Given a positive integer N, the task is to find the permutation of the first N natural numbers, say arr[] such that (arr[i] != i + 1) and the sum of the absolute difference between arr[i] and (i + 1) is minimum.
Examples:
Input: N = 4
Output: 2 1 4 3
Explanation:
Consider the permutation {2, 1, 4, 3}, Now, the sum is abs(2 – 1) + abs(1 – 2) + abs(4 – 3) + abs(3 – 4) = 1 + 1 + 1 + 1 = 4, which is minimum.Input: N = 7
Output: 2 1 4 3 6 7 5
Naive Approach: The simplest approach to solve the given problem is to generate all possible permutations of the first N Natural Numbers and print that permutation that satisfies the given criteria.
Time Complexity: O(N!)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by observing the fact that the resultant array can be formed by swapping alternating adjacent pairs in order to allow the new position with the minimum sum of the absolute difference between arr[i] and (i +1). In case when N is greater than 1 and N is odd then the last element can be swapped by any of the second last or third last elements of the permutation. Follow the steps below to solve the given problem:
- Initialize an array, say arr[] with the first N natural number arranged in ascending order.
- Traverse the array and swap all the adjacent element as swap(arr[i], arr[i – 1]).
- Now, if the value of N is greater than 1 and N is odd then swap(arr[N – 1], arr[N – 2]).
- After completing the above steps, print the array arr[] as the resultant permutation.
Below is the implementation of the above approach:
C++
// C++ program for the above approach// Function to generate the permutation// of the first N natural numbers having// sum of absolute difference between// element and indices as minimum#include <iostream>using namespace std;void swap(int& a, int& b){ int temp = a; a = b; b = temp;}void findPermutation(int N){ // Initialize array arr[] from 1 to N int arr[N]; for (int i = 0; i < N; i++) { arr[i] = i + 1; } for (int i = 1; i < N; i += 2) { // Swap alternate positions swap(arr[i], arr[i - 1]); } // Check N is greater than 1 and // N is odd if (N % 2 == 1 && N > 1) { // Swapping last two positions swap(arr[N - 1], arr[N - 2]); } // Print the permutation for (int i = 0; i < N; i++) { cout << arr[i] << " "; }}// Driver codeint main(){ int N = 7; findPermutation(N); return 0;}// This code is contributed by Parth Manchanda |
Java
// Java program for the above approach// Function to generate the permutation// of the first N natural numbers having// sum of absolute difference between// element and indices as minimumimport java.util.*;class GFG{static void findPermutation(int N){ // Initialize array arr[] from 1 to N int[] arr = new int[N]; int temp; for(int i = 0; i < N; i++) { arr[i] = i + 1; } for(int i = 1; i < N; i += 2) { // Swap alternate positions temp = arr[i]; arr[i] = arr[i - 1]; arr[i - 1] = temp; } // Check N is greater than 1 and // N is odd if (N % 2 == 1 && N > 1) { // Swapping last two positions temp = arr[N - 1]; arr[N - 1] = arr[N - 2]; arr[N - 2] = temp; } // Print the permutation for(int i = 0; i < N; i++) { System.out.print(arr[i] + " "); }}// Driver codepublic static void main(String[] args){ int N = 7; findPermutation(N);}}// This code is contributed by subhammahato348 |
Python3
# Python3 program for the above approach# Function to generate the permutation# of the first N natural numbers having# sum of absolute difference between# element and indices as minimumdef findPermutation(N): # Initialize array arr[] from 1 to N arr = [i + 1 for i in range(N)] # Swap alternate positions for i in range(1, N, 2): arr[i], arr[i-1] = arr[i-1], arr[i] # Check N is greater than 1 and # N is odd if N % 2 and N > 1: # Swapping last two positions arr[-1], arr[-2] = arr[-2], arr[-1] # Print the permutation for i in arr: print(i, end = " ")# Driver Codeif __name__ == '__main__': N = 7 findPermutation(N) |
C#
// C# program for the above approach// Function to generate the permutation// of the first N natural numbers having// sum of absolute difference between// element and indices as minimumusing System;class GFG { static void findPermutation(int N) { // Initialize array arr[] from 1 to N int[] arr = new int[N]; int temp; for (int i = 0; i < N; i++) { arr[i] = i + 1; } for (int i = 1; i < N; i += 2) { // Swap alternate positions temp = arr[i]; arr[i] = arr[i - 1]; arr[i - 1] = temp; } // Check N is greater than 1 and // N is odd if (N % 2 == 1 && N > 1) { // Swapping last two positions temp = arr[N - 1]; arr[N - 1] = arr[N - 2]; arr[N - 2] = temp; } // Print the permutation for (int i = 0; i < N; i++) { Console.Write(arr[i] + " "); } } // Driver code public static void Main() { int N = 7; findPermutation(N); }}// This code is contributed by ukasp. |
Javascript
<script>// Javascript program for the above approach// Function to generate the permutation// of the first N natural numbers having// sum of absolute difference between// element and indices as minimumfunction findPermutation(N){ var i; // Initialize array arr[] from 1 to N var arr = new Array(N); for(i = 0; i < N; i++) { arr[i] = i + 1; } for(i = 1; i < N; i += 2) { // Swap alternate positions var temp = arr[i]; arr[i] = arr[i - 1]; arr[i - 1] = temp; } // Check N is greater than 1 and // N is odd if (N % 2 == 1 && N > 1) { // Swapping last two positions var temp = arr[N - 1]; arr[N - 1] = arr[N - 2]; arr[N - 2] = temp; } // Print the permutation for(i = 0; i < N; i++) { document.write(arr[i] + " "); }}// Driver codevar N = 7;findPermutation(N);// This code is contributed by SURENDRA_GANGWAR</script> |
2 1 4 3 6 7 5
Time Complexity: O(N)
Auxiliary Space: O(N)
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