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)
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