You are given an integer N. Print numbers from N to 1 without the help of loops.
Examples:
Input: N = 5
Output: 5 4 3 2 1
Explanation: We have to print numbers from 5 to 1.Input: N = 10
Output: 10 9 8 7 6 5 4 3 2 1
Explanation: We have to print numbers from 10 to 1.
Approach: If we take a look at this problem carefully, we can see that the idea of “loop” is to track some counter value, e.g., “i = 0” till “i <= 100”. So, if we aren’t allowed to use loops, how can we track something?
Well, one possibility is the use of ‘recursion‘, provided we use the terminating condition carefully. Here is a solution that prints numbers using recursion.
C++
// C++ program to How will you print
// numbers from N to 1 without using a loop?
#include <iostream>
using namespace std;
class gfg {
// It prints numbers from N to 1
public:
void printNos(unsigned int n)
{
if (n > 0) {
cout << n << " ";
printNos(n - 1);
}
return;
}
};
// Driver code
int main()
{
int n = 10;
gfg g;
g.printNos(n);
return 0;
}
C
#include <stdio.h>
// Prints numbers from N to 1
void printNos(unsigned int n)
{
if (n > 0) {
printf("%d ", n);
printNos(n - 1);
}
return;
}
// Driver code
int main()
{
int n = 10;
printNos(n);
getchar();
return 0;
}
Java
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
class GFG {
// Prints numbers from N to 1
static void printNos(int n)
{
if (n > 0) {
System.out.print(n + " ");
printNos(n - 1);
}
return;
}
// Driver Code
public static void main(String[] args)
{
int n = 10;
printNos(n);
}
}
Python3
# Python3 program to Print
# numbers from N to 1
def printNos(n):
if n > 0:
print(n, end=' ')
printNos(n - 1)
# Driver code
n = 10
printNos(n)
C#
// C# code for print numbers from
// N to 1 without using loop
using System;
class GFG {
// Prints numbers from N to 1
static void printNos(int n)
{
if (n > 0) {
Console.Write(n + " ");
printNos(n - 1);
}
return;
}
// Driver Code
public static void Main()
{
int n = 10;
printNos(n);
}
}
PHP
<?php
// PHP program print numbers
// from N to 1 without
// using loop
// Prints numbers from N to 1
function printNos($n)
{
if($n > 0)
{
echo $n, " ";
printNos($n - 1);
}
return;
}
// Driver code
$n=10;
printNos($n);
?>
Javascript
// Javascript code for print numbers from
// N to 1 without using loop
// Prints numbers from N to 1
function printNos(n)
{
if(n > 0)
{
console.log(n + " ");
printNos(n - 1);
}
return;
}
var n = 10;
printNos(n);
Output
10 9 8 7 6 5 4 3 2 1
Time Complexity: O(n)
Auxiliary Space: O(n)
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