Given two arrays, we have to check whether we can sort two arrays in strictly ascending order by swapping A[i] and B[i].
Examples:
Input : A[ ]={ 1, 4, 3, 5, 7}, B[ ]={ 2, 2, 5, 8, 9}
Output : True
After swapping A[1] and B[1], both the arrays are sorted.Input : A[ ]={ 1, 4, 5, 5, 7}, B[ ]={ 2, 2, 5, 8, 9}
Output : False
It is not possible to make both the arrays sorted with any number of swaps.
We are given two arrays, we can swap A[i] with B[i] so that we can sort both the array in strictly ascending order so we have to sort the array in such a way that A[i] < A[i+1] and B[i] < B[i+1].
We will use a greedy approach and solve the problem.
We will get the minimum and maximum of A[i] and B[i] and assign minimum to B[i] and maximum to A[i].
Now, we will check that array A and array B is strictly increasing or not.
Let us consider our approach is incorrect, (there is possibility to arrange but our approach gives false), that means any one or more position is switched.
That means a[i-1] is not less than a[i] or a[i+1] is not greater than a[i] . Now if a[i] is not greater than a[i-1] we cannot switch a[i] with b[i] as b[i] is always less than a[i]. Now let us take a[i+1] is not greater than a[i] so we can switch a[i] with b[i] as a[i] > b[i], but as a[i] > b[i] and a[i+1]> b[i+1] and a[i]>a[i+1] so a[i] can never be less than b[i+1] so there is no possible switch. We can similarly prove for b[i].
So it is proved that there might be more possible combinations for arranging the array when the output is YES but there is no possible way of arranging the array according to constraints when output is NO.
Below is the implementation of the above approach:
// C++ implementation of above approach #include <iostream> using namespace std;
// Function to check whether both the array can be // sorted in (strictly increasing ) ascending order bool IsSorted( int A[], int B[], int n)
{ // Traverse through the array
// and find out the min and max
// variable at each position
// make one array of min variables
// and another of maximum variable
for ( int i = 0; i < n; i++) {
int x, y;
// Maximum and minimum variable
x = max(A[i], B[i]);
y = min(A[i], B[i]);
// Assign min value to
// B[i] and max value to A[i]
A[i] = x;
B[i] = y;
}
// Now check whether the array is
// sorted or not
for ( int i = 1; i < n; i++) {
if (A[i] <= A[i - 1] || B[i] <= B[i - 1])
return false ;
}
return true ;
} // Driver code int main()
{ int A[] = { 1, 4, 3, 5, 7 };
int B[] = { 2, 2, 5, 8, 9 };
int n = sizeof (A) / sizeof ( int );
cout << (IsSorted(A, B, n) ? "True" : "False" );
return 0;
} |
// Java implementation of above approach import java.io.*;
class GFG
{ // Function to check whether both the array can be // sorted in (strictly increasing ) ascending order static boolean IsSorted( int []A, int []B, int n)
{ // Traverse through the array
// and find out the min and max
// variable at each position
// make one array of min variables
// and another of maximum variable
for ( int i = 0 ; i < n; i++)
{
int x, y;
// Maximum and minimum variable
x = Math.max(A[i], B[i]);
y = Math.min(A[i], B[i]);
// Assign min value to
// B[i] and max value to A[i]
A[i] = x;
B[i] = y;
}
// Now check whether the array is
// sorted or not
for ( int i = 1 ; i < n; i++)
{
if (A[i] <= A[i - 1 ] || B[i] <= B[i - 1 ])
return false ;
}
return true ;
} // Driver code public static void main (String[] args)
{ int []A = { 1 , 4 , 3 , 5 , 7 };
int []B = { 2 , 2 , 5 , 8 , 9 };
int n = A.length;
