给定一个二元数组arr[] ，任务是在删除最多1 个空单元后找到最长的非空单元子数组。
The array indices filled with 0 are known as empty cell whereas the indices filled with 1 are known as non-empty cells.
例子：
Input: arr[] = {1, 1, 0, 1}
Output: 3
Explanation:
Removal of 0 modifies the array to {1, 1, 1}, thus maximizing the length of the subarray to 3.
Input: arr[] = {1, 1, 1, 1, 1}
Output: 5
方法：
这个想法是在每个索引的前缀和后缀中存储 1 的频率，以计算来自特定索引的两个方向上 1 的最长连续序列。请按照以下步骤解决问题：
- 初始化两个数组l[]和r[] ，它们分别存储数组arr[] 中从数组左侧和右侧开始的最长连续1的长度。
- 在索引(0, N) 上迭代输入数组，并为每个arr[i] = 1增加 1计数。否则，将count的值存储到l[i] 中的第(i – 1)个索引将 count 重置为零。
- 类似地，通过遍历索引[N – 1, 0]重复上述步骤，将计数从右侧存储在r[] 中。
- 对于包含0 的每个第i个索引索引，通过删除该0来计算可能的非空子数组的长度，它等于l[i] + r[i] 。
- 计算所有这些长度的最大值并打印结果。
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 length
// of a subarray of 1s after removing
// at most one 0
int longestSubarray(int a[], int n)
{
// Stores the count of consecutive
// 1's from left
int l[n];
// Stores the count of consecutive
// 1's from right
int r[n];
// Traverse left to right
for (int i = 0, count = 0;
i < n; i++) {
// If cell is non-empty
if (a[i] == 1)
// Increase count
count++;
// If cell is empty
else {
// Store the count of
// consecutive 1's
// till (i - 1)-th index
l[i] = count;
count = 0;
}
}
// Traverse from right to left
for (int i = n - 1, count = 0;
i >= 0; i--) {
if (a[i] == 1)
count++;
else {
// Store the count of
// consecutive 1s
// till (i + 1)-th index
r[i] = count;
count = 0;
}
}
// Stores the length of
// longest subarray
int ans = -1;
for (int i = 0; i < n; ++i) {
if (a[i] == 0)
// Store the maximum
ans = max(ans, l[i] + r[i]);
}
// If array a contains only 1s
// return n else return ans
return ans < 0 ? n : ans;
}
// Driver Code
int main()
{
int arr[] = { 0, 1, 1, 1, 0, 1,
0, 1, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << longestSubarray(arr, n);
return 0;
} Java
// Java program for the above approach
class GFG{
// Function to find the maximum length
// of a subarray of 1s after removing
// at most one 0
public static int longestSubarray(int[] a,
int n)
{
// Stores the count of consecutive
// 1's from left
int[] l = new int[n];
// Stores the count of consecutive
// 1's from right
int[] r = new int[n];
// Traverse left to right
for(int i = 0, count = 0;
i < n; i++)
{
// If cell is non-empty
if (a[i] == 1)
// Increase count
count++;
// If cell is empty
else
{
// Store the count of
// consecutive 1's
// till (i - 1)-th index
l[i] = count;
count = 0;
}
}
// Traverse from right to left
for(int i = n - 1, count = 0;
i >= 0; i--)
{
if (a[i] == 1)
count++;
else
{
// Store the count of
// consecutive 1s
// till (i + 1)-th index
r[i] = count;
count = 0;
}
}
// Stores the length of
// longest subarray
int ans = -1;
for(int i = 0; i < n; ++i)
{
if (a[i] == 0)
// Store the maximum
ans = Math.max(ans, l[i] + r[i]);
}
// If array a contains only 1s
// return n else return ans
return ans < 0 ? n : ans;
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 0, 1, 1, 1, 0,
1, 0, 1, 1 };
int n = arr.length;
System.out.println(longestSubarray(arr, n));
}
}
// This code is contributed by divyeshrabadiya07Python3
# Python3 program for the above approach
# Function to find the maximum length
# of a subarray of 1s after removing
# at most one 0
def longestSubarray(a, n):
# Stores the count of consecutive
# 1's from left
l = [0] * (n)
# Stores the count of consecutive
# 1's from right
r = [0] * (n)
count = 0
# Traverse left to right
for i in range(n):
# If cell is non-empty
if (a[i] == 1):
# Increase count
count += 1
# If cell is empty
else:
# Store the count of
# consecutive 1's
# till (i - 1)-th index
l[i] = count
count = 0
count = 0
# Traverse from right to left
for i in range(n - 1, -1, -1):
if (a[i] == 1):
count += 1
else:
# Store the count of
# consecutive 1s
# till (i + 1)-th index
r[i] = count
count = 0
# Stores the length of
# longest subarray
ans = -1
for i in range(n):
if (a[i] == 0):
# Store the maximum
ans = max(ans, l[i] + r[i])
# If array a contains only 1s
# return n else return ans
return ans < 0 and n or ans
# Driver code
arr = [ 0, 1, 1, 1, 0, 1, 0, 1, 1 ]
n = len(arr)
print(longestSubarray(arr, n))
# This code is contributed by sanjoy_62C#
// C# program for the above approach
using System;
class GFG{
// Function to find the maximum length
// of a subarray of 1s after removing
// at most one 0
public static int longestSubarray(int[] a,
int n)
{
// Stores the count of consecutive
// 1's from left
int[] l = new int[n];
// Stores the count of consecutive
// 1's from right
int[] r = new int[n];
// Traverse left to right
for(int i = 0, count = 0; i < n; i++)
{
// If cell is non-empty
if (a[i] == 1)
// Increase count
count++;
// If cell is empty
else
{
// Store the count of
// consecutive 1's
// till (i - 1)-th index
l[i] = count;
count = 0;
}
}
// Traverse from right to left
for(int i = n - 1, count = 0;
i >= 0; i--)
{
if (a[i] == 1)
count++;
else
{
// Store the count of
// consecutive 1s
// till (i + 1)-th index
r[i] = count;
count = 0;
}
}
// Stores the length of
// longest subarray
int ans = -1;
for(int i = 0; i < n; ++i)
{
if (a[i] == 0)
// Store the maximum
ans = Math.Max(ans, l[i] + r[i]);
}
// If array a contains only 1s
// return n else return ans
return ans < 0 ? n : ans;
}
// Driver code
public static void Main()
{
int[] arr = { 0, 1, 1, 1, 0,
1, 0, 1, 1 };
int n = arr.Length;
Console.Write(longestSubarray(arr, n));
}
}
// This code is contributed by sanjoy_62Javascript
输出：
4时间复杂度： O(N)
辅助空间： O(N)