Check if two arrays can be made equal by swapping pairs of one of the arrays

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Given two binary arrays arr1[] and arr2[] of the same size, the task is to make both the arrays equal by swapping pairs of arr1[ ] only if arr1[i] = 0 and arr1[j] = 1 (0 ? i < j < N)). If it is possible to make both the arrays equal, print “Yes”. Otherwise, print “No”.

Examples:

Input: arr1[] = {0, 0, 1, 1}, arr2[] = {1, 1, 0, 0}
Output: Yes
Explanation:
Swap arr1[1] and arr1[3], it becomes arr1[] = {0, 1, 1, 0}.
Swap arr1[0] and arr1[2], it becomes arr1[] = {1, 1, 0, 0}.

Input: arr1[] = {1, 0, 1, 0, 1}, arr2[] = {0, 1, 0, 0, 1}
Output: No

Approach: Follow the steps below to solve the problem:

Initialize two variable, say count and flag (= true).
Traverse the array and for every array element, perform the following operations:
- If arr1[i] != arr2[i]:
  - If arr1[i] == 0, increment count by 1.
  - Otherwise, decrement count by 1 and if count < 0, update flag = false.
If flag is equal to true, print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if two arrays
// can be made equal or not by swapping
// pairs of only one of the arrays
void checkArrays(int arr1[], int arr2[], int N)
{
    // Stores elements required
    // to be replaced
    int count = 0;
 
    // To check if the arrays
    // can be made equal or not
    bool flag = true;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If array elements are not equal
        if (arr1[i] != arr2[i]) {
 
            if (arr1[i] == 0)
 
                // Increment count by 1
                count++;
            else {
 
                // Decrement count by 1
                count--;
                if (count < 0) {
                    flag = 0;
                    break;
                }
            }
        }
    }
 
    // If flag is true and count is 0,
    // print "Yes". Otherwise "No"
    if (flag && count == 0)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
 
// Driver Code
int main()
{
    // Given arrays
    int arr1[] = { 0, 0, 1, 1 };
    int arr2[] = { 1, 1, 0, 0 };
 
    // Size of the array
    int N = sizeof(arr1) / sizeof(arr1[0]);
    checkArrays(arr1, arr2, N);
 
    return 0;
}

Java

// Java program for above approach
public class GFG 
{
 
  // Function to check if two arrays
  // can be made equal or not by swapping
  // pairs of only one of the arrays
  static void checkArrays(int arr1[], int arr2[], int N)
  {
 
    // Stores elements required
    // to be replaced
    int count = 0;
 
    // To check if the arrays
    // can be made equal or not
    boolean flag = true;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // If array elements are not equal
      if (arr1[i] != arr2[i])
      {
        if (arr1[i] == 0)
 
          // Increment count by 1
          count++;
        else
        {
 
          // Decrement count by 1
          count--;
          if (count < 0)
          {
            flag = false;
            break;
          }
        }
      }
    }
 
    // If flag is true and count is 0,
    // print "Yes". Otherwise "No"
    if ((flag && (count == 0)) == true)
      System.out.println("Yes");
    else
      System.out.println("No");
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    // Given arrays
    int arr1[] = { 0, 0, 1, 1 };
    int arr2[] = { 1, 1, 0, 0 };
 
    // Size of the array
    int N = arr1.length;   
    checkArrays(arr1, arr2, N);
  }
}
 
// This code is contributed by AnkThon

Python3

# Python3 program for above approach
 
# Function to check if two arrays
# can be made equal or not by swapping
# pairs of only one of the arrays
def checkArrays(arr1, arr2, N):
   
    # Stores elements required
    # to be replaced
    count = 0
 
    # To check if the arrays
    # can be made equal or not
    flag = True
 
    # Traverse the array
    for i in range(N):
 
        # If array elements are not equal
        if (arr1[i] != arr2[i]):
 
            if (arr1[i] == 0):
 
                # Increment count by 1
                count += 1
            else:
 
                # Decrement count by 1
                count -= 1
                if (count < 0):
                    flag = 0
                    break
 