if (IsSorted(A, B, n) == true )
{
System.out.println( "True" );
}
else
{
System.out.println( "False" );
}
} } // This code is contributed by ajit |
# Python3 implementation of above approach # Function to check whether both the array can be # sorted in (strictly increasing ) ascending order def IsSorted(A, B, n) :
# Traverse through the array
# and find out the min and max
# variable at each position
# make one array of min variables
# and another of maximum variable
for i in range (n) :
# Maximum and minimum variable
x = max (A[i], B[i]);
y = min (A[i], B[i]);
# Assign min value to
# B[i] and max value to A[i]
A[i] = x;
B[i] = y;
# Now check whether the array is
# sorted or not
for i in range ( 1 , n) :
if (A[i] < = A[i - 1 ] or B[i] < = B[i - 1 ]) :
return False ;
return True ;
# Driver code if __name__ = = "__main__" :
A = [ 1 , 4 , 3 , 5 , 7 ];
B = [ 2 , 2 , 5 , 8 , 9 ];
n = len (A);
if (IsSorted(A, B, n)) :
print ( True )
else :
print ( False )
# This code is contributed by AnkitRai01 |
// C# implementation of above approach using System;
class GFG
{ // Function to check whether both the array can be // sorted in (strictly increasing ) ascending order static bool IsSorted( int []A, int []B, int n)
{ // Traverse through the array
// and find out the min and max
// variable at each position
// make one array of min variables
// and another of maximum variable
for ( int i = 0; i < n; i++) {
int x, y;
// Maximum and minimum variable
x = Math.Max(A[i], B[i]);
y = Math.Min(A[i], B[i]);
// Assign min value to
// B[i] and max value to A[i]
A[i] = x;
B[i] = y;
}
// Now check whether the array is
// sorted or not
for ( int i = 1; i < n; i++) {
if (A[i] <= A[i - 1] || B[i] <= B[i - 1])
return false ;
}
return true ;
} // Driver code public static void Main()
{ int []A = { 1, 4, 3, 5, 7 };
int []B = { 2, 2, 5, 8, 9 };
int n = A.Length;
if (IsSorted(A, B, n) == true )
{
Console.Write( "True" );
}
else
{
Console.Write( "False" );
}
} } // This code is contributed // by Akanksha Rai |
<script> // Javascript implementation of the approach // Function to check whether both the // array can be sorted in (strictly // increasing) ascending order function IsSorted(A, B, n)
{ // Traverse through the array
// and find out the min and max
// variable at each position
// make one array of min variables
// and another of maximum variable
for ( var i = 0; i < n; i++)
{
var x, y;
// Maximum and minimum variable
x = Math.max(A[i], B[i]);
y = Math.min(A[i], B[i]);
// Assign min value to
// B[i] and max value to A[i]
A[i] = x;
B[i] = y;
}
// Now check whether the array is
// sorted or not
for ( var i = 1; i < n; i++)
{
if (A[i] <= A[i - 1] ||
B[i] <= B[i - 1])
return false ;
}
return true ;
} // Driver Code var A = [ 1, 4, 3, 5, 7 ];
var B = [ 2, 2, 5, 8, 9 ];
var n = A.length;
document.write(IsSorted(A, B, n) ? "True" : "False" );
// This code is contributed by SoumikMondal </script> |
True
Time Complexity: O(N)
Space Complexity: O(1) as no extra space has been used.
Approach(Brute force approach): Try all possible swaps of elements between the two arrays and check if the resulting arrays are sorted
- The approach to solve this problem is to iterate over the arrays and check if swapping A[i] and B[i] at the same index results in sorted arrays.
- f A[i] and B[i] are already sorted, continue. If swapping A[i] and B[i] results in sorted arrays, continue.
- If none of the swaps result in sorted arrays, return false. If all swaps result in sorted arrays,
- return true.
