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Check whether we can sort two arrays by swapping A[i] and B[i]

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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++

// 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

// 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

# 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#

// 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

                    

Javascript

<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>

                    

Output
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

  1. 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. 
  2. f A[i] and B[i] are already sorted, continue. If swapping A[i] and B[i] results in sorted arrays, continue.
  3.  If none of the swaps result in sorted arrays, return false. If all swaps result in sorted arrays, 
  4.  return true.

C++

#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;
}

                    

Java

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");
        }
    }
}

                    

Python3

# 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")

                    

C#

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");
        }
    }
}

                    

Javascript

// 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");
}

                    

Output
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.



Last Updated : 23 Aug, 2023
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