给定一个大小为N的数组arr[] ，具有相等数量的偶数和奇数整数，任务是找到需要反转的最小子数组数，以使相邻元素对的总和为奇数。
例子：
Input: arr[] = {13, 2, 6, 8, 3, 5, 7, 10, 14, 15}
Output: 3
Explanation:
Step 1: Reverse subarray [2, 4] to modify the array to {13, 2, 3, 8, 6, 5, 7, 10, 14, 15}
Step 2: Reverse subarray [4, 5] to modify the array to {13, 2, 3, 8, 5, 6, 7, 10, 14, 15}
Step 3: Reverse subarray [8, 9] to modify the array to {13, 2, 3, 8, 5, 6, 7, 10, 15, 14}
After the above reversals, the sum of all adjacent element pairs is odd.
Therefore, the minimum reversals required is 3.
Input: arr[] = {1, 3, 4, 5, 9, 6, 8}
Output: 2
处理方法：使相邻元素之和为奇数，其思想是将元素以奇偶或偶奇的方式排列。要最小化反转操作的次数，请观察给定数组的以下属性：
- 如果有任何长度为M的子数组具有相同奇偶校验的所有元素，那么也存在一个长度为M的子数组，它具有相反奇偶校验的所有元素，因为数组中偶数和奇数元素的计数相同。
- 因此，生成交替奇偶校验长度为M的子数组所需的反转操作次数为(M – 1) 。
请按照以下步骤解决问题：
- 遍历给定数组，求数组中每个连续相同的奇偶元素所需的子数组反转次数。
- 令连续奇数和偶数元素的反转次数分别为cntOdd和cntEven 。
- 最小反转计数由cntOdd和cntEven的最大值给出，因为任务是删除所有连续的相同元素和需要删除的最大连续元素。
下面是上述方法的实现：
C++14
// C++ program for the above approach
#include
using namespace std;
// Function to count reversals to
// seperate elements with same parity
int separate(int arr[], int n, int parity)
{
int count = 1, res = 0;
// Traverse the given array
for (int i = 1; i < n; i++) {
// Count size of subarray having
// integers with same parity only
if (((arr[i] + parity) & 1)
&& ((arr[i - 1] + parity) & 1))
count++;
// Otherwise
else {
// Reversals required is equal
// to one less than subarray size
if (count > 1)
res += count - 1;
count = 1;
}
}
// Return the total reversals
return res;
}
// Function to print the array elements
void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Function to count the mimimum reversals
// required to make make sum
// of all adjacent elements odd
void requiredOps(int arr[], int N)
{
// Stores operations required for
// separating adjacent odd elements
int res1 = separate(arr, N, 0);
// Stores operations required for
// separating adjacent even elements
int res2 = separate(arr, N, 1);
// Maximum of two is the return
cout << max(res1, res2);
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 13, 2, 6, 8, 3, 5,
7, 10, 14, 15 };
// Size of array
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
requiredOps(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to count reversals to
// seperate elements with same parity
static int separate(int arr[], int n,
int parity)
{
int count = 1, res = 0;
// Traverse the given array
for(int i = 1; i < n; i++)
{
// Count size of subarray having
// integers with same parity only
if (((arr[i] + parity) & 1) != 0 &&
((arr[i - 1] + parity) & 1) != 0)
count++;
// Otherwise
else
{
// Reversals required is equal
// to one less than subarray size
if (count > 1)
res += count - 1;
count = 1;
}
}
// Return the total reversals
return res;
}
// Function to print the array elements
void printArray(int arr[], int n)
{
for(int i = 0; i < n; i++)
System.out.println(arr[i] + " ");
}
// Function to count the mimimum reversals
// required to make make sum
// of all adjacent elements odd
static void requiredOps(int arr[], int N)
{
// Stores operations required for
// separating adjacent odd elements
int res1 = separate(arr, N, 0);
// Stores operations required for
// separating adjacent even elements
int res2 = separate(arr, N, 1);
// Maximum of two is the return
System.out.print(Math.max(res1, res2));
}
// Driver Code
public static void main(String args[])
{
// Given array arr[]
int arr[] = { 13, 2, 6, 8, 3, 5,
7, 10, 14, 15 };
// Size of array
int N = arr.length;
// Function Call
requiredOps(arr, N);
}
}
// This code is contributed by ipg2016107Python3
# Python3 program for the
# above approach
# Function to count reversals
# to seperate elements with
# same parity
def separate(arr, n, parity):
count = 1
res = 0
# Traverse the given array
for i in range(1, n):
# Count size of subarray
# having integers with
# same parity only
if (((arr[i] + parity) & 1) and
((arr[i - 1] + parity) & 1)):
count += 1
# Otherwise
else:
# Reversals required is
# equal to one less than
# subarray size
if (count > 1):
res += count - 1
count = 1
# Return the total
# reversals
return res
# Function to print the
# array elements
def printArray(arr, n):
for i in range(n):
print(arr[i],
end = " ")
# Function to count the mimimum
# reversals required to make
# make sum of all adjacent
# elements odd
def requiredOps(arr, N):
# Stores operations required
# for separating adjacent
# odd elements
res1 = separate(arr, N, 0)
# Stores operations required
# for separating adjacent
# even elements
res2 = separate(arr, N, 1)
# Maximum of two is the
# return
print(max(res1, res2))
# Driver Code
if __name__ == "__main__":
# Given array arr[]
arr = [13, 2, 6, 8, 3, 5,
7, 10, 14, 15]
# Size of array
N = len(arr)
# Function Call
requiredOps(arr, N)
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
class GFG{
// Function to count reversals to
// seperate elements with same parity
static int separate(int[] arr, int n,
int parity)
{
int count = 1, res = 0;
// Traverse the given array
for(int i = 1; i < n; i++)
{
// Count size of subarray having
// integers with same parity only
if (((arr[i] + parity) & 1) != 0 &&
((arr[i - 1] + parity) & 1) != 0)
count++;
// Otherwise
else
{
// Reversals required is equal
// to one less than subarray size
if (count > 1)
res += count - 1;
count = 1;
}
}
// Return the total reversals
return res;
}
// Function to print the array elements
void printArray(int[] arr, int n)
{
for(int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
// Function to count the mimimum reversals
// required to make make sum
// of all adjacent elements odd
static void requiredOps(int[] arr, int N)
{
// Stores operations required for
// separating adjacent odd elements
int res1 = separate(arr, N, 0);
// Stores operations required for
// separating adjacent even elements
int res2 = separate(arr, N, 1);
// Maximum of two is the return
Console.Write(Math.Max(res1, res2));
}
// Driver code
public static void Main()
{
// Given array arr[]
int[] arr = { 13, 2, 6, 8, 3, 5,
7, 10, 14, 15 };
// Size of array
int N = arr.Length;
// Function Call
requiredOps(arr, N);
}
}
// This code is contributed by sanjoy_62Javascript
输出：
3时间复杂度： O(N)
辅助空间： O(1)
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