Given a number N, the task is to print the numbers from N to 1.
Examples:
Input: N = 10
Output: 10 9 8 7 6 5 4 3 2 1
Input: N = 7
Output: 7 6 5 4 3 2 1
Approach 1: Run a loop from N to 1 and print the value of N for each iteration. Decrement the value of N by 1 after each iteration.
Below is the implementation of the above approach.
C++
// C++ program to print all numbers between 1// to N in reverse order#include <bits/stdc++.h>using namespace std;// Recursive function to print from N to 1void PrintReverseOrder(int N){ for (int i = N; i > 0; i--) cout << i << " ";}// Driven Codeint main(){ int N = 5; PrintReverseOrder(N); return 0;} |
Java
// Java program to print all numbers between 1// to N in reverse orderimport java.util.*;class GFG {// Recursive function to print from N to 1static void PrintReverseOrder(int N){ for (int i = N; i > 0; i--) System.out.print( +i + " ");}// Driver codepublic static void main(String[] args){ int N = 5; PrintReverseOrder(N);}}// This code is contributed by shivanisinghss2110 |
Python3
# Python3 program to print all numbers # between 1 to N in reverse order# Recursive function to print # from N to 1def PrintReverseOrder(N): for i in range(N, 0, -1): print(i, end = " ");# Driver codeif __name__ == '__main__': N = 5; PrintReverseOrder(N);# This code is contributed by 29AjayKumar |
C#
// C# program to print all numbers // between 1 to N in reverse orderusing System;class GFG{// Recursive function to print// from N to 1static void PrintReverseOrder(int N){ for(int i = N; i > 0; i--) Console.Write(i + " ");}// Driver codepublic static void Main(String[] args){ int N = 5; PrintReverseOrder(N);}}// This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program to print all numbers between 1 // to N in reverse order // Recursive function to print from N to 1 function PrintReverseOrder(N) { for (let i = N; i > 0; i--) document.write(i + " "); } let N = 5; PrintReverseOrder(N); </script> |
Output:
5 4 3 2 1
Time Complexity: O(N)
Auxiliary Space: O(1)
Approach 2: We will use recursion to solve this problem.
- Check for the base case. Here it is N<=0.
- If base condition satisfied, return to the main function.
- If base condition not satisfied, print N and call the function recursively with value (N – 1) until base condition satisfies.
Below is the implementation of the above approach.
C++
// C++ program to print all numbers between 1// to N in reverse order#include <bits/stdc++.h>using namespace std;// Recursive function to print from N to 1void PrintReverseOrder(int N){ // if N is less than 1 // then return void function if (N <= 0) { return; } else { cout << N << " "; // recursive call of the function PrintReverseOrder(N - 1); }}// Driven Codeint main(){ int N = 5; PrintReverseOrder(N); return 0;} |
Java
// Java program to print all numbers // between 1 to N in reverse orderclass GFG{// Recursive function to print // from N to 1static void PrintReverseOrder(int N){ // If N is less than 1 then // return static void function if (N <= 0) { return; } else { System.out.print(N + " "); // Recursive call of the function PrintReverseOrder(N - 1); }}// Driver codepublic static void main(String[] args){ int N = 5; PrintReverseOrder(N);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to print all numbers between 1# to N in reverse order# Recursive function to print from N to 1def PrintReverseOrder(N): # if N is less than 1 # then return void function if (N <= 0): return; else: print(N, end = " "); # recursive call of the function PrintReverseOrder(N - 1); # Driver CodeN = 5;PrintReverseOrder(N);# This code is contributed by Nidhi_biet |
C#
// C# program to print all numbers // between 1 to N in reverse orderusing System;class GFG{// Recursive function to print // from N to 1static void PrintReverseOrder(int N){ // If N is less than 1 then // return static void function if (N <= 0) { return; } else { Console.Write(N + " "); // Recursive call of the function PrintReverseOrder(N - 1); }}// Driver codepublic static void Main(){ int N = 5; PrintReverseOrder(N);}}// This code is contributed by Code_Mech |
Javascript
<script> // Javascript program to print all numbers between 1 // to N in reverse order // Recursive function to print from N to 1 function PrintReverseOrder(N) { // if N is less than 1 // then return void function if (N <= 0) { return; } else { document.write(N + " "); // recursive call of the function PrintReverseOrder(N - 1); } } // Driver code let N = 5; PrintReverseOrder(N); // This code is contributed by suresh07.</script> |
Output:
5 4 3 2 1
Time Complexity: O(N)
Auxiliary Space: O(N)
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