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# Print the Pattern

Last Updated on March 25, 2022 by Ria Pathak Recursion

Easy

#### PROBLEM STATEMENT(SIMPLIFIED):

Given a number N, print its decreasing sequence (i.e. keep subtracting by 5) till it reaches (<=0) and then print its increasing sequence(i.e. keep adding by 5) till it reaches N again.

See original problem statement here

#### For Example:

``````Input : 12

Output : 12 7 2 -3 2 7 12``````

### SOLVING APPROACH:

1. Recursively keep printing value of N and decrementing it by 5 until it becomes less than equal to 0.

2. Then print the same values in reverse fashion using `Tail Recursion`.

### STATE SPACE TREE: ### SOLUTIONS:

```
#include <stdio.h>

void print_pattern(int n){
if(n <= 0){
printf("%d ",n);
return;
}
printf("%d ",n);
print_pattern(n-5);
printf("%d ",n);
}

int main(){
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
print_pattern(n);
printf("\n");
}
return 0;
}
```
```
#include <bits/stdc++.h>
using namespace std;

void print_pattern(int n){
if(n <= 0){
cout<<n<<" ";
return;
}
cout<<n<<" ";
print_pattern(n-5);
cout<<n<<" ";
}

int main(){
int t;cin>>t;
while(t--){
int n;cin>>n;
print_pattern(n);
cout<<"\n";
}
return 0;
}
```
```import java.util.*;
import java.io.*;

public class Main {

static void print_pattern(int n){
if(n <= 0){
System.out.print(n + " ");
return;
}
System.out.print(n + " ");
print_pattern(n-5);
System.out.print(n + " ");
}
public static void main(String args[]) throws IOException {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t!=0){
int n = sc.nextInt();
print_pattern(n);
System.out.println();
t--;
}
}
}
```
```def print_pattern(n):

if n <= 0:
print(n, end = " ")
return

print(n, end = " ")
print_pattern(n - 5)
print(n, end = " ")

for _ in range(int(input())):
n = int(input())
print_pattern(n)
print()

```

[forminator_quiz id="968"]

This article tried to discuss Recursion. Hope this blog helps you understand and solve the problem. To practice more problems on Recursion you can check out MYCODE | Competitive Programming.