通过交换其中一个数组的对来检查两个数组是否相等
原文:https://www . geeksforgeeks . org/check-if-two-arrays-可以通过交换阵列中的一对来使其相等/
给定两个大小相同的二进制数组 arr1[] 和 arr2[] ，任务是仅当 arr1[i] = 0 和arr 1[j]= 1(0≤I<j<N))时，通过交换成对的 arr1[ ] 来使两个数组相等。如果可以使两个阵列相等，打印“是”。否则，打印“否”。
示例:
输入: arr1[] = {0，0，1，1}，arr2[] = {1，1，0，0} 输出:是 解释: 交换 arr1[1]和 arr1[3]，变成 arr1[] = {0，1，1，0}。 交换 arr1[0]和 arr1[2]，变成 arr1[] = {1，1，0，0}。
输入: arr1[] = {1，0，1，0，1}，arr2[] = {0，1，0，0，1} 输出:否
方法:按照以下步骤解决问题:
- 初始化两个变量，说计数和标志 (= 真)。
- 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 ，将计数增加 1 。
- 否则，用 1 递减计数，如果计数< 0 ，则更新标志=假。
- 如果 arr1[i]！= arr2[i]:
- 如果标志等于真，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 <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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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.
java 描述语言
<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
时间复杂度:O(N) T5辅助空间:** O(1)
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