Given two arrays arr1[] and arr2[] of N integers. We can choose any two adjacent elements from array arr1[] and swap them if they are of opposite parity, the task is to check if it is possible to convert array arr1[] to array arr2[] by performing the given operation on arr1[]. Print “Yes” if it is possible to convert array arr1[] to arr2[] Else print “No”.
Examples:
Input: arr1[] = {5, 7, 8, 2, 10, 13}, arr2[] = {8, 5, 2, 7, 13, 10}
Output: Yes
Explanation:
At first, swap 10 and 13 so arr1[] = [5, 7, 8, 2, 13, 10].
Now, swap 7 and 8 so arr1[] = [5, 8, 7, 2, 13, 10].
Now, swap 5 and 8 so arr1[] = [8, 5, 7, 2, 13, 10].
Now, swap 7 and 2 so arr1[] = [8, 5, 2, 7, 13, 10] = arr2[].
In each operation, we swap adjacent elements with different parity.
Input: arr1[] = {0, 1, 13, 3, 4, 14, 6}, arr2[] = {0, 1, 14, 3, 4, 13, 6}
Output: No
Explanation:
It is not possible to swap 13, 14 because they are not adjacent.
Approach: The problem can be solved using Greedy Approach. Since we cannot swap any two even or odd numbers. So the relative position of both even and odd numbers in the arrays arr1[] and arr2[] must be exactly the same to make both the arrays equal with the given operation. Below are the steps:
- Create two arrays(say even[] and odd[]) insert all the even and odd numbers from arr1[] in even[] and odd[] respectively.
- Now check whether the even and odd numbers in arr2[] are in the same order as in even[] and odd[].
- If the above steps doesn’t gives any number from arr2[] which are not in the order of numbers in even[] and odd[] arrays respectively then, print “Yes” else print “No”.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function which checks if it is // possible to convert arr1[] to // arr2[] by given operations void convert( int a[], int b[], int n)
{ // even[] will store the even
// elements of a[]
// odd[] will store the odd
// elements of a[]
vector< int > even, odd;
// Traverse a[] and insert the even
// and odd element respectively
for ( int x = 0; x < n; x++) {
if (a[x] % 2 == 0)
even.push_back(a[x]);
else
odd.push_back(a[x]);
}
// ei points to the next
// available even element
// oi points to the next
// available odd element
int ei = 0, oi = 0;
// poss will store whether the
// given transformation
// of a[] to b[] is possible
bool poss = true ;
// Traverse b[]
for ( int x = 0; x < n; x++) {
if (b[x] % 2 == 0) {
// Check if both even
// elements are equal
if (ei < even.size()
&& b[x] == even[ei]) {
ei++;
}
else {
poss = false ;
break ;
}
}
else {
// Check if both odd
// elements are equal
if (oi < odd.size()
&& b[x] == odd[oi]) {
oi++;
}
else {
poss = false ;
break ;
}
}
}
// If poss is true, then we can
// transform a[] to b[]
if (poss)
cout << "Yes" << endl;
else
cout << "No" << endl;
} // Driver Code int main()
{ // Given arrays
int arr1[] = { 0, 1, 13, 3, 4, 14, 6 };
int arr2[] = { 0, 1, 14, 3, 4, 13, 6 };
int N = sizeof (arr1) / sizeof (arr1[0]);
// Function Call
convert(arr1, arr2, N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function which checks if it is // possible to convert arr1[] to // arr2[] by given operations static void convert( int a[], int b[], int n)
{ // even[] will store the even
// elements of a[]
// odd[] will store the odd
// elements of a[]
Vector<Integer> even = new Vector<Integer>(),
odd = new Vector<Integer>();
// Traverse a[] and insert the even
// and odd element respectively
for ( int x = 0 ; x < n; x++)
{
if (a[x] % 2 == 0 )
even.add(a[x]);
else
odd.add(a[x]);
}
// ei points to the next
// available even element
// oi points to the next
// available odd element
int ei = 0 , oi = 0 ;
// poss will store whether the
// given transformation
// of a[] to b[] is possible
boolean poss = true ;
// Traverse b[]
for ( int x = 0 ; x < n; x++)
{
if (b[x] % 2 == 0 )
{
// Check if both even
// elements are equal
if (ei < even.size() &&
b[x] == even.get(ei))
{
ei++;
}
else
{
poss = false ;
break ;
}
}
else
{
// Check if both odd
// elements are equal
if (oi < odd.size() &&
b[x] == odd.get(oi))
{
oi++;
}
else
{
poss = false ;
break ;
}
}
}
// If poss is true, then we can
// transform a[] to b[]
if (poss)
System.out.print( "Yes" + "\n" );
else
System.out.print( "No" + "\n" );
} // Driver Code public static void main(String[] args)
{ // Given arrays
int arr1[] = { 0 , 1 , 13 , 3 , 4 , 14 , 6 };
