Given a positive integer N, the task is to find the permutation of first N natural numbers such that the product of Bitwise AND(&) of pairs of adjacent elements is greater than 0. If no such permutation is found, then print “Not Possible”.
Examples:
Input: N = 3
Output: 1 3 2
Explanation:
1 & 3 = 1
3 & 2 = 2
Product of bitwise AND of adjacent elements = 1 * 2 = 2(>0)
Therefore, the required output is 1 3 2Input: 2
Output: Not Possible
Explanation:
Possible permutations of first N(= 2) natural numbers are: { { 1, 2 }, { 2, 1 } }
1 & 2 = 0
2 & 1 = 0
Therefore, the required output is Not Possible.
Naive Approach: The simplest approach to solve this problem is to generate all possible permutations of the first N natural numbers. For every permutation check if the product of bitwise AND of its adjacent elements is greater than 0 or not. If found to be true, then print that permutation. Otherwise, print “Not Possible”.
Time Complexity: O(N * N!)
Auxiliary Space: O(1)
Efficient Approach: The problem can be solved based on the following observations:
If N is a power of 2, then the bitwise AND of N with any number less than N must be 0.
If bitwise AND of adjacent elements is equal to 0, then the product of bitwise AND of its adjacent element must be 0.
Follow the steps below to solve the problem:
- If N is a power of 2, then print “Not Possible”.
- Initialize an array say, arr[] to store the permutation of first N natural numbers that satisfy the given condition.
- Iterate over the range [1, N], and update arr[i] = i
- Update first three elements of the array such that bitwise AND of its adjacent elements must be greater than 0, i.e. arr[1] = 2, arr[2] = 3 and arr[3] = 1
- Iterate over the range [4, N]. For every ith value, check if i is a power of 2 or not. If found to be true, then swap(arr[i], arr[i+1]).
- Finally, print the arr[].
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to check if a number// is a power of 2 or notbool isPowerOfTwo(int n){ if (n == 0) return false; return (ceil(log2(n)) == floor(log2(n)));}// Function to find the permutation of first N// natural numbers such that the product of// bitwise AND of adjacent elements is > 0void findThePermutation(int N){ // Base Case, If N = 1, print 1 if (N == 1) { cout << "1"; return; } // If N is a power of 2, // print "Not Possible" if (isPowerOfTwo(N)) { cout << "Not Possible"; return; } // Stores permutation of first N // natural numbers that satisfy // the condition int arr[N + 1]; // Iterate over the range [1, N] for (int i = 1; i <= N; i++) { // Update arr[i] arr[i] = i; } // Update arr[1], arr[2] and arr[3] arr[1] = 2, arr[2] = 3, arr[3] = 1; // Iterate over the range [4, N] for (int i = 4; i <= N; i++) { // If i is a power of 2 if (isPowerOfTwo(i)) { // Swap adjacent elements swap(arr[i], arr[i + 1]); // Update i i++; } } // Print the array for (int i = 1; i <= N; i++) cout << arr[i] << " ";}// Driver Codeint main(){ // Input int N = 5; // Function Call findThePermutation(N); return 0;} |
Java
// Java program to implement the above approachclass GFG { // Function to calculate the // log base 2 of an integer public static int log2(int N) { // calculate log2 N indirectly // using log() method int result = (int)(Math.log(N) / Math.log(2)); return result; } // Function to check if a number // is a power of 2 or not static boolean isPowerOfTwo(int n) { if (n == 0) return false; return Math.ceil(log2(n)) == Math.floor(log2(n)); } // Function to find the permutation of first N // natural numbers such that the product of // bitwise AND of adjacent elements is > 0 static void findThePermutation(int N) { // Base Case, If N = 1, print 1 if (N == 1) { System.out.print("1"); return; } // If N is a power of 2, // print "Not Possible" if (isPowerOfTwo(N) == false) { System.out.print("Not Possible"); return; } // Stores permutation of first N // natural numbers that satisfy // the condition int arr[] = new int[N + 1]; // Iterate over the range [1, N] for (int i = 1; i <= N; i++) { // Update arr[i] arr[i] = i; } // Update arr[1], arr[2] and arr[3] arr[1] = 2; arr[2] = 3; arr[3] = 1; int temp; // Iterate over the range [4, N] for (int i = 4; i <= N; i++) { // If i is a power of 2 if (isPowerOfTwo(i) == true) { // Swap adjacent elements temp = arr[i]; arr[i] = arr[i + 1]; arr[i + 1] = temp ; // Update i i++; } } // Print the array for (int i = 1; i <= N; i++) System.out.print(arr[i] + " "); } // Driver Code public static void main(String[] args) { // Input int N = 5; // Function Call findThePermutation(N); }}// This code is contributed by AnkThon |