#include <iostream> using namespace std;
// Function to check if two arrays can be sorted by swapping elements bool canSortBySwapping( int A[], int B[], int n)
{ // Iterate over the array
for ( int i = 1; i < n; i++) {
// If A[i] and B[i] are already sorted, continue
if (A[i] > A[i - 1] && B[i] > B[i - 1]) {
continue ;
}
// If swapping A[i] and B[i] results in sorted arrays, continue
if (A[i] > B[i - 1] && B[i] > A[i - 1]) {
continue ;
}
// If swapping A[i] and B[i] does not result in sorted arrays, return false
return false ;
}
// If all swaps result in sorted arrays, return true
return true ;
} // Driver code int main()
{ int A[] = { 1, 4, 3, 5, 7 };
int B[] = { 2, 2, 5, 8, 9 };
int n = sizeof (A) / sizeof ( int );
if (canSortBySwapping(A, B, n)) {
cout << "True" << endl;
} else {
cout << "False" << endl;
}
return 0;
} |
import java.util.Arrays;
public class GFG {
// Function to check if two arrays can be sorted by swapping elements
public static boolean canSortBySwapping( int [] A, int [] B, int n) {
// Iterate over the array
for ( int i = 1 ; i < n; i++) {
// If A[i] and B[i] are already sorted, continue
if (A[i] > A[i - 1 ] && B[i] > B[i - 1 ]) {
continue ;
}
// If swapping A[i] and B[i] results in sorted arrays, continue
if (A[i] > B[i - 1 ] && B[i] > A[i - 1 ]) {
continue ;
}
// If swapping A[i] and B[i] does not result in sorted arrays, return false
return false ;
}
// If all swaps result in sorted arrays, return true
return true ;
}
// Driver code
public static void main(String[] args) {
int [] A = { 1 , 4 , 3 , 5 , 7 };
int [] B = { 2 , 2 , 5 , 8 , 9 };
int n = A.length;
if (canSortBySwapping(A, B, n)) {
System.out.println( "True" );
} else {
System.out.println( "False" );
}
}
} |
# Function to check if two arrays can be sorted by swapping elements def canSortBySwapping(A, B, n):
# Iterate over the array
for i in range ( 1 , n):
#If A[i] and B[i] are already sorted, continue
if A[i] > A[i - 1 ] and B[i] > B[i - 1 ]:
continue
# If swapping A[i] and B[i] results in sorted arrays, continue
if A[i] > B[i - 1 ] and B[i] > A[i - 1 ]:
continue
#If swapping A[i] and B[i] does not result in sorted arrays, return false
return False
#If all swaps result in sorted arrays, return true
return True
A = [ 1 , 4 , 3 , 5 , 7 ]
B = [ 2 , 2 , 5 , 8 , 9 ]
n = len (A)
if canSortBySwapping(A, B, n):
print ( "True" )
else :
print ( "False" )
|
using System;
class GFG
{ // Function to check if two arrays can be sorted by swapping elements
static bool CanSortBySwapping( int [] A, int [] B, int n)
{
// Iterate over the arrays
for ( int i = 1; i < n; i++)
{
// If A[i] and B[i] are already sorted, continue
if (A[i] > A[i - 1] && B[i] > B[i - 1])
{
continue ;
}
// If swapping A[i] and B[i] results in sorted arrays, continue
if (A[i] > B[i - 1] && B[i] > A[i - 1])
{
continue ;
}
// If swapping A[i] and B[i] does not result in sorted arrays, return false
return false ;
}
// If all swaps result in sorted arrays, return true
return true ;
}
static void Main()
{
int [] A = { 1, 4, 3, 5, 7 };
int [] B = { 2, 2, 5, 8, 9 };
int n = A.Length;
if (CanSortBySwapping(A, B, n))
{
Console.WriteLine( "True" );
}
else
{
Console.WriteLine( "False" );
}
}
} |
// Function to check if two arrays can be sorted by swapping elements function canSortBySwapping(A, B, n) {
// Iterate over the array for (let i = 1; i < n; i++) {
// If A[i] and B[i] are already sorted, continue
if (A[i] > A[i - 1] && B[i] > B[i - 1]) {
continue ;
}
// If swapping A[i] and B[i] results in sorted arrays, continue
if (A[i] > B[i - 1] && B[i] > A[i - 1]) {
continue ;
}
// If swapping A[i] and B[i] does not result in sorted arrays,
// return false
return false ;
}
// If all swaps result in sorted arrays, return true
return true ;
} const A = [1, 4, 3, 5, 7]; const B = [2, 2, 5, 8, 9]; const n = A.length; if (canSortBySwapping(A, B, n)) {
console.log( "True" );
} else {
console.log( "False" );
} |
True
Time Complexity: O(n), where n is the size of the arrays. This is because the code iterates over the arrays only once
Space Complexity: O(1), which is constant as the code uses only a few variables to keep track of the maximum and minimum values. Therefore, the space used does not depend on the size of the input.