给定大小为N的二进制数组arr [] ，任务是找到可以转换为1 s的最大计数0 s，以使相邻的数组元素对都不为1 。
例子：
Input: arr[] = { 1, 0, 0, 0, 1 }
Output: 1
Explanation:
Updating arr[2] to 1 modifies arr[] to { 1, 0, 1, 0, 1 }
Therefore, the required output is 1
Input: arr[] = { 0, 0, 1, 0, 0, 0, 0, 1, 0, 0 }
Output: 3
Explanation:
Updating arr[0], arr[5] and arr[9] modifies arr[] to { 1, 0, 1, 0, 0, 1, 0, 1, 0, 1 }
Therefore, the required output is 3
方法：可以使用贪婪技术解决问题。请按照以下步骤解决问题：
- 在数组的开头和结尾处插入0 。
- 遍历数组arr []。在第ith次迭代中，检查arr [i] ， arr [i + 1]和arr [i + 2]是否为0 s。如果发现为真，则增加计数并更新arr [i + 1] = 1 。
- 最后，打印获得的计数。
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 maximum count of 0s
// required to be converted into 1s such
// no pair of adjacent elements are 1
void maxPositionsOccupied(vector& arr)
{
// Base Case
if (arr.size() == 0) {
cout << 0;
return;
}
// Insert 0 at the end
// of the array
arr.push_back(0);
// Insert 0 at the front
// of the array
arr.insert(arr.begin(), 0);
// Stores the maximum count of of 0s
// that can be converted into 1s
int ans = 0;
// Stores index of array elements
int i = 0;
// Traverse the array
while ((i < arr.size() - 2)) {
// If adjacent elements are 0s
if ((arr[i] == 0) && (arr[i + 1] == 0)
&& (arr[i + 2] == 0)) {
// Update ans
ans++;
// Update arr[i + 1]
arr[i + 1] = 1;
}
// Update i
i++;
}
// Print the answer
cout << ans;
}
// Driver Code
int main()
{
// Given binary array
vector arr = { 1, 0, 0, 0, 1 };
// Prints the maximum 0 to 1
// conversions required
maxPositionsOccupied(arr);
return 0;
} Java
// Java program for the above approach
public class GFG
{
// Function to find the maximum count of 0s
// required to be converted into 1s such
// no pair of adjacent elements are 1
static void maxPositionsOccupied(int[] arr)
{
// Base Case
if (arr.length == 0)
{
System.out.print(0);
return;
}
// Stores the maximum count of of 0s
// that can be converted into 1s
int ans = 0;
// Stores index of array elements
int i = 0;
// Traverse the array
while ((i < arr.length - 2))
{
// If adjacent elements are 0s
if ((arr[i] == 0) &&
(arr[i + 1] == 0) &&
(arr[i + 2] == 0))
{
// Update ans
ans++;
// Update arr[i + 1]
arr[i + 1] = 1;
}
// Update i
i++;
}
// Print the answer
System.out.print(ans);
}
// Driver code
public static void main(String[] args)
{
// Given binary array
int[] arr = { 1, 0, 0, 0, 1 };
// Prints the maximum 0 to 1
// conversions required
maxPositionsOccupied(arr);
}
}
// This code is contributed by divyeshrabadiya07.Python3
# Python3 program for the above approach
# Function to find the maximum count of 0s
# required to be converted into 1s such
# no pair of adjacent elements are 1
def maxPositionsOccupied(arr):
# Base Case
if (len(arr) == 0):
print(0)
# Insert 0 at the end
# of the array
arr.append(0)
# Insert 0 at the front
# of the array
arr.insert(0, 0)
# Stores the maximum count of of 0s
# that can be converted into 1s
ans = 0
# Stores index of array elements
i = 0
# Traverse the array
while((i < len(arr) - 2)):
# If adjacent elements are 0s
if ((arr[i] == 0) and
(arr[i + 1] == 0) and
(arr[i + 2] == 0)):
# Update ans
ans += 1
# Update arr[i + 1]
arr[i + 1] = 1
# Update i
i += 1
# Print the answer
print(ans)
# Driver Code
if __name__ == '__main__':
# Given binary array
arr = [ 1, 0, 0, 0, 1 ]
# Prints the maximum 0 to 1
# conversions required
maxPositionsOccupied(arr)
# This code is contributed by SURENDRA_GANGWARC#
// C# program for the above approach
using System;
class GFG{
// Function to find the maximum count of 0s
// required to be converted into 1s such
// no pair of adjacent elements are 1
static void maxPositionsOccupied(int[] arr)
{
// Base Case
if (arr.Length == 0)
{
Console.Write(0);
return;
}
// Stores the maximum count of of 0s
// that can be converted into 1s
int ans = 0;
// Stores index of array elements
int i = 0;
// Traverse the array
while ((i < arr.Length - 2))
{
// If adjacent elements are 0s
if ((arr[i] == 0) &&
(arr[i + 1] == 0) &&
(arr[i + 2] == 0))
{
// Update ans
ans++;
// Update arr[i + 1]
arr[i + 1] = 1;
}
// Update i
i++;
}
// Print the answer
Console.Write(ans);
}
// Driver code
static void Main()
{
// Given binary array
int[] arr = { 1, 0, 0, 0, 1 };
// Prints the maximum 0 to 1
// conversions required
maxPositionsOccupied(arr);
}
}
// This code is contributed by divyesh072019输出：
1时间复杂度： O(N)
辅助空间： O(1)