Python3
# Python3 program to implement# the above approachfrom math import ceil, floor, log2# Function to check if a number# is a power of 2 or notdef isPowerOfTwo(n): if (n == 0): return False return (ceil(log2(n)) == floor(log2(n)))# Function to find the permutation of first N# natural numbers such that the product of# bitwise AND of adjacent elements is > 0def findThePermutation(N): # Base Case, If N = 1, pr1 if (N == 1): print("1") return # If N is a power of 2, # print "Not Possible" if (isPowerOfTwo(N)): print("Not Possible") return # Stores permutation of first N # natural numbers that satisfy # the condition arr = [i for i in range(N + 1)] # Iterate over the range [1, N] for i in range(1, N + 1): # Update arr[i] arr[i] = i # Update arr[1], arr[2] and arr[3] arr[1], arr[2], arr[3] = 2, 3, 1 # Iterate over the range [4, N] for i in range(4, N + 1): # If i is a power of 2 if (isPowerOfTwo(i)): # Swap adjacent elements arr[i], arr[i + 1] = arr[i + 1], arr[i] # Update i i += 1 # Print the array for i in range(1, N + 1): print(arr[i], end = " ")# Driver Codeif __name__ == '__main__': # Input N = 5 # Function Call findThePermutation(N)# This code is contributed by mohit kumar 29 |
C#
// C# program to implement the above approachusing System;class GFG { // Function to calculate the // log base 2 of an integer public static int log2(int N) { // calculate log2 N indirectly // using log() method int result = (int)(Math.Log(N) / Math.Log(2)); return result; } // Function to check if a number // is a power of 2 or not static bool isPowerOfTwo(int n) { if (n == 0) return false; return Math.Ceiling((double)log2(n)) == Math.Floor((double)log2(n)); } // Function to find the permutation of first N // natural numbers such that the product of // bitwise AND of adjacent elements is > 0 static void findThePermutation(int N) { // Base Case, If N = 1, print 1 if (N == 1) { Console.Write("1"); return; } // If N is a power of 2, // print "Not Possible" if (isPowerOfTwo(N) == false) { Console.Write("Not Possible"); return; } // Stores permutation of first N // natural numbers that satisfy // the condition int []arr = new int[N + 1]; // Iterate over the range [1, N] for (int i = 1; i <= N; i++) { // Update arr[i] arr[i] = i; } // Update arr[1], arr[2] and arr[3] arr[1] = 2; arr[2] = 3; arr[3] = 1; int temp; // Iterate over the range [4, N] for (int i = 4; i <= N; i++) { // If i is a power of 2 if (isPowerOfTwo(i) == true) { // Swap adjacent elements temp = arr[i]; arr[i] = arr[i + 1]; arr[i + 1] = temp ; // Update i i++; } } // Print the array for (int i = 1; i <= N; i++) Console.Write(arr[i] + " "); } // Driver Code public static void Main(String[] args) { // Input int N = 5; // Function Call findThePermutation(N); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript program to implement // the above approach // Function to calculate the // log base 2 of an integer function log2(N) { // calculate log2 N indirectly // using log() method var result = parseInt( (Math.log(N) / Math.log(2))); return result; } // Function to check if a number // is a power of 2 or not function isPowerOfTwo(n) { if (n == 0) return false; return Math.ceil(log2(n)) == Math.floor(log2(n)); } // Function to find the permutation of first N // natural numbers such that the product of // bitwise AND of adjacent elements is > 0 function findThePermutation(N) { // Base Case, If N = 1, print 1 if (N == 1) { document.write("1"); return; } // If N is a power of 2, // print "Not Possible" if (isPowerOfTwo(N) == false) { document.write("Not Possible"); return; } // Stores permutation of first N // natural numbers that satisfy // the condition var arr = Array(N + 1).fill(0); // Iterate over the range [1, N] for (i = 1; i <= N; i++) { // Update arr[i] arr[i] = i; } // Update arr[1], arr[2] and arr[3] arr[1] = 2; arr[2] = 3; arr[3] = 1; var temp; // Iterate over the range [4, N] for (i = 4; i <= N; i++) { // If i is a power of 2 if (isPowerOfTwo(i) == true) { // Swap adjacent elements temp = arr[i]; arr[i] = arr[i + 1]; arr[i + 1] = temp; // Update i i++; } } // Print the array for (i = 1; i <= N; i++) document.write(arr[i] + " "); } // Driver Code // Input var N = 5; // Function Call findThePermutation(N);// This code is contributed by todaysgaurav</script> |
Output:
2 3 1 5 4
Time Complexity: O(N)
Auxiliary Space: O(N)
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