给定一个数组arr [] ,任务是删除最多一个元素并计算严格增加的子数组的最大长度。
例子:
Input: arr[] = {1, 2, 5, 3, 4}
Output: 4
After deleting 5, the resulting array will be {1, 2, 3, 4}
and the maximum length of its strictly increasing subarray is 4.
Input: arr[] = {1, 2}
Output: 2
The complete array is already strictly increasing.
方法:
- 创建两个大小为N的数组pre []和pos [] 。
- 从(0,N)遍历输入数组arr []以找出当前元素arr [i]在数组中的贡献,直到现在[0,i)并更新pre []数组(如果它对严格意义上的贡献)增加子数组。
- 从[N – 2,0]遍历输入数组arr []以找出当前元素arr [j]到现在(N,j)在数组中的贡献,如果arr [j则更新pos []数组]贡献了最长的增长子数组。
- 在不删除任何元素的情况下,计算严格增加的子数组的最大长度。
- 遍历数组pre []和pos []以通过排除该元素来找出当前元素的贡献。
- 维护变量ans以找到到目前为止找到的最大值。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the maximum length of
// strictly increasing subarray after
// removing atmost one element
int maxIncSubarr(int a[], int n)
{
// Create two arrays pre and pos
int pre[n] = { 0 };
int pos[n] = { 0 };
pre[0] = 1;
pos[n - 1] = 1;
int l = 0;
// Find out the contribution of the current
// element in array[0, i] and update pre[i]
for (int i = 1; i < n; i++) {
if (a[i] > a[i - 1])
pre[i] = pre[i - 1] + 1;
else
pre[i] = 1;
}
// Find out the contribution of the current
// element in array[N - 1, i] and update pos[i]
l = 1;
for (int i = n - 2; i >= 0; i--) {
if (a[i] < a[i + 1])
pos[i] = pos[i + 1] + 1;
else
pos[i] = 1;
}
// Calculate the maximum length of the
// stricly increasing subarray without
// removing any element
int ans = 0;
l = 1;
for (int i = 1; i < n; i++) {
if (a[i] > a[i - 1])
l++;
else
l = 1;
ans = max(ans, l);
}
// Calculate the maximum length of the
// strictly increasing subarray after
// removing the current element
for (int i = 1; i <= n - 2; i++) {
if (a[i - 1] < a[i + 1])
ans = max(pre[i - 1] + pos[i + 1], ans);
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 1, 2 };
int n = sizeof(arr) / sizeof(int);
cout << maxIncSubarr(arr, n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the maximum length of
// strictly increasing subarray after
// removing atmost one element
static int maxIncSubarr(int a[], int n)
{
// Create two arrays pre and pos
int pre[] = new int[n] ;
int pos[] = new int[n] ;
pre[0] = 1;
pos[n - 1] = 1;
int l = 0;
// Find out the contribution of the current
// element in array[0, i] and update pre[i]
for (int i = 1; i < n; i++)
{
if (a[i] > a[i - 1])
pre[i] = pre[i - 1] + 1;
else
pre[i] = 1;
}
// Find out the contribution of the current
// element in array[N - 1, i] and update pos[i]
l = 1;
for (int i = n - 2; i >= 0; i--)
{
if (a[i] < a[i + 1])
pos[i] = pos[i + 1] + 1;
else
pos[i] = 1;
}
// Calculate the maximum length of the
// stricly increasing subarray without
// removing any element
int ans = 0;
l = 1;
for (int i = 1; i < n; i++)
{
if (a[i] > a[i - 1])
l++;
else
l = 1;
ans = Math.max(ans, l);
}
// Calculate the maximum length of the
// strictly increasing subarray after
// removing the current element
for (int i = 1; i <= n - 2; i++)
{
if (a[i - 1] < a[i + 1])
ans = Math.max(pre[i - 1] +
pos[i + 1], ans);
}
return ans;
}
// Driver code
public static void main (String[] args)
{
int arr[] = {1, 2};
int n = arr.length;
System.out.println(maxIncSubarr(arr, n));
}
}
// This code is contributed by AnkitRai01
Python3
# Python implementation of the approach
# Function to return the maximum length of
# strictly increasing subarray after
# removing atmost one element
def maxIncSubarr(a, n):
# Create two arrays pre and pos
pre = [0] * n;
pos = [0] * n;
pre[0] = 1;
pos[n - 1] = 1;
l = 0;
# Find out the contribution of the current
# element in array[0, i] and update pre[i]
for i in range(1, n):
if (a[i] > a[i - 1]):
pre[i] = pre[i - 1] + 1;
else:
pre[i] = 1;
# Find out the contribution of the current
# element in array[N - 1, i] and update pos[i]
l = 1;
for i in range(n - 2, -1, -1):
if (a[i] < a[i + 1]):
pos[i] = pos[i + 1] + 1;
else:
pos[i] = 1;
# Calculate the maximum length of the
# stricly increasing subarray without
# removing any element
ans = 0;
l = 1;
for i in range(1, n):
if (a[i] > a[i - 1]):
l += 1;
else:
l = 1;
ans = max(ans, l);
# Calculate the maximum length of the
# strictly increasing subarray after
# removing the current element
for i in range(1, n - 1):
if (a[i - 1] < a[i + 1]):
ans = max(pre[i - 1] + pos[i + 1], ans);
return ans;
# Driver code
if __name__ == '__main__':
arr = [ 1, 2 ];
n = len(arr);
print(maxIncSubarr(arr, n));
# This code is contributed by PrinciRaj1992
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the maximum length of
// strictly increasing subarray after
// removing atmost one element
static int maxIncSubarr(int []a, int n)
{
// Create two arrays pre and pos
int []pre = new int[n] ;
int []pos = new int[n] ;
pre[0] = 1;
pos[n - 1] = 1;
int l = 0;
// Find out the contribution of the current
// element in array[0, i] and update pre[i]
for (int i = 1; i < n; i++)
{
if (a[i] > a[i - 1])
pre[i] = pre[i - 1] + 1;
else
pre[i] = 1;
}
// Find out the contribution of the current
// element in array[N - 1, i] and update pos[i]
l = 1;
for (int i = n - 2; i >= 0; i--)
{
if (a[i] < a[i + 1])
pos[i] = pos[i + 1] + 1;
else
pos[i] = 1;
}
// Calculate the maximum length of the
// stricly increasing subarray without
// removing any element
int ans = 0;
l = 1;
for (int i = 1; i < n; i++)
{
if (a[i] > a[i - 1])
l++;
else
l = 1;
ans = Math.Max(ans, l);
}
// Calculate the maximum length of the
// strictly increasing subarray after
// removing the current element
for (int i = 1; i <= n - 2; i++)
{
if (a[i - 1] < a[i + 1])
ans = Math.Max(pre[i - 1] +
pos[i + 1], ans);
}
return ans;
}
// Driver code
public static void Main()
{
int []arr = {1, 2};
int n = arr.Length;
Console.WriteLine(maxIncSubarr(arr, n));
}
}
// This code is contributed by AnkitRai01
Javascript
输出:
2
时间复杂度: O(N)
空间复杂度: O(N)