给定相同大小的整数数组arr1和二进制数组arr2 ，任务是找到这些数组的乘积的最大可能和，即(arr1 [0] * arr2 [0])+(arr1 [1] * arr2 [ 1])+…..(arr1 [N-1] * arr2 [N-1])通过切换数组arr2中的任意两个相邻位获得。此切换次数可以无限次。
数组arr2中2个相邻位的切换如下：
- 00切换到11
- 01切换到10
- 10切换到01
- 11切换到00
例子：
Input: arr1 = {2, 3, 1}, arr2 = {0, 0, 1}
Output: 6
Explanation:
if we put 1 corresponding to a positive integer
then arr2 will be {1, 1, 1}
No. of 1's initially and now are odd
It means parity is same
so this arrangement is fine
Hence sum will be 2 + 3 + 1 = 6.
Input: arr1 = {2, -4, 5, 3}, arr2 = {0, 1, 0, 1}
Output: 8
方法 ：
- 每次操作后，奇偶校验保持不变，即不。如果最初是偶数，则1的个数是偶数；如果最初是奇数，则1的奇数是奇数。
- 第二个观察结果是，可以通过从第i个位切换来完成第i个和第j个位的切换，例如(i，i + 1)，(i + 1，i + 2)…。 (j-1，j)在这里，每个位都切换两次(如果位被切换两次，则返回其初始值)，除了i和j之外，最终第i个和第j个位切换。
- 从这两个观察结果可以看出，我们可以在数组arr2中进行1和0的任何排列。我们唯一需要注意的是1和0的奇偶性。最终奇偶校验必须与初始奇偶校验相同。
- 要获得最大总和，如果arr1中位于第i个位置的元素为正，则将1置于arr2的第i个位置，否则置0。
下面是上述方法的实现：
C++
// C++ program to find the maximum SoP of two arrays
// by toggling adjacent bits in the second array
#include
using namespace std;
// Function to return Max Sum
int maxSum(int arr1[], int arr2[], int n)
{
// intialParity and finalParity are 0
// if total no. of 1's is even else 1
int initialParity = 0, finalParity = 0;
// minPositive and maxNegative will store
// smallest positive and smallest negative
// integer respectively.
int sum = 0,
minPositive = INT_MAX,
maxNegative = INT_MIN;
for (int i = 0; i < n; i++) {
// Count of Initial Parity
initialParity += arr2[i];
// if arr1[i] is positive then add 1
// in finalParity to get 1 at arr2[i]
if (arr1[i] >= 0) {
finalParity += 1;
sum += arr1[i];
minPositive = min(minPositive, arr1[i]);
}
else {
maxNegative = max(maxNegative, arr1[i]);
}
}
// if both parity are odd or even
// then return sum
if (initialParity % 2 == finalParity % 2) {
return sum;
}
// else add one more 1 or remove 1
else {
// if minPositive > maxNegative,
// put 1 at maxNegative
// and add it to our sum
if (minPositive + maxNegative >= 0) {
return sum + maxNegative;
}
// else remove minPositive no.
else {
return sum - minPositive;
}
}
}
// Driver code
int main()
{
int arr1[] = { 2, -4, 5, 3 };
int arr2[] = { 0, 1, 0, 1 };
int n = sizeof(arr1) / sizeof(arr1[0]);
cout << maxSum(arr1, arr2, n) << endl;
return 0;
} Java
// Java program to find the maximum SoP
// of two arrays by toggling adjacent bits
// in the second array
class GFG
{
// Function to return Max Sum
static int maxSum(int arr1[],
int arr2[], int n)
{
// intialParity and finalParity are 0
// if total no. of 1's is even else 1
int initialParity = 0, finalParity = 0;
// minPositive and maxNegative will store
// smallest positive and smallest negative
// integer respectively.
