给定两个二进制阵列ARR1 []和ARR2 []相同的尺寸，该任务是使两个阵列通过交换对ARR1的等于[]仅当ARR1 [I] = 0和ARR1 [J] = 1(0≤ i
例子：
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
方法：请按照以下步骤解决问题：
- 初始化两个变量，例如count和flag (= true )。
- 遍历数组，并对每个数组元素执行以下操作：
- 如果arr1 [i]！= arr2 [i]：
- 如果arr1 [i] == 0 ，则将计数递增1 。
- 否则，将count减1 ，如果count <0 ，则更新标志= false 。
- 如果arr1 [i]！= arr2 [i]：
- 如果标志等于true ，则打印“是” 。否则，打印“否” 。
下面是上述方法的实现：
C++
// C++ program for above approach
#include
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 AnkThonPython3
# 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.输出：
Yes时间复杂度： O(N)
辅助空间： O(1)