给定大小N和大小为N的二进制数组B []的[]数组A中，任务是检查，如果数组A []可以通过交换对被转换成一个排序后的数组(A [I]，A [J ])如果B [i]不等于B [j] 。如果可以对数组A []进行排序，则打印“是”。否则，打印“ No ”。
例子：
Input: A[] = {3, 1, 2}, B[] = {0, 1, 1}
Output: Yes
Explanation:
Swap element at position 1 and position 2 of A[] since B[1]!=B[2]. So, A[] = {1, 3, 2}, B[] = {1, 0, 1}
Now, swap element at position 2 and position 3 of A[] since B[2]!=B[3]. So, A[] = {1, 2, 3}. Hence, it is sorted.
Input: A[] = {5, 15, 4}, B[] = {0, 0, 0}
Output: No
方法：可以根据以下观察结果解决问题：
If at least two elements of the array, B[] are different, then it is possible to swap any two elements of the array A[].
请按照以下步骤解决问题：
- 检查给定数组A []是否已按升序排序。如果发现是真的，则打印“是” 。
- 否则，计算数组B []中存在的1 s和0 s的数量。
- 如果数组B []至少包含一个0和一个1 ，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 array, A[] can be converted
// into sorted array by swapping (A[i], A[j]) if B[i]
// not equal to B[j]
bool checkifSorted(int A[], int B[], int N)
{
// Stores if array A[] is sorted
// in descending order or not
bool flag = false;
// Traverse the array A[]
for (int i = 0; i < N - 1; i++) {
// If A[i] is greater than A[i + 1]
if (A[i] > A[i + 1]) {
// Update flag
flag = true;
break;
}
}
// If array is sorted
// in ascending order
if (!flag) {
return true;
}
// count = 2: Check if 0s and 1s
// both present in the B[]
int count = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// If current element is 0
if (B[i] == 0) {
// Update count
count++;
break;
}
}
// Traverse the array B[]
for (int i = 0; i < N; i++) {
// If current element is 1
if (B[i] == 1) {
count++;
break;
}
}
// If both 0s and 1s are present
// in the array
if (count == 2) {
return true;
}
return false;
}
// Driver Code
int main()
{
// Input array A[]
int A[] = { 3, 1, 2 };
// Input array B[]
int B[] = { 0, 1, 1 };
int N = sizeof(A) / sizeof(A[0]);
// Function call
bool check = checkifSorted(A, B, N);
// If true,print YES
if (check) {
cout << "YES" < Java
// Java program of the above approach
import java.io.*;
class GFG {
// Function to check if array, A[] can be converted
// into sorted array by swapping (A[i], A[j]) if B[i]
// not equal to B[j]
static boolean checkifSorted(int A[], int B[], int N)
{
// Stores if array A[] is sorted
// in descending order or not
boolean flag = false;
// Traverse the array A[]
for (int i = 0; i < N - 1; i++) {
// If A[i] is greater than A[i + 1]
if (A[i] > A[i + 1]) {
// Update flag
flag = true;
break;
}
}
// If array is sorted
// in ascending order
if (!flag) {
return true;
}
// count = 2: Check if 0s and 1s
// both present in the B[]
int count = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// If current element is 0
if (B[i] == 0) {
// Update count
count++;
break;
}
}
// Traverse the array B[]
for (int i = 0; i < N; i++) {
// If current element is 1
if (B[i] == 1) {
count++;
break;
}
}
// If both 0s and 1s are present
// in the array
if (count == 2) {
return true;
}
return false;
}
// Driver Code
public static void main(String[] args)
{
// Input array A[]
int A[] = { 3, 1, 2 };
// Input array B[]
int B[] = { 0, 1, 1 };
int N = A.length;
// Function call
boolean check = checkifSorted(A, B, N);
// If true,print YES
if (check) {
System.out.println("YES");
}
// Else print NO
else {
System.out.println("NO");
}
}
}Python3
# Python program of the above approach
# Function to check if array, A[] can be converted
# into sorted array by swapping (A[i], A[j]) if B[i]
# not equal to B[j]
def checkifSorted(A, B, N):
# Stores if array A[] is sorted
# in descending order or not
flag = False
# Traverse the array A[]
for i in range( N - 1):
# If A[i] is greater than A[i + 1]
if (A[i] > A[i + 1]):
# Update flag
flag = True
break
# If array is sorted
# in ascending order
if (not flag):
return True
# count = 2: Check if 0s and 1s
# both present in the B[]
count = 0
# Traverse the array
for i in range(N):
# If current element is 0
if (B[i] == 0):
# Update count
count += 1
break
# Traverse the array B[]
for i in range(N):
# If current element is 1
if B[i]:
count += 1
break
# If both 0s and 1s are present
# in the array
if (count == 2):
return True
return False
# Driver Code
# Input array A[]
A = [ 3, 1, 2 ]
# Input array B[]
B = [ 0, 1, 1 ]
N = len(A)
# Function call
check = checkifSorted(A, B, N)
# If true,print YES
if (check):
print("YES")
# Else print NO
else:
print("NO")
# This code is contributed by rohitsingh07052C#
// C# program of the above approach
using System;
public class GFG
{
// Function to check if array, A[] can be converted
// into sorted array by swapping (A[i], A[j]) if B[i]
// not equal to B[j]
static bool checkifSorted(int []A, int []B, int N)
{
// Stores if array A[] is sorted
// in descending order or not
bool flag = false;
// Traverse the array A[]
for (int i = 0; i < N - 1; i++) {
// If A[i] is greater than A[i + 1]
if (A[i] > A[i + 1]) {
// Update flag
flag = true;
break;
}
}
// If array is sorted
// in ascending order
if (!flag) {
return true;
}
// count = 2: Check if 0s and 1s
// both present in the B[]
int count = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// If current element is 0
if (B[i] == 0) {
// Update count
count++;
break;
}
}
// Traverse the array B[]
for (int i = 0; i < N; i++)
{
// If current element is 1
if (B[i] == 1)
{
count++;
break;
}
}
// If both 0s and 1s are present
// in the array
if (count == 2)
{
return true;
}
return false;
}
// Driver Code
public static void Main(string[] args)
{
// Input array A[]
int []A = { 3, 1, 2 };
// Input array B[]
int []B = { 0, 1, 1 };
int N = A.Length;
// Function call
bool check = checkifSorted(A, B, N);
// If true,print YES
if (check) {
Console.WriteLine("YES");
}
// Else print NO
else {
Console.WriteLine("NO");
}
}
}
// This code is contributed by AnkThon输出：
YES时间复杂度： O(N)
辅助空间： O(1)