Given a positive integer n, find the lexicographically smallest permutation p of {1, 2, .. n} such that p_{i} != i. i.e., i should not be there at i-th position where i varies from 1 to n.

Examples:

Input : 5 Output : 2 1 4 5 3 Consider the two permutations that follow the requirement that position and numbers should not be same. p = (2, 1, 4, 5, 3) and q = (2, 4, 1, 5, 3). Since p is lexicographically smaller, our output is p. Input : 6 Output : 2 1 4 3 6 5

Since we need lexicographically smallest (and 1 cannot come at position 1), we put 2 at first position. After 2, we put the next smallest element i.e., 1. After that the next smallest considering it does not violates our condition of pi != i.

Now, if our n is even we simply take two variables one which will contain our count of even numbers and one which will contain our count of odd numbers and then we will keep them adding in the vector till we reach n.

But, if our n is odd, we do the same task till we reach n-1 because if we add till n then in the end we will end up having p_{i} = i. So when we reach n-1, we first add n to the position n-1 and then on position n we will put n-2.

The C++ implementation of the above program is given below.

#include <bits/stdc++.h> using namespace std; // Function to print the permutation void findPermutation(vector<int> a, int n) { vector<int> res; // Initial numbers to be pushed to result int en = 2, on = 1; // If n is even if (n % 2 == 0) { for (int i = 0; i < n; i++) { if (i % 2 == 0) { res.push_back(en); en += 2; } else { res.push_back(on); on += 2; } } } // If n is odd else { for (int i = 0; i < n - 2; i++) { if (i % 2 == 0) { res.push_back(en); en += 2; } else { res.push_back(on); on += 2; } } res.push_back(n); res.push_back(n - 2); } // Print result for (int i = 0; i < n; i++) cout << res[i] << " "; cout << "\n"; } // Driver Code int main() { long long int n = 9; findPermutation(n); return 0; }

Output:

2 1 4 3 6 5 8 9 7

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