Interview problems

Using Java Programming Language.All test cases passed

import java.util.* ;

import java.io.*;

public class Solution {

public static void NumberPattern(int n) {

// Write your code here

for(int i=0;i<=n;i++)

{

for(int j=n-i-1;j>=0;j--)

{

System.out.print(" ");

}

for(int j=0;j<=n-(n-i);j++)

{

System.out.print(Math.abs(j-i));

}

for(int j=1;j<=i;j++)

{

System.out.print(j);

}

System.out.println();

}

65 views

0 replies

1 upvote

Durgesh Thakur

Interview problems

Number Pattern

import java.util.* ;
import java.io.*; 

public class Solution {
	public static void NumberPattern(int n) {
		// Write your code here
        int count =0;
        for(int i=0;i<=n;i++){
            for(int j=0;j<n-i;j++){
                System.out.print(" ");
            }
            if(i==0){
                System.out.print('0');
            }else{
                for(int k=i;k>=0;k--){
                    System.out.print(k);
                }
                for(int k =1;k<=i;k++){
                    System.out.print(k);
                }
            }
          
            System.out.println();
        }
	}

}

49 views

0 replies

0 upvotes

Vinay Singh Chouhan

Interview problems

Mirror image of triangle

#include <bits/stdc++.h> 
void pattern(int n)
{
    for(int i=0;i<=n;i++){
        for(int j=i;j<n;j++){
            cout<<" ";
        }
        for(int j=i;j>=0;j--){
            cout<<j;
        }
        for(int j=1;j<=i;j++){
            cout<<j;
        }
        cout<<endl;
    }
}