Check if array can be sorted by swapping pairs having GCD equal to the smallest element in the array

Last Updated : 04 Nov, 2023

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# 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>

Output

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))

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));

Output

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)

Check if an array of pairs can be sorted by swapping pairs with different first elements

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Check if array can be sorted by swapping pairs having GCD equal to the smallest element in the array

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