# Number pattern 30 in C

Write a C program to print the given number pattern using for loop. How to print the given triangular number pattern using for loop in C programming. Logic to print the given number pattern using loop in C program.

Example

Input

`Input N: 5`

Output

## Logic to print the given number pattern 1

Before we get into this pattern I recommend you to look to the given pattern carefully for a minute. The pattern contains N rows (where N is the total rows to be printed). Each row contains exactly N - i + 1 columns (where i is the current row number).
Step-by-step descriptive logic:

1. To iterate though rows, run an outer loop from 1 to N.
2. To print columns, run an inner loop from i to N. Inside this loop print the value of j (where j is the current column number).

## Program to print the given number pattern 1

``````/**
* C program to print number pattern
*/

#include <stdio.h>

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

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

for(i=1; i<=N; i++)
{
// Logic to print numbers
for(j=i; j<=N; j++)
{
printf("%d", j);
}

printf("\n");
}
return 0;
}``````

Output

```Enter N: 5
12345
2345
345
45
5```

## Logic to print the given number pattern 2

The above pattern is similar to pattern we just printed except the trailing spaces. The logic of this pattern would be similar as first pattern we did, we only need to add logic to print trailing spaces before printing numbers. There are i - 1 spaces per row (where i is the current row number). To see or count total spaces per row you can hover mouse pointer to the pattern above.
Step-by-step descriptive logic:

1. To print spaces, run an inner loop from 1 to i - 1 (where i is the current row number). Inside this loop print single blank space.

## Program to print the given number pattern 2

``````/**
* C program to print number pattern
*/

#include <stdio.h>

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

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

for(i=1; i<=N; i++)
{
// Logic to print spaces
for(j=1; j<i; j++)
{
printf(" ");
}

// Logic to print numbers
for(j=i; j<=N; j++)
{
printf("%d", j);
}

printf("\n");
}

return 0;
}``````

Happy coding 😉 