Triangle number pattern using 0, 1 in C – 2

Write a C program to print the given triangle number pattern using 0, 1. How to print the given triangle number pattern with 0, 1 using for loop in C programming. Logic to print the given triangle number pattern with 0, 1 in C program.

Example

Input

Input N: 5

Output

Required knowledge

Basic C programming, If else, Loop

Logic to print the given triangle number pattern

  1. The pattern consists of N rows (where N is the total number of rows to be printed). Hence the outer loop formation to iterate through rows will be for(i=1; i<=N; i++).
  2. Each row contains exactly i number of columns (where i is the current row number) i.e. first row contains 1 column, 2 row contains 2 column and so on.
    Hence the inner loop formation to iterate through columns will be for(j=1; j<=i; j++).
  3. Once you are done with the loop semantics, you just need to print the number. If you notice the numbers are in special order. For each odd rows 1 gets printed and for each even rows 0 gets printed. Now to implement this we need to check a simple condition if(i % 2 == 1) before printing 1s or 0s.

Lets, now implement this in C program.

Program to print given triangle number pattern

/**
 * C program to print the given number pattern
 */

#include <stdio.h>

int main()
{
    int i, j, N;

    printf("Enter N: ");
    scanf("%d", &N);

    for(i=1; i<=N; i++)
    {
        for(j=1; j<=i; j++)
        {
            // Print 1s for every odd rows
            if(i % 2 == 1)
            {
                printf("1");
            }
            else
            {
                printf("0");
            }
        }

        printf("\n");
    }

    return 0;
}

Output

Enter N: 5
1
00
111
0000
11111

You can also print this pattern without using if else. Below is a tricky approach to print the given number pattern without using if else. It uses bitwise operator to check even odd.

Program to print the given number pattern without using if

/**
 * C program to print the given number pattern
 */

#include <stdio.h>

int main()
{
    int i, j, N;

    printf("Enter N: ");
    scanf("%d", &N);

    for(i=1; i<=N; i++)
    {
        for(j=1; j<=i; j++)
        {
            printf("%d", (i & 1));
        }

        printf("\n");
    }

    return 0;
}

Note: You can also get the below pattern with the same logic

What you need to do is, swap the two printf(); statements. Replace the printf(“1”); with printf(“0”); and vice versa.
Happy coding 😉