    # If flag is true and count is 0,
    # pr"Yes". Otherwise "No"
    if (flag and count == 0):
        print("Yes")
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
     
    # Given arrays
    arr1 = [0, 0, 1, 1]
    arr2 = [1, 1, 0, 0]
 
    # Size of the array
    N = len(arr1)
 
    checkArrays(arr1, arr2, N)
 
    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if two arrays
  // can be made equal or not by swapping
  // pairs of only one of the arrays
  static void checkArrays(int[] arr1, int[] arr2, int N)
  {
 
    // Stores elements required
    // to be replaced
    int count = 0;
 
    // To check if the arrays
    // can be made equal or not
    bool flag = true;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // If array elements are not equal
      if (arr1[i] != arr2[i])
      {
        if (arr1[i] == 0)
 
          // Increment count by 1
          count++;
        else
        {
 
          // Decrement count by 1
          count--;
          if (count < 0)
          {
            flag = false;
            break;
          }
        }
      }
    }
 
    // If flag is true and count is 0,
    // print "Yes". Otherwise "No"
    if ((flag && (count == 0)) == true)
      Console.WriteLine("Yes");
    else
      Console.WriteLine("No");
  }
 
// Driver Code
static public void Main()
{
    // Given arrays
    int[] arr1 = { 0, 0, 1, 1 };
    int[] arr2 = { 1, 1, 0, 0 };
 
    // Size of the array
    int N = arr1.Length;   
    checkArrays(arr1, arr2, N);
}
}
 
// This code is contributed by susmitakundugoaldanga.

Javascript

<script>
// Java script program for above approach
 
 
// Function to check if two arrays
// can be made equal or not by swapping
// pairs of only one of the arrays
function checkArrays(arr1,arr2,N)
{
 
    // Stores elements required
    // to be replaced
    let count = 0;
 
    // To check if the arrays
    // can be made equal or not
    let flag = true;
 
    // Traverse the array
    for (let i = 0; i < N; i++) {
 
    // If array elements are not equal
    if (arr1[i] != arr2[i])
    {
        if (arr1[i] == 0)
 
        // Increment count by 1
        count++;
        else
        {
 
        // Decrement count by 1
        count--;
        if (count < 0)
        {
            flag = false;
            break;
        }
        }
    }
    }
 
    // If flag is true and count is 0,
    // print "Yes". Otherwise "No"
    if ((flag && (count == 0)) == true)
    document.write("Yes");
    else
    document.write("No");
}
 
// Driver Code
 
    // Given arrays
    let arr1 = [ 0, 0, 1, 1 ];
    let arr2 = [ 1, 1, 0, 0 ];
 
    // Size of the array
    let N = arr1.length;
    checkArrays(arr1, arr2, N);
 
// This code is contributed by Gottumukkala Sravan Kumar (171fa07058)
</script>

Output

Yes

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

Another Approach:

Define two binary arrays arr1[] and arr2[] of the same size.
Calculate the size of the arrays using the sizeof() operator and store it in the variable n.
Initialize three integer variables zero_count, one_count, and mismatch_count to zero.
Use a for loop to iterate through the arrays from 0 to n-1.
Inside the for loop, check if the current element of arr1 is zero. If yes, increment the zero_count variable. Otherwise, increment the one_count variable.
Also inside the for loop, check if the current element of arr1 is not equal to the current element of arr2. If yes, increment the mismatch_count variable.
After the for loop, check if the zero_count and one_count variables are equal and the mismatch_count variable is even. If yes, print “Yes”. Otherwise, print “No”.
End the program.

