Given an integer N(N > 1), the task is to arrange all integers from the range [1, N] in an array such that none of the elements are same as the index (1-based indexing) in which they are present in the array.
Examples:
Input: N = 2
Output: 2 1
Explanation: Only possible arrangement of an array of size 2 is 2 1.Input: N=5
Output: 2 1 5 3 4
Explanation: One possible arrangement of an array of size 5 is 2 1 5 3 4l.
Approach: The simplest idea is to place N at the first index and place the remaining elements [1, N – 1] in the remaining indices.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to place first N natural// numbers in an array such that none// of the values are equal to its indicesvoid generatepermutation(int N){ // Stores the required array vector<int> answer; // Place N at the first position answer.push_back(N); // Iterate the range [1, N) for (int i = 1; i < N; i++) { // Append elements to the sequence answer.push_back(i); } // Print the sequence for(int i:answer) cout << i << " ";}// Driver Codeint main(){ int N = 4; generatepermutation(N); return 0;}// This code is contributed by mohit kumar 29. |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to place first N natural// numbers in an array such that none// of the values are equal to its indicesstatic void generatepermutation(int N){ // Stores the required array Vector<Integer> answer = new Vector<Integer>(); // Place N at the first position answer.add(N); // Iterate the range [1, N) for (int i = 1; i < N; i++) { // Append elements to the sequence answer.add(i); } // Print the sequence for(int i:answer) System.out.print(i+ " ");}// Driver Codepublic static void main(String[] args){ int N = 4; generatepermutation(N);}}// This code is contributed by 29AjayKumar |
Python3
# Python program for the above approach# Function to place first N natural# numbers in an array such that none# of the values are equal to its indicesdef generatepermutation(N): # Stores the required array answer = [] # Place N at the first position answer.append(N) # Iterate the range [1, N) for i in range(1, N): # Append elements to the sequence answer.append(i) # Print the sequence print(*answer)# Driver CodeN = 4generatepermutation(N) |
C#
// C# program for the above approachusing System;class GFG{// Function to place first N natural// numbers in an array such that none// of the values are equal to its indicesstatic void generatepermutation(int N){ // Stores the required array int[] answer = new int[N]; // Place N at the first position answer[0] = N; // Iterate the range [1, N) for (int i = 1; i < N; i++) { // Append elements to the sequence answer[i] = i; } // Print the sequence foreach(int i in answer) Console.Write(i+ " ");}// Driver Codestatic public void Main (){ int N = 4; generatepermutation(N);}}// This code is contributed by Dharanendra L V |
Javascript
<script> // JavaScript program for the above approach // Function to place first N natural // numbers in an array such that none // of the values are equal to its indices function generatepermutation(N) { // Stores the required array var answer = []; // Place N at the first position answer.push(N); console.log(answer); // Iterate the range [1, N) for (var i = 1; i < N; i++) { console.log(answer); // Append elements to the sequence answer.push(i); } // Print the sequence for (var i in answer) document.write(answer[i] + " "); } // Driver Code var N = 4; generatepermutation(N); </script> |
Output
4 1 2 3
Time Complexity: O(N)
Auxiliary Space: O(N) because using extra space for vector
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