int sum = 0,
minPositive = Integer.MAX_VALUE,
maxNegative = Integer.MIN_VALUE;
for (int i = 0; i < n; i++)
{
// Count of Initial Parity
initialParity += arr2[i];
// if arr1[i] is positive then add 1
// in finalParity to get 1 at arr2[i]
if (arr1[i] >= 0)
{
finalParity += 1;
sum += arr1[i];
minPositive = Math.min(minPositive,
arr1[i]);
}
else
{
maxNegative = Math.max(maxNegative,
arr1[i]);
}
}
// if both parity are odd or even
// then return sum
if (initialParity % 2 == finalParity % 2)
{
return sum;
}
// else add one more 1 or remove 1
else
{
// if minPositive > maxNegative,
// put 1 at maxNegative
// and add it to our sum
if (minPositive + maxNegative >= 0)
{
return sum + maxNegative;
}
// else remove minPositive no.
else
{
return sum - minPositive;
}
}
}
// Driver code
public static void main(String []args)
{
int arr1[] = { 2, -4, 5, 3 };
int arr2[] = { 0, 1, 0, 1 };
int n = arr1.length;
System.out.println(maxSum(arr1, arr2, n));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program to find the
# maximum SoP of two arrays by
# toggling adjacent bits#
# in the second array
import sys
# Function to return Max Sum
def maxSum(arr1, arr2, n) :
# intialParity and finalParity are 0
# if total no. of 1's is even else 1
initialParity, finalParity = 0, 0
# minPositive and maxNegative will store
# smallest positive and smallest negative
# integer respectively.
sum = 0
minPositive = sys.maxsize
maxNegative = -sys.maxsize - 1
for i in range(n) :
# Count of Initial Parity
initialParity += arr2[i];
# if arr1[i] is positive then add 1
# in finalParity to get 1 at arr2[i]
if (arr1[i] >= 0) :
finalParity += 1
sum += arr1[i]
minPositive = min(minPositive, arr1[i])
else :
maxNegative = max(maxNegative, arr1[i])
# if both parity are odd or even
# then return sum
if (initialParity % 2 == finalParity % 2) :
return sum
# else add one more 1 or remove 1
else :
# if minPositive > maxNegative,
# put 1 at maxNegative
# and add it to our sum
if (minPositive + maxNegative >= 0) :
return sum + maxNegative
# else remove minPositive no.
else :
return sum - minPositive
# Driver code
arr1 = [ 2, -4, 5, 3 ]
arr2 = [ 0, 1, 0, 1 ]
n = len(arr1)
print(maxSum(arr1, arr2, n))
# This code is contributed by divyamohan123C#
// C# program to find the maximum SoP
// of two arrays by toggling adjacent bits
// in the second array
using System;
class GFG
{
// Function to return Max Sum
static int maxSum(int []arr1,
int []arr2, int n)
{
// intialParity and finalParity are 0
// if total no. of 1's is even else 1
int initialParity = 0, finalParity = 0;
// minPositive and maxNegative will store
// smallest positive and smallest negative
// integer respectively.
int sum = 0,
minPositive = int.MaxValue,
maxNegative = int.MinValue;
for (int i = 0; i < n; i++)
{
// Count of Initial Parity
initialParity += arr2[i];
// if arr1[i] is positive then add 1
// in finalParity to get 1 at arr2[i]
if (arr1[i] >= 0)
{
finalParity += 1;
sum += arr1[i];
minPositive = Math.Min(minPositive,
arr1[i]);
}
else
{
maxNegative = Math.Max(maxNegative,
arr1[i]);
}
}
// if both parity are odd or even
// then return sum
if (initialParity % 2 == finalParity % 2)
{
return sum;
}
// else add one more 1 or remove 1
else
{
// if minPositive > maxNegative,
// put 1 at maxNegative
// and add it to our sum
if (minPositive + maxNegative >= 0)
{
return sum + maxNegative;
}
// else remove minPositive no.
else
{
return sum - minPositive;
}
}
}
// Driver code
public static void Main(String []args)
{
int []arr1 = { 2, -4, 5, 3 };
int []arr2 = { 0, 1, 0, 1 };
int n = arr1.Length;
Console.WriteLine(maxSum(arr1, arr2, n));
}
}
// This code is contributed by 29AjayKumar输出：
8
时间复杂度： 
，其中n是数组的大小。