给定两个大小为N 的二进制数组X[]和Y[] ，任务是通过选择任何奇数长度的子数组并翻转所有奇数索引元素的最少操作次数将数组X[]转换为数组Y[]子阵列。
例子：
Input: X[] = {1, 0, 0, 0, 0, 1}, Y[] = {1, 1, 0, 1, 1, 1}
Output: 2
Explanation:
Initially, X[] is {1, 0, 0, 0, 0, 1}.
Operation 1: Choose the sub-array {0, 0, 0} from array X[] and change the 2nd and 4th character and convert it to {1, 0, 1}.
Now X becomes {1, 1, 0, 1, 0, 1}.
Operation 2: Choose the sub-array {0} containing only the 5th character and convert it to 1.
Finally, X becomes {1, 1, 0, 1, 1, 1}, which is equal to Y.
Therefore, the count of operations is 2.
Input: X[] = {0, 1, 0}, Y[] = {0, 1, 0}
Output: 0
Explanation:
Since both the arrays X and Y are equal thus the minimum operations would be 0.
方法：这个想法是分别计算偶数和奇数位置的操作。请按照以下步骤解决问题：
- 将变量C初始化为0以存储操作计数。
- 在奇数位置遍历数组X[]元素并取计数器count = 0。
- 检查X[i]和Y[i]之间的连续不相等元素，并将计数器计数每次增加1 。
- 当X[i]和Y[i]相等时，增加全局C以将操作增加 1，因为在该操作中，所有奇数位置都可以与Y 中的一样。
- 类似地，在偶数位置上遍历数组X[]元素并再次取计数器count = 0。
- 检查X[i]和Y[i]之间的连续不相等元素，并将计数器计数每次增加1 。
- 当X[i]和Y[i]相等时，增加全局计数器C以将操作增加1 。
- 完成上述步骤后，打印C的值作为所需操作的结果计数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the minimum flip
// of subarrays required at alternate
// index to make binary arrays equals
void minOperation(int X[], int Y[],
int n)
{
// Stores count of total operations
int C = 0;
// Stores count of consecutive
// unequal elements
int count = 0;
// Loop to run on odd positions
for (int i = 1; i < n; i = i + 2) {
if (X[i] != Y[i]) {
count++;
}
else {
// Incrementing the
// global counter
if (count != 0)
C++;
// Change count to 0
count = 0;
}
}
// If all last elements are equal
if (count != 0)
C++;
count = 0;
// Loop to run on even positions
for (int i = 0; i < n; i = i + 2) {
if (X[i] != Y[i]) {
count++;
}
else {
// Incrementing the
// global counter
if (count != 0)
C++;
// Change count to 0
count = 0;
}
}
if (count != 0)
C++;
// Print the minimum operations
cout << C;
}
// Driver Code
int main()
{
int X[] = { 1, 0, 0, 0, 0, 1 };
int Y[] = { 1, 1, 0, 1, 1, 1 };
int N = sizeof(X) / sizeof(X[0]);
// Function Call
minOperation(X, Y, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the minimum flip
// of subarrays required at alternate
// index to make binary arrays equals
static void minOperation(int X[], int Y[],
int n)
{
// Stores count of total operations
int C = 0;
// Stores count of consecutive
// unequal elements
int count = 0;
// Loop to run on odd positions
for(int i = 1; i < n; i = i + 2)
{
if (X[i] != Y[i])
{
count++;
}
else
{
// Incrementing the
// global counter
if (count != 0)
C++;
// Change count to 0
count = 0;
}
}
// If all last elements are equal
if (count != 0)
C++;
count = 0;
// Loop to run on even positions
for(int i = 0; i < n; i = i + 2)
{
if (X[i] != Y[i])
{
count++;
}
else
{
// Incrementing the
// global counter
if (count != 0)
C++;
// Change count to 0
count = 0;
}
}
if (count != 0)
C++;
// Print the minimum operations
System.out.print(C);
}
// Driver Code
public static void main(String[] args)
{
int X[] = { 1, 0, 0, 0, 0, 1 };
int Y[] = { 1, 1, 0, 1, 1, 1 };
int N = X.length;
// Function Call
minOperation(X, Y, N);
}
}
// This code is contributed by susmitakundugoaldangaPython3
# Python program for the above approach
# Function to find the minimum flip
# of subarrays required at alternate
# index to make binary arrays equals
def minOperation(X, Y, n):
# Stores count of total operations
C = 0;
# Stores count of consecutive
# unequal elements
count = 0;
# Loop to run on odd positions
for i in range(1, n, 2):
if (X[i] != Y[i]):
count += 1;
else:
# Incrementing the
# global counter
if (count != 0):
C += 1;
# Change count to 0
count = 0;
# If all last elements are equal
if (count != 0):
C += 1;
count = 0;
# Loop to run on even positions
for i in range(0, n, 2):
if (X[i] != Y[i]):
count += 1;
else:
# Incrementing the
# global counter
if (count != 0):
C += 1;
# Change count to 0
count = 0;
if (count != 0):
C += 1;
# Prthe minimum operations
print(C);
# Driver Code
if __name__ == '__main__':
X = [1, 0, 0, 0, 0, 1];
Y = [1, 1, 0, 1, 1, 1];
N = len(X);
# Function Call
minOperation(X, Y, N);
# This code is contributed by 29AjayKumarC#
// C# program for the above approach
using System;
class GFG{
// Function to find the minimum flip
// of subarrays required at alternate
// index to make binary arrays equals
static void minOperation(int []X, int []Y,
int n)
{
// Stores count of total operations
int C = 0;
// Stores count of consecutive
// unequal elements
int count = 0;
// Loop to run on odd positions
for(int i = 1; i < n; i = i + 2)
{
if (X[i] != Y[i])
{
count++;
}
else
{
// Incrementing the
// global counter
if (count != 0)
C++;
// Change count to 0
count = 0;
}
}
// If all last elements are equal
if (count != 0)
C++;
count = 0;
// Loop to run on even positions
for(int i = 0; i < n; i = i + 2)
{
if (X[i] != Y[i])
{
count++;
}
else
{
// Incrementing the
// global counter
if (count != 0)
C++;
// Change count to 0
count = 0;
}
}
if (count != 0)
C++;
// Print the minimum operations
Console.Write(C);
}
// Driver Code
public static void Main(String[] args)
{
int []X = { 1, 0, 0, 0, 0, 1 };
int []Y = { 1, 1, 0, 1, 1, 1 };
int N = X.Length;
// Function Call
minOperation(X, Y, N);
}
}
// This code is contributed by Amit KatiyarJavascript
输出：
2时间复杂度： O(N)
辅助空间： O(1)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live