int arr2[] = { 0 , 1 , 14 , 3 , 4 , 13 , 6 };
int N = arr1.length;
// Function Call
convert(arr1, arr2, N);
} } // This code is contributed by gauravrajput1 |
# Python3 program for the above approach # Function which checks if it is # possible to convert arr1[] to # arr2[] by given operations def convert(a, b, n):
# even[] will store the even
# elements of a[]
# odd[] will store the odd
# elements of a[]
even = []
odd = []
# Traverse a[] and insert the even
# and odd element respectively
for x in range (n):
if (a[x] % 2 = = 0 ):
even.append(a[x])
else :
odd.append(a[x])
# ei points to the next
# available even element
# oi points to the next
# available odd element
ei, oi = 0 , 0
# poss will store whether the
# given transformation
# of a[] to b[] is possible
poss = True
# Traverse b[]
for x in range (n):
if (b[x] % 2 = = 0 ):
# Check if both even
# elements are equal
if (ei < len (even) and
b[x] = = even[ei]):
ei + = 1
else :
poss = False
break
else :
# Check if both odd
# elements are equal
if (oi < len (odd) and
b[x] = = odd[oi]):
oi + = 1
else :
poss = False
break
# If poss is true, then we can
# transform a[] to b[]
if (poss):
print ( "Yes" )
else :
print ( "No" )
# Driver Code if __name__ = = "__main__" :
# Given arrays
arr1 = [ 0 , 1 , 13 , 3 , 4 , 14 , 6 ]
arr2 = [ 0 , 1 , 14 , 3 , 4 , 13 , 6 ]
N = len (arr1)
# Function call
convert(arr1, arr2, N)
# This code is contributed by chitranayal |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Function which checks if it is // possible to convert arr1[] to // arr2[] by given operations static void convert( int []a, int []b, int n)
{ // even[] will store the even
// elements of []a
// odd[] will store the odd
// elements of []a
List< int > even = new List< int >(),
odd = new List< int >();
// Traverse []a and insert the even
// and odd element respectively
for ( int x = 0; x < n; x++)
{
if (a[x] % 2 == 0)
even.Add(a[x]);
else
odd.Add(a[x]);
}
// ei points to the next
// available even element
// oi points to the next
// available odd element
int ei = 0, oi = 0;
// poss will store whether the
// given transformation
// of []a to []b is possible
bool poss = true ;
// Traverse []b
for ( int x = 0; x < n; x++)
{
if (b[x] % 2 == 0)
{
// Check if both even
// elements are equal
if (ei < even.Count &&
b[x] == even[ei])
{
ei++;
}
else
{
poss = false ;
break ;
}
}
else
{
// Check if both odd
// elements are equal
if (oi < odd.Count &&
b[x] == odd[oi])
{
oi++;
}
else
{
poss = false ;
break ;
}
}
}
// If poss is true, then we can
// transform []a to []b
if (poss)
Console.Write( "Yes" + "\n" );
else
Console.Write( "No" + "\n" );
} // Driver Code public static void Main(String[] args)
{ // Given arrays
int []arr1 = { 0, 1, 13, 3, 4, 14, 6 };
int []arr2 = { 0, 1, 14, 3, 4, 13, 6 };
int N = arr1.Length;
// Function call
convert(arr1, arr2, N);
} } // This code is contributed by gauravrajput1 |
<script> // JavaScript program for the above approach // Function which checks if it is // possible to convert arr1[] to // arr2[] by given operations function convert(a, b, n)
{ // even[] will store the even
// elements of a[]
// odd[] will store the odd
// elements of a[]
var even = [], odd = [];
// Traverse a[] and insert the even
// and odd element respectively
for ( var x = 0; x < n; x++) {
if (a[x] % 2 == 0)
even.push(a[x]);
else
odd.push(a[x]);
}
// ei points to the next
// available even element
// oi points to the next
// available odd element
var ei = 0, oi = 0;
// poss will store whether the
// given transformation
// of a[] to b[] is possible
var poss = true ;
// Traverse b[]
for ( var x = 0; x < n; x++) {
if (b[x] % 2 == 0) {
// Check if both even
// elements are equal
if (ei < even.length
&& b[x] == even[ei]) {
ei++;
}
else {
poss = false ;
break ;
}
}
else {
// Check if both odd
// elements are equal
if (oi < odd.length
&& b[x] == odd[oi]) {
oi++;
}
else {
poss = false ;
break ;
}
}
}
// If poss is true, then we can
// transform a[] to b[]
if (poss)
document.write( "Yes" );
else
document.write( "No" );
} // Driver Code // Given arrays var arr1 = [0, 1, 13, 3, 4, 14, 6 ];
var arr2 = [0, 1, 14, 3, 4, 13, 6 ];
var N = arr1.length;
// Function Call convert(arr1, arr2, N); </script> |
No
Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(N), where N is the number of elements in the array.