Check if array can be sorted by swapping pairs having GCD equal to the smallest element in the array
Given an array arr[] of size N, the task is to check if an array can be sorted by swapping only the elements whose GCD (greatest common divisor) is equal to the smallest element of the array. Print “Yes” if it is possible to sort the array. Otherwise, print “No”.
Examples:
Input: arr[] = {4, 3, 6, 6, 2, 9}
Output: Yes
Explanation:
Smallest element in the array = 2
Swap arr[0] and arr[2], since they have gcd equal to 2. Therefore, arr[] = {6, 3, 4, 6, 2, 9}
Swap arr[0] and arr[4], since they have gcd equal to 2. Therefore, arr[] = {2, 3, 4, 6, 6, 9}
Input: arr[] = {2, 6, 2, 4, 5}
Output: No
Approach: The idea is to find the smallest element from the array and check if it is possible to sort the array. Below are the steps:
- Find the smallest element of the array and store it in a variable, say mn.
- Store the array in a temporary array, say B[].
- Sort the original array arr[].
- Now, iterate over the array, if there is an element which is not divisible by mn and whose position is changed after sorting, then print “NO”. Otherwise, rearrange the elements divisible by mn by swapping it with other elements.
- If the position of all elements which are not divisible by mn remains unchanged, then print “YES”.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to check if it is // possible to sort array or not void isPossible( int arr[], int N) { // Store the smallest element int mn = INT_MAX; // Copy the original array int B[N]; // Iterate over the given array for ( int i = 0; i < N; i++) { // Update smallest element mn = min(mn, arr[i]); // Copy elements of arr[] // to array B[] B[i] = arr[i]; } // Sort original array sort(arr, arr + N); // Iterate over the given array for ( int i = 0; i < N; i++) { // If the i-th element is not // in its sorted place if (arr[i] != B[i]) { // Not possible to swap if (B[i] % mn != 0) { cout << "No" ; return ; } } } cout << "Yes" ; return ; } // Driver Code int main() { // Given array int N = 6; int arr[] = { 4, 3, 6, 6, 2, 9 }; // Function Call isPossible(arr, N); return 0; } |
Java
// Java implementation of // the above approach import java.util.*; class GFG{ // Function to check if it is // possible to sort array or not static void isPossible( int arr[], int N) { // Store the smallest element int mn = Integer.MAX_VALUE; // Copy the original array int []B = new int [N]; // Iterate over the given array for ( int i = 0 ; i < N; i++) { // Update smallest element mn = Math.min(mn, arr[i]); // Copy elements of arr[] // to array B[] B[i] = arr[i]; } // Sort original array Arrays.sort(arr); // Iterate over the given array for ( int i = 0 ; i < N; i++) { // If the i-th element is not // in its sorted place if (arr[i] != B[i]) { // Not possible to swap if (B[i] % mn != 0 ) { System.out.print( "No" ); return ; } } } System.out.print( "Yes" ); return ; } // Driver Code public static void main(String[] args) { // Given array int N = 6 ; int arr[] = { 4 , 3 , 6 , 6 , 2 , 9 }; // Function Call isPossible(arr, N); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of # the above approach import sys # Function to check if it is # possible to sort array or not def isPossible(arr, N): # Store the smallest element mn = sys.maxsize; # Copy the original array B = [ 0 ] * N; # Iterate over the given array for i in range (N): # Update smallest element mn = min (mn, arr[i]); # Copy elements of arr # to array B B[i] = arr[i]; # Sort original array arr.sort(); # Iterate over the given array for i in range (N): # If the i-th element is not # in its sorted place if (arr[i] ! = B[i]): # Not possible to swap if (B[i] % mn ! = 0 ): print ( "No" ); return ; print ( "Yes" ); return ; # Driver Code if __name__ = = '__main__' : # Given array N = 6 ; arr = [ 4 , 3 , 6 , 6 , 2 , 9 ]; # Function Call isPossible(arr, N); # This code is contributed by 29AjayKumar |
C#
// C# implementation of // the above approach using System; class GFG{ // Function to check if it is // possible to sort array or not static void isPossible( int []arr, int N) { // Store the smallest element int mn = int .MaxValue; // Copy the original array int []B = new int [N]; // Iterate over the given array for ( int i = 0; i < N; i++) { // Update smallest element mn = Math.Min(mn, arr[i]); // Copy elements of []arr // to array []B B[i] = arr[i]; } // Sort original array Array.Sort(arr); // Iterate over the given array for ( int i = 0; i < N; i++) { // If the i-th element is not // in its sorted place if (arr[i] != B[i]) { // Not possible to swap if (B[i] % mn != 0) { Console.Write( "No" ); return ; } } } Console.Write( "Yes" ); return ; } // Driver Code public static void Main(String[] args) { // Given array int N = 6; int []arr = {4, 3, 6, 6, 2, 9}; // Function Call isPossible(arr, N); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript program for // the above approach // Function to check if it is // possible to sort array or not function isPossible(arr, N) { // Store the smallest element let mn = Number.MAX_VALUE; // Copy the original array let B = []; // Iterate over the given array for (let i = 0; i < N; i++) { // Update smallest element mn = Math.min(mn, arr[i]); // Copy elements of arr[] // to array B[] B[i] = arr[i]; } // Sort original array arr.sort(); // Iterate over the given array for (let i = 0; i < N; i++) { // If the i-th element is not // in its sorted place if (arr[i] != B[i]) { // Not possible to swap if (B[i] % mn != 0) { document.write( "No" ); return ; } } } document.write( "Yes" ); return ; } // Driver code // Given array let N = 6; let arr = [4, 3, 6, 6, 2, 9]; // Function Call isPossible(arr, N); </script> |
Yes
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
Approach 2:
Here’s another approach to check if array can be sorted by swapping pairs having GCD equal to the smallest element in the array:
- Initialize two variables, say left and right, with the value 0.
