给定两个阵列ARR1 []和ARR2 [大小为N]，由二进制整数的，任务是，以检查是否ARR1 []可以通过交换任何对的数组元素的(ARR1转换为ARR2 [] [I]中，ARR1 [ j]) ，使得i
例子：
Input: arr1[] = {0, 1, 1, 0}, arr2[] = {0, 0 1, 1}
Output: Yes
Explanation:
The array arr1[] can be made equal to arr2[] by swapping arr1[1] and arr1[3].
Input: arr1[] = {1, 0, 1}, arr2[] = {0, 1, 0}
Output: No
方法：解决此问题的想法是基于以下观察结果：
- 该操作不会更改数组arr1 []中1和0的数量的频率，因此，如果数组之间的0和1的数量不同，则它们永远不能与上述操作相等。
- 如果arr2 []的某个前缀比相同长度的arr1 []的前缀包含更多的1 ，则不可能使arr1 []和arr2 []相等，因为1只能右移。
- 否则，在所有其他情况下，可以使数组相等。
请按照以下步骤解决问题：
- 初始化一个变量，比如说count为0 ，以存储arr1 []和arr2 []的前缀和的差。
- 数1 和在两个阵列和校验0如果在ARR1 1s和0s []的数目不等于1和0的数量ARR2 []，然后打印“否”。
- 使用变量i遍历[1，N – 1]范围，并执行以下操作：
- 将值(arr1 [i] – arr2 [i])添加到变量count 。
- 如果count的值小于0 ，则打印“ No”，否则继续对下一对元素进行操作。
- 完成上述步骤后，如果在任何步骤中计数都不为负，则打印“是” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if arr1[] can be
// converted to arr2[] by swapping pair
// (i, j) such that i < j and arr[i] is
// 1 and arr[j] is 0
void canMakeEqual(int arr1[], int arr2[], int N)
{
// Stores the differences of prefix
// sum of arr1 and arr2
int count = 0;
// Stores the count of 1 and
// zero of arr1
int arr1_one = 0, arr1_zero = 0;
// Stores the count of 1 and
// zero of arr2
int arr2_one = 0, arr2_zero = 0;
// Iterate in the range [0, N - 1]
for (int i = 0; i < N; i++) {
// If arr1[i] is 1, then
// increment arr1_one by one
if (arr1[i] == 1) {
arr1_one++;
}
// Otherwise increment
// arr1_zero by one
else if (arr1[i] == 0) {
arr1_zero++;
}
// If arr2[i] is 1, then
// increment arr2_one by one
if (arr2[i] == 1) {
arr2_one++;
}
// Otherwise increment
// arr2_zero by one
else if (arr2[i] == 0) {
arr2_zero++;
}
}
// Check if number of 1s and 0s
// of arr1 is equal to number of
// 1s and 0s of arr2 respectievly
if (arr1_one != arr2_one || arr1_zero != arr2_zero) {
cout << "No";
return;
}
// Iterate over the range [0, N-1]
for (int i = 0; i < N; i++) {
// Increment count by differences
// arr1[i] and arr2[i]
count = count + (arr1[i] - arr2[i]);
// Check if number of 1's in
// arr2 are more than arr1 and
// then print "No"
if (count < 0) {
cout << "No";
return;
}
}
// Finally, print "Yes"
cout << "Yes";
}
// Driver Code
int main()
{
// Given input arrays
int arr1[] = { 0, 1, 1, 0 };
int arr2[] = { 0, 0, 1, 1 };
// Size of the array
int N = sizeof(arr1) / sizeof(arr1[0]);
// Function Call
canMakeEqual(arr1, arr2, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to check if arr1[] can be
// converted to arr2[] by swapping pair
// (i, j) such that i < j and arr[i] is
// 1 and arr[j] is 0
static void canMakeEqual(int []arr1, int []arr2, int N)
{
// Stores the differences of prefix
// sum of arr1 and arr2
int count = 0;
// Stores the count of 1 and
// zero of arr1
int arr1_one = 0, arr1_zero = 0;
// Stores the count of 1 and
// zero of arr2
int arr2_one = 0, arr2_zero = 0;
// Iterate in the range [0, N - 1]
for (int i = 0; i < N; i++) {
// If arr1[i] is 1, then
// increment arr1_one by one
if (arr1[i] == 1) {
arr1_one++;
}
// Otherwise increment
// arr1_zero by one
else if (arr1[i] == 0) {
arr1_zero++;
}
// If arr2[i] is 1, then
// increment arr2_one by one
if (arr2[i] == 1) {
arr2_one++;
}
// Otherwise increment
// arr2_zero by one
else if (arr2[i] == 0) {
arr2_zero++;
}
}
// Check if number of 1s and 0s
// of arr1 is equal to number of
// 1s and 0s of arr2 respectievly
if (arr1_one != arr2_one || arr1_zero != arr2_zero) {
System.out.print("No");
return;
}
// Iterate over the range [0, N-1]
for (int i = 0; i < N; i++) {
// Increment count by differences
// arr1[i] and arr2[i]
count = count + (arr1[i] - arr2[i]);
// Check if number of 1's in
