Find permutation of [1, N] such that (arr[i] != i+1) and sum of absolute difference between arr[i] and (i+1) is minimum
Last Updated :
17 Jul, 2021
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++
#include <iostream>
using namespace std;
void swap(int& a, int& b)
{
int temp = a;
a = b;
b = temp;
}
void findPermutation(int N)
{
int arr[N];
for (int i = 0; i < N; i++) {
arr[i] = i + 1;
}
for (int i = 1; i < N; i += 2) {
swap(arr[i], arr[i - 1]);
}
if (N % 2 == 1 && N > 1) {
swap(arr[N - 1], arr[N - 2]);
}
for (int i = 0; i < N; i++) {
cout << arr[i] << " ";
}
}
int main()
{
int N = 7;
findPermutation(N);
return 0;
}
|
Java
import java.util.*;
class GFG{
static void findPermutation(int 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)
{
temp = arr[i];
arr[i] = arr[i - 1];
arr[i - 1] = temp;
}
if (N % 2 == 1 && N > 1)
{
temp = arr[N - 1];
arr[N - 1] = arr[N - 2];
arr[N - 2] = temp;
}
for(int i = 0; i < N; i++)
{
System.out.print(arr[i] + " ");
}
}
public static void main(String[] args)
{
int N = 7;
findPermutation(N);
}
}
|
Python3
def findPermutation(N):
arr = [i + 1 for i in range(N)]
for i in range(1, N, 2):
arr[i], arr[i-1] = arr[i-1], arr[i]
if N % 2 and N > 1:
arr[-1], arr[-2] = arr[-2], arr[-1]
for i in arr:
print(i, end = " ")
if __name__ == '__main__':
N = 7
findPermutation(N)
|
C#
using System;
class GFG {
static void findPermutation(int 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) {
temp = arr[i];
arr[i] = arr[i - 1];
arr[i - 1] = temp;
}
if (N % 2 == 1 && N > 1) {
temp = arr[N - 1];
arr[N - 1] = arr[N - 2];
arr[N - 2] = temp;
}
for (int i = 0; i < N; i++) {
Console.Write(arr[i] + " ");
}
}
public static void Main()
{
int N = 7;
findPermutation(N);
}
}
|
Javascript
<script>
function findPermutation(N)
{
var i;
var arr = new Array(N);
for(i = 0; i < N; i++)
{
arr[i] = i + 1;
}
for(i = 1; i < N; i += 2)
{
var temp = arr[i];
arr[i] = arr[i - 1];
arr[i - 1] = temp;
}
if (N % 2 == 1 && N > 1)
{
var temp = arr[N - 1];
arr[N - 1] = arr[N - 2];
arr[N - 2] = temp;
}
for(i = 0; i < N; i++)
{
document.write(arr[i] + " ");
}
}
var N = 7;
findPermutation(N);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
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