Below is the implementation of the above approach:

C++

#include <iostream>
using namespace std;
 
int main() {
    int arr1[] = {0, 0, 1, 1};
    int arr2[] = {1, 1, 0, 0};
    int n = sizeof(arr1) / sizeof(arr1[0]); // calculate size of arrays
 
    int zero_count = 0, one_count = 0, mismatch_count = 0;
    for (int i = 0; i < n; i++) {
        if (arr1[i] == 0) {
            zero_count++; // count the number of zeros in arr1
        } else {
            one_count++; // count the number of ones in arr1
        }
        if (arr1[i] != arr2[i]) {
            mismatch_count++; // count the number of mismatches between arr1 and arr2
        }
    }
 
    if (zero_count == one_count && mismatch_count % 2 == 0) {
        cout << "Yes"; // if the number of zeros and ones in arr1 are equal and the number of mismatches is even, we can make both arrays equal by swapping pairs of arr1
    } else {
        cout << "No"; // otherwise, we cannot make both arrays equal by swapping pairs of arr1
    }
    return 0;
}
// This code is contributed by rudra1807raj

Java

public class Main {
    public static void main(String[] args) {
        int[] arr1 = {0, 0, 1, 1};
        int[] arr2 = {1, 1, 0, 0};
        int n = arr1.length; // Calculate the size of arrays
 
        int zeroCount = 0, oneCount = 0, mismatchCount = 0;
        for (int i = 0; i < n; i++) {
            if (arr1[i] == 0) {
                zeroCount++; // Count the number of zeros in arr1
            } else {
                oneCount++; // Count the number of ones in arr1
            }
            if (arr1[i] != arr2[i]) {
                mismatchCount++; // Count the number of mismatches between arr1 and arr2
            }
        }
 
        if (zeroCount == oneCount && mismatchCount % 2 == 0) {
            System.out.println("Yes"); // If the number of zeros and ones in arr1 are equal and the number of mismatches is even, we can make both arrays equal by swapping pairs of arr1
        } else {
            System.out.println("No"); // Otherwise, we cannot make both arrays equal by swapping pairs of arr1
        }
    }
}
//This code is Contributed by chinmaya121221

Python3

# Given arrays
arr1 = [0, 0, 1, 1]
arr2 = [1, 1, 0, 0]
n = len(arr1)  # calculate size of arrays
 
zero_count = 0
one_count = 0
mismatch_count = 0
 
# Loop through each element of the arrays
for i in range(n):
    # Count the number of zeros in arr1
    if arr1[i] == 0:
        zero_count += 1
    else:
        one_count += 1  # Count the number of ones in arr1
 
    # Count the number of mismatches between arr1 and arr2
    if arr1[i] != arr2[i]:
        mismatch_count += 1
 
# Check the conditions to determine if it's possible to make both arrays equal
if zero_count == one_count and mismatch_count % 2 == 0:
    print("Yes")  # If conditions are met, we can make both arrays equal by swapping pairs of arr1
else:
    print("No")  # Otherwise, we cannot make both arrays equal by swapping pairs of arr1

C#

using System;
 
class Program
{
    static void Main()
    {
        int[] arr1 = { 0, 0, 1, 1 };
        int[] arr2 = { 1, 1, 0, 0 };
        int n = arr1.Length; // Calculate the size of arrays
 
        int zeroCount = 0, oneCount = 0, mismatchCount = 0;
 
        for (int i = 0; i < n; i++)
        {
            if (arr1[i] == 0)
            {
                zeroCount++; // Count the number of zeros in arr1
            }
            else
            {
                oneCount++; // Count the number of ones in arr1
            }
 
            if (arr1[i] != arr2[i])
            {
                mismatchCount++; // Count the number of mismatches between arr1 and arr2
            }
        }
 
        if (zeroCount == oneCount && mismatchCount % 2 == 0)
        {
            Console.WriteLine("Yes"); // If the number of zeros and ones 
                                      // in arr1 are equal and the number of mismatches is
                                      // even, we can make both arrays equal by swapping
                                      // pairs of arr1
        }
        else
        {
            Console.WriteLine("No"); // Otherwise, we cannot make both arrays 
                                     // equal by swapping pairs of arr1
        }
    }
}

Output:

Yes

Time Complexity: The time complexity of the given code is O(n), where n is the size of the arrays. This is because the code uses a single loop to iterate through both arrays, and each operation inside the loop takes constant time.

Auxiliary Space: The space complexity of the code is O(1), as it uses only a few variables to store the counts and does not allocate any additional memory based on the input size.

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