Given two integers N and K, the task is to find a permutation of integers from the range [1, N] such that the number of indices (1-based indexing) where gcd(p[i], i) > 1 is exactly K. Print -1 if such permutation is not possible.
Examples:
Input: N = 4, K = 3
Output: 1 2 3 4
gcd(1, 1) = 1
gcd(2, 2) = 2
gcd(3, 3) = 3
gcd(4, 4) = 4
Therefore, there are exactly 3 indices where gcd(p[i], i) > 1Input: N = 1, K = 1
Output: -1
Approach: A couple of observations can be made here:
- gcd(i, i + 1) = 1
- gcd(1, i) = 1
- gcd(i, i) = i
Since gcd of 1 with any other number is always going to be one, K can almost be N – 1. Consider the permutation where p[i] = i, The number of indices where gcd(p[i], i) > 1 will be N – 1. Notice that swapping 2 continuous elements (excluding 1) will reduce the count of such indices by exactly 2. And swapping with 1 will reduce the count by exactly 1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to print the required permutationvoid printPermutation(int n, int k){ // Not possible if (k >= n || (n % 2 == 0 && k == 0)) { cout << -1; return; } int per[n + 1], i; // Initial permutation for (i = 1; i <= n; i++) per[i] = i; // Indices for which gcd(p[i], i) > 1 // in the initial permutation int cnt = n - 1; i = 2; while (i < n) { // Reduce the count by 2 if (cnt - 1 > k) { swap(per[i], per[i + 1]); cnt -= 2; } // Reduce the count by 1 else if (cnt - 1 == k) { swap(per[1], per[i]); cnt--; } else break; i += 2; } // Print the permutation for (i = 1; i <= n; i++) cout << per[i] << " ";}// Driver codeint main(){ int n = 4, k = 3; printPermutation(n, k); return 0;} |
Java
// Java implementation of the approach class GFG{// Function to print the required permutation static void printPermutation(int n, int k) { // Not possible if (k >= n || (n % 2 == 0 && k == 0)) { System.out.print(-1); return; } int per[] = new int[n + 1], i; // Initial permutation for (i = 1; i <= n; i++) { per[i] = i; } // Indices for which gcd(p[i], i) > 1 // in the initial permutation int cnt = n - 1; i = 2; while (i < n) { // Reduce the count by 2 if (cnt - 1 > k) { swap(per, i, i + 1); cnt -= 2; } // Reduce the count by 1 else if (cnt - 1 == k) { swap(per, 1, i); cnt--; } else { break; } i += 2; } // Print the permutation for (i = 1; i <= n; i++) { System.out.print(per[i] + " "); } } static int[] swap(int[] arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; return arr; } // Driver code public static void main(String[] args) { int n = 4, k = 3; printPermutation(n, k); }}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 implementation of the approach# Function to print the required permutationdef printPermutation(n, k): # Not possible if (k >= n or (n % 2 == 0 and k == 0)): print(-1); return; per = [0] * (n + 1); # Initial permutation for i in range(1, n + 1): per[i] = i; # Indices for which gcd(p[i], i) > 1 # in the initial permutation cnt = n - 1; i = 2; while (i < n): # Reduce the count by 2 if (cnt - 1 > k): t = per[i]; per[i] = per[i + 1]; per[i + 1] = t; # swap(per[i], per[i + 1]); cnt -= 2; # Reduce the count by 1 elif (cnt - 1 == k): # swap(per[1], per[i]); t = per[1]; per[1] = per[i]; per[i] = t; cnt -= 1; else: break; i += 2; # Print the permutation for i in range(1, n + 1): print(per[i], end = " ");# Driver coden = 4;k = 3;printPermutation(n, k);# This code is contributed by mits |
C#
// C# implementation of the approach using System;class GFG { // Function to print the required permutation static void printPermutation(int n, int k) { // Not possible if (k >= n || (n % 2 == 0 && k == 0)) { Console.Write(-1); return; } int []per = new int[n + 1] ; int i ; // Initial permutation for (i = 1; i <= n; i++) { per[i] = i; } // Indices for which gcd(p[i], i) > 1 // in the initial permutation int cnt = n - 1; i = 2; while (i < n) { // Reduce the count by 2 if (cnt - 1 > k) { swap(per, i, i + 1); cnt -= 2; } // Reduce the count by 1 else if (cnt - 1 == k) { swap(per, 1, i); cnt--; } else { break; } i += 2; } // Print the permutation for (i = 1; i <= n; i++) { Console.Write(per[i] + " "); } } static int[] swap(int[] arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; return arr; } // Driver code public static void Main() { int n = 4, k = 3; printPermutation(n, k); } } /* This code contributed by Ryuga */ |
PHP
<?php// PHP implementation of the approach// Function to print the required permutationfunction printPermutation($n, $k){ // Not possible if ($k >= $n || ($n % 2 == 0 && $k == 0)) { echo -1; return; } $per[$n + 1] = array(); $i; // Initial permutation for ($i = 1; $i <= $n; $i++) $per[$i] = $i; // Indices for which gcd(p[i], i) > 1 // in the initial permutation $cnt = $n - 1; $i = 2; while ($i < $n) { // Reduce the count by 2 if ($cnt - 1 > $k) { list($per[$i], $per[$i + 1]) = array($per[$i + 1], $per[$i]); //swap(per[i], per[i + 1]); $cnt -= 2; } // Reduce the count by 1 else if ($cnt - 1 == $k) { // swap(per[1], per[i]); list($per[1], $per[$i]) = array($per[$i], $per[1]); $cnt--; } else break; $i += 2; } // Print the permutation for ($i = 1; $i <= $n; $i++) echo $per[$i], " ";}// Driver code$n = 4;$k = 3;printPermutation($n, $k);// This code is contributed by ajit.?> |
Javascript
<script> // Javascript implementation of the approach // Function to print the required permutation function printPermutation(n, k) { // Not possible if (k >= n || (n % 2 == 0 && k == 0)) { document.write(-1); return; } let per = new Array(n + 1); per.fill(0); let i ; // Initial permutation for (i = 1; i <= n; i++) { per[i] = i; } // Indices for which gcd(p[i], i) > 1 // in the initial permutation let cnt = n - 1; i = 2; while (i < n) { // Reduce the count by 2 if (cnt - 1 > k) { swap(per, i, i + 1); cnt -= 2; } // Reduce the count by 1 else if (cnt - 1 == k) { swap(per, 1, i); cnt--; } else { break; } i += 2; } // Print the permutation for (i = 1; i <= n; i++) { document.write(per[i] + " "); } } function swap(arr, i, j) { let temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; return arr; } let n = 4, k = 3; printPermutation(n, k); </script> |
Output:
1 2 3 4
Time Complexity: O(N)
Auxiliary Space: O(N)
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