Check if two arrays can be made equal by swapping pairs of one of the arrays
Last Updated :
26 Oct, 2023
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 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 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# 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.
<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>
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
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
# 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
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
}
}
}
// Define the main function
function main() {
// Define two arrays
let arr1 = [0, 0, 1, 1];
let arr2 = [1, 1, 0, 0];
// Calculate the size of the arrays
let n = arr1.length;
// Initialize counters
let zero_count = 0;
let one_count = 0;
let mismatch_count = 0;
// Loop through the arrays
for (let 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
}
}
// Check conditions
if (zero_count === one_count && mismatch_count % 2 === 0) {
console.log("Yes"); // If conditions are met, print "Yes"
} else {
console.log("No"); // Otherwise, print "No"
}
}
// Call the main function
main();
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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