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 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 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
<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>
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;
}
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));
}
}
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));
}
}
// 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));
Time Complexity: O(N)
Auxiliary Space: O(1)
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