// arr2 are more than arr1 and
// then print "No"
if (count < 0) {
System.out.print("No");
return;
}
}
// Finally, print "Yes"
System.out.print("Yes");
}
// Driver Code
public static void main(String[] args)
{
// Given input arrays
int []arr1 = { 0, 1, 1, 0 };
int []arr2 = { 0, 0, 1, 1 };
// Size of the array
int N = arr1.length;
// Function Call
canMakeEqual(arr1, arr2, N);
}
}
// This code is contributed by code_hunt.Python3
# Python 3 program for the above approach
# Function to check if arr1[] can be
# converted to arr2[] by swapping pair
# (i, j) such that i < j and arr[i] is
# 1 and arr[j] is 0
def canMakeEqual(arr1, arr2, N):
# Stores the differences of prefix
# sum of arr1 and arr2
count = 0
# Stores the count of 1 and
# zero of arr1
arr1_one = 0
arr1_zero = 0
# Stores the count of 1 and
# zero of arr2
arr2_one = 0
arr2_zero = 0
# Iterate in the range [0, N - 1]
for i in range(N):
# If arr1[i] is 1, then
# increment arr1_one by one
if (arr1[i] == 1):
arr1_one += 1
# Otherwise increment
# arr1_zero by one
elif(arr1[i] == 0):
arr1_zero += 1
# If arr2[i] is 1, then
# increment arr2_one by one
if (arr2[i] == 1):
arr2_one += 1
# Otherwise increment
# arr2_zero by one
elif (arr2[i] == 0):
arr2_zero += 1
# Check if number of 1s and 0s
# of arr1 is equal to number of
# 1s and 0s of arr2 respectievly
if (arr1_one != arr2_one or arr1_zero != arr2_zero):
print("No")
return
# Iterate over the range [0, N-1]
for i in range(N):
# Increment count by differences
# arr1[i] and arr2[i]
count = count + (arr1[i] - arr2[i])
# Check if number of 1's in
# arr2 are more than arr1 and
# then print "No"
if (count < 0):
print("No")
return
# Finally, print "Yes"
print("Yes")
# Driver Code
if __name__ == '__main__':
# Given input a
arr1 = [0, 1, 1, 0]
arr2 = [0, 0, 1, 1]
# Size of the array
N = len(arr1)
# Function Call
canMakeEqual(arr1, arr2, N)
# This code is contributed by ipg2016107.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check if arr1[] can be
// converted to arr2[] by swapping pair
// (i, j) such that i < j and arr[i] is
// 1 and arr[j] is 0
static void canMakeEqual(int []arr1, int []arr2, int N)
{
// Stores the differences of prefix
// sum of arr1 and arr2
int count = 0;
// Stores the count of 1 and
// zero of arr1
int arr1_one = 0, arr1_zero = 0;
// Stores the count of 1 and
// zero of arr2
int arr2_one = 0, arr2_zero = 0;
// Iterate in the range [0, N - 1]
for (int i = 0; i < N; i++) {
// If arr1[i] is 1, then
// increment arr1_one by one
if (arr1[i] == 1) {
arr1_one++;
}
// Otherwise increment
// arr1_zero by one
else if (arr1[i] == 0) {
arr1_zero++;
}
// If arr2[i] is 1, then
// increment arr2_one by one
if (arr2[i] == 1) {
arr2_one++;
}
// Otherwise increment
// arr2_zero by one
else if (arr2[i] == 0) {
arr2_zero++;
}
}
// Check if number of 1s and 0s
// of arr1 is equal to number of
// 1s and 0s of arr2 respectievly
if (arr1_one != arr2_one || arr1_zero != arr2_zero) {
Console.WriteLine("No");
return;
}
// Iterate over the range [0, N-1]
for (int i = 0; i < N; i++) {
// Increment count by differences
// arr1[i] and arr2[i]
count = count + (arr1[i] - arr2[i]);
// Check if number of 1's in
// arr2 are more than arr1 and
// then print "No"
if (count < 0) {
Console.WriteLine("No");
return;
}
}
// Finally, print "Yes"
Console.WriteLine("Yes");
}
// Driver Code
public static void Main()
{
// Given input arrays
int []arr1 = { 0, 1, 1, 0 };
int []arr2 = { 0, 0, 1, 1 };
// Size of the array
int N = arr1.Length;
// Function Call
canMakeEqual(arr1, arr2, N);
}
}
// This code is contributed by bgangwar59.输出：
Yes时间复杂度： O(N)
辅助空间： O(1)