给定一个数组A [] ，任务是找到将数组转换为B []所需的最少操作数，以使得对于B中的每个索引(最后一个除外)， parity(b [i])！= parity(b [i + 1])其中parity(x)= x％3 。
以下是要执行的操作：
Any element from the set {1, 2} can be added to any element of the array.
例子：
Input: A[] = {2, 1, 3, 0}
Output: 1
1 can be added to 0 in a single operation and the array becomes {2, 1, 3, 1}.
Input: A[] = {2, 2, 2, 2}
Output: 2
方法：一种最佳方法是更新两个其他元素之间的元素，以便可以对其进行更新，以使其在同一个操作中，奇偶校验可以与之前的元素和紧随其后的元素不同(以数字表示)操作的数量需要最小化)。
因此，开始遍历数组从A [1]到A [n – 2] ，并对每个元素A [i]进行奇偶校验(A [i])==奇偶校验(A [i – 1])或奇偶校验(A [i])== parity(A [i + 1]) ，更新A [i]使其奇偶性不同于其周围的两个数字，并跟踪执行的操作数。最后打印计数。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the parity of a number
int parity(int a)
{
return a % 3;
}
// Function to return the minimum
// number of operations required
int solve(int array[], int size)
{
int operations = 0;
for (int i = 0; i < size - 1; i++) {
// Operation needs to be performed
if (parity(array[i]) == parity(array[i + 1])) {
operations++;
if (i + 2 < size) {
// Parity of previous element
int pari1 = parity(array[i]);
// Parity of next element
int pari2 = parity(array[i + 2]);
// Update parity of current element to be other than
// the parities of the pervious and the next number
if (pari1 == pari2) {
if (pari1 == 0)
array[i + 1] = 1;
else if (pari1 == 1)
array[i + 1] = 0;
else
array[i + 1] = 1;
}
else {
if ((pari1 == 0 && pari2 == 1)
|| (pari1 == 1 && pari2 == 0))
array[i + 1] = 2;
if ((pari1 == 1 && pari2 == 2)
|| (pari1 == 2 && pari2 == 1))
array[i + 1] = 0;
if ((pari1 == 2 && pari2 == 0)
|| (pari1 == 0 && pari2 == 2))
array[i + 1] = 1;
}
}
}
}
return operations;
}
// Driver Code
int main()
{
int array[] = { 2, 1, 3, 0 };
int size = sizeof(array) / sizeof(array[0]);
cout << solve(array, size) << endl;
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to return the parity of a number
static int parity(int a)
{
return a % 3;
}
// Function to return the minimum
// number of operations required
static int solve(int []array, int size)
{
int operations = 0;
for (int i = 0; i < size - 1; i++)
{
// Operation needs to be performed
if (parity(array[i]) == parity(array[i + 1]))
{
operations++;
if (i + 2 < size)
{
// Parity of previous element
int pari1 = parity(array[i]);
// Parity of next element
int pari2 = parity(array[i + 2]);
// Update parity of current
// element to be other than
// the parities of the pervious
// and the next number
if (pari1 == pari2)
{
if (pari1 == 0)
array[i + 1] = 1;
else if (pari1 == 1)
array[i + 1] = 0;
else
array[i + 1] = 1;
}
else
{
if ((pari1 == 0 && pari2 == 1) ||
(pari1 == 1 && pari2 == 0))
array[i + 1] = 2;
if ((pari1 == 1 && pari2 == 2)
|| (pari1 == 2 && pari2 == 1))
array[i + 1] = 0;
if ((pari1 == 2 && pari2 == 0)
|| (pari1 == 0 && pari2 == 2))
array[i + 1] = 1;
}
}
}
}
return operations;
}
// Driver Code
public static void main (String arg[])
{
int []array = { 2, 1, 3, 0 };
int size = array.length;
System.out.println(solve(array, size));
}
}
// This code is contributed by RyugaPython3
# Python3 implementation of the approach
# Function to return the parity
# of a number
def parity(a):
return a % 3
# Function to return the minimum
# number of operations required
def solve(array, size):
operations = 0
for i in range(0, size - 1):
# Operation needs to be performed
if parity(array[i]) == parity(array[i + 1]):
operations += 1
if i + 2 < size:
# Parity of previous element
pari1 = parity(array[i])
# Parity of next element
pari2 = parity(array[i + 2])
# Update parity of current element to
# be other than the parities of the
# previous and the next number
if pari1 == pari2:
if pari1 == 0:
array[i + 1] = 1
elif pari1 == 1:
array[i + 1] = 0
else:
array[i + 1] = 1
else:
if ((pari1 == 0 and pari2 == 1) or
(pari1 == 1 and pari2 == 0)):
array[i + 1] = 2
if ((pari1 == 1 and pari2 == 2) or
(pari1 == 2 and pari2 == 1)):
array[i + 1] = 0
if ((pari1 == 2 and pari2 == 0) and
(pari1 == 0 and pari2 == 2)):
array[i + 1] = 1
return operations
# Driver Code
if __name__ == "__main__":
array = [2, 1, 3, 0]
size = len(array)
print(solve(array, size))
# This code is contributed by Rituraj JainC#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the parity of a number
static int parity(int a)
{
return a % 3;
}
// Function to return the minimum
// number of operations required
static int solve(int []array, int size)
{
int operations = 0;
for (int i = 0; i < size - 1; i++)
{
// Operation needs to be performed
if (parity(array[i]) == parity(array[i + 1]))
{
operations++;
if (i + 2 < size)
{
// Parity of previous element
int pari1 = parity(array[i]);
// Parity of next element
int pari2 = parity(array[i + 2]);
// Update parity of current
// element to be other than
// the parities of the pervious
// and the next number
if (pari1 == pari2)
{
if (pari1 == 0)
array[i + 1] = 1;
else if (pari1 == 1)
array[i + 1] = 0;
else
array[i + 1] = 1;
}
else
{
if ((pari1 == 0 && pari2 == 1) ||
(pari1 == 1 && pari2 == 0))
array[i + 1] = 2;
if ((pari1 == 1 && pari2 == 2)
|| (pari1 == 2 && pari2 == 1))
array[i + 1] = 0;
if ((pari1 == 2 && pari2 == 0)
|| (pari1 == 0 && pari2 == 2))
array[i + 1] = 1;
}
}
}
}
return operations;
}
// Driver Code
public static void Main()
{
int []array = { 2, 1, 3, 0 };
int size = array.Length;
Console.WriteLine(solve(array, size));
}
}
// This code is contributed
// by Akanksha RaiPHP
Javascript
输出：
1时间复杂度： O(N)
辅助空间： O(1)