- Traverse the array from left to right and do the following:
- a. If the current element arr[i] is greater than the next element arr[i+1], set left to i.
- Traverse the array from right to left and do the following:
- a. If the current element arr[i] is less than the previous element arr[i-1], set right to i.
- If either left or right is still 0, then the array is already sorted and can be swapped using adjacent elements.
- Otherwise, swap arr[left] with arr[right] and check if the resulting array is sorted.
- If the array is sorted, return “Yes”, otherwise return “No”.
Here’s the C++ code for the above approach:
C++
#include <bits/stdc++.h> using namespace std; // Function to check if an array can be sorted // by swapping adjacent elements string isPossible( int arr[], int n) { int left = 0, right = 0; // Find the left index where the array is not sorted for ( int i = 0; i < n - 1; i++) { if (arr[i] > arr[i+1]) { left = i; break ; } } // If array is already sorted if (left == 0) { return "Yes" ; } // Find the right index where the array is not sorted for ( int i = n - 1; i >= left; i--) { if (arr[i] < arr[i-1]) { right = i; break ; } } // Swap adjacent elements swap(arr[left], arr[right]); // Check if array is sorted for ( int i = 0; i < n - 1; i++) { if (arr[i] > arr[i+1]) { return "No" ; } } return "Yes" ; } // Driver code int main() { int arr[] = {4, 2, 3, 1}; int n = sizeof (arr)/ sizeof (arr[0]); cout << isPossible(arr, n); return 0; } |
Java
public class GFG { // Function to check if an array can be sorted // by swapping adjacent elements public static String isPossible( int [] arr) { int n = arr.length; int left = 0 , right = 0 ; // Find the left index where the array is not sorted for ( int i = 0 ; i < n - 1 ; i++) { if (arr[i] > arr[i + 1 ]) { left = i; break ; } } // If the array is already sorted if (left == 0 ) { return "Yes" ; } // Find the right index where the array is not sorted for ( int i = n - 1 ; i >= left; i--) { if (arr[i] < arr[i - 1 ]) { right = i; break ; } } // Swap adjacent elements int temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; // Check if the array is sorted for ( int i = 0 ; i < n - 1 ; i++) { if (arr[i] > arr[i + 1 ]) { return "No" ; } } return "Yes" ; } public static void main(String[] args) { int [] arr = { 4 , 2 , 3 , 1 }; // Function Call System.out.println(isPossible(arr)); } } |
Python3
def isPossible(arr, n): left = 0 right = 0 # Find the left index where the array is not sorted for i in range (n - 1 ): if arr[i] > arr[i + 1 ]: left = i break # If array is already sorted if left = = 0 : return "Yes" # Find the right index where the array is not sorted for i in range (n - 1 , left, - 1 ): if arr[i] < arr[i - 1 ]: right = i break # Swap adjacent elements arr[left], arr[right] = arr[right], arr[left] # Check if array is sorted for i in range (n - 1 ): if arr[i] > arr[i + 1 ]: return "No" return "Yes" # Driver code arr = [ 4 , 2 , 3 , 1 ] n = len (arr) print (isPossible(arr, n)) |
C#
using System; class Program { // Function to check if an array can be sorted // by swapping adjacent elements static string IsPossible( int [] arr) { int n = arr.Length; int left = 0, right = 0; // Find the left index where the array is not sorted for ( int i = 0; i < n - 1; i++) { if (arr[i] > arr[i + 1]) { left = i; break ; } } // If array is already sorted if (left == 0) { return "Yes" ; } // Find the right index where the array is not sorted for ( int i = n - 1; i >= left; i--) { if (arr[i] < arr[i - 1]) { right = i; break ; } } // Swap adjacent elements int temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; // Check if array is sorted for ( int i = 0; i < n - 1; i++) { if (arr[i] > arr[i + 1]) { return "No" ; } } return "Yes" ; } // Driver code static void Main() { int [] arr = { 4, 2, 3, 1 }; Console.WriteLine(IsPossible(arr)); } } |
Javascript
// Function to check if an array can be sorted // by swapping adjacent elements function isPossible(arr, n) { let left = 0, right = 0; // Find the left index where the array is not sorted for (let i = 0; i < n - 1; i++) { if (arr[i] > arr[i + 1]) { left = i; break ; } } // If array is already sorted if (left == 0) { return "Yes" ; } // Find the right index where the array is not sorted for (let i = n - 1; i >= left; i--) { if (arr[i] < arr[i - 1]) { right = i; break ; } } // Swap adjacent elements [arr[left], arr[right]] = [arr[right], arr[left]]; // Check if array is sorted for (let i = 0; i < n - 1; i++) { if (arr[i] > arr[i + 1]) { return "No" ; } } return "Yes" ; } let arr = [4, 2, 3, 1]; let n = arr.length; console.log(isPossible(arr, n)); |
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)