给定N个整数的数组arr [] 。任务是找到从数组arr []中删除最多一个元素后可以删除的连续严格递减序列的长度。
例子
Input: arr[] = {8, 7, 3, 5, 2, 9}
Output: 4
Explanation:
If we remove 3, The maximum length of decreasing sequence is 4 and the sequence is { 8, 7, 5, 2 }
If we remove 5, The maximum length of decreasing sequence is 4 and the sequence is { 8, 7, 3, 2 }
In both removal we get 4 as the maximum length.
Input: arr[] = {1, 2, 9, 8, 3, 7, 6, 4}
Output: 5
方法：
- 创建两个数组， left []存储从左到右的递减序列的长度，而right []存储从右到左的递减序列的长度。
- 遍历给定数组arr [] 。
- 如果前一个元素( arr [i-1] )大于下一个元素( arr [i + 1] )，则检查删除该元素是否会给出最大的递减子序列长度。
- 更新递减子序列的最大长度。
下面是上述方法的实现：
C++
// C++ program to find maximum length
// of decreasing sequence by removing
// at most one element
#include
using namespace std;
// Function to find the maximum length
int maxLength(int* a, int n)
{
// Intialise maximum length to 1
int maximum = 1;
// Intialise left[] to find the
// length of decreasing sequence
// from left to right
int left[n];
// Intialise right[] to find the
// length of decreasing sequence
// from right to left
int right[n];
// Initially store 1 at each index of
// left and right array
for (int i = 0; i < n; i++) {
left[i] = 1;
right[i] = 1;
}
// Iterate over the array arr[] to
// store length of decreasing
// sequence that can be obtained
// at every index in the right array
for (int i = n - 2; i >= 0; i--) {
if (a[i] > a[i + 1]) {
right[i] = right[i + 1] + 1;
}
// Store the length of longest
// continuous decreasing
// sequence in maximum
maximum = max(maximum, right[i]);
}
// Iterate over the array arr[] to
// store length of decreasing
// sequence that can be obtained
// at every index in the left array
for (int i = 1; i < n; i++) {
if (a[i] < a[i - 1]) {
left[i] = left[i - 1] + 1;
}
}
if (n > 2) {
// Check if we can obtain a
// longer decreasing sequence
// after removal of any element
// from the array arr[] with
// the help of left[] & right[]
for (int i = 1; i < n - 1; i++) {
if (a[i - 1] > a[i + 1]) {
maximum = max(maximum,
left[i - 1] + right[i + 1]);
}
}
}
// Return maximum length of sequence
return maximum;
}
// Driver code
int main()
{
int arr[6] = { 8, 7, 3, 5, 2, 9 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function calling
cout << maxLength(arr, n) << endl;
return 0;
} Java
// Java program to find maximum length
// of decreasing sequence by removing
// at most one element
class GFG {
// Function to find the maximum length
static int maxLength(int []a, int n)
{
// Intialise maximum length to 1
int maximum = 1;
// Intialise left[] to find the
// length of decreasing sequence
// from left to right
int left [] = new int[n];
// Intialise right[] to find the
// length of decreasing sequence
// from right to left
int right[] = new int[n];
// Initially store 1 at each index of
// left and right array
for (int i = 0; i < n; i++) {
left[i] = 1;
right[i] = 1;
}
// Iterate over the array arr[] to
// store length of decreasing
// sequence that can be obtained
// at every index in the right array
for (int i = n - 2; i >= 0; i--) {
if (a[i] > a[i + 1]) {
right[i] = right[i + 1] + 1;
}
// Store the length of longest
// continuous decreasing
// sequence in maximum
maximum = Math.max(maximum, right[i]);
}
// Iterate over the array arr[] to
// store length of decreasing
// sequence that can be obtained
// at every index in the left array
for (int i = 1; i < n; i++) {
if (a[i] < a[i - 1]) {
left[i] = left[i - 1] + 1;
}
}
if (n > 2) {
// Check if we can obtain a
// longer decreasing sequence
// after removal of any element
// from the array arr[] with
// the help of left[] & right[]
for (int i = 1; i < n - 1; i++) {
if (a[i - 1] > a[i + 1]) {
maximum = Math.max(maximum, left[i - 1] + right[i + 1]);
}
}
}
// Return maximum length of sequence
return maximum;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 8, 7, 3, 5, 2, 9 };
int n = arr.length;
// Function calling
System.out.println(maxLength(arr, n));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 program to find maximum length
# of decreasing sequence by removing
# at most one element
# Function to find the maximum length
def maxLength(a, n) :
# Intialise maximum length to 1
maximum = 1;
# Intialise left[] to find the
# length of decreasing sequence
# from left to right
left = [0]*n;
# Intialise right[] to find the
# length of decreasing sequence
# from right to left
right = [0]*n;
# Initially store 1 at each index of
# left and right array
for i in range(n) :
left[i] = 1;
right[i] = 1;
# Iterate over the array arr[] to
# store length of decreasing
# sequence that can be obtained
# at every index in the right array
for i in range(n - 2, -1, -1) :
if (a[i] > a[i + 1]) :
right[i] = right[i + 1] + 1;
# Store the length of longest
# continuous decreasing
# sequence in maximum
maximum = max(maximum, right[i]);
# Iterate over the array arr[] to
# store length of decreasing
# sequence that can be obtained
# at every index in the left array
for i in range(1, n) :
if (a[i] < a[i - 1]) :
left[i] = left[i - 1] + 1;
if (n > 2) :
# Check if we can obtain a
# longer decreasing sequence
# after removal of any element
# from the array arr[] with
# the help of left[] & right[]
for i in range(1, n -1) :
if (a[i - 1] > a[i + 1]) :
maximum = max(maximum, left[i - 1] + right[i + 1]);
# Return maximum length of sequence
return maximum;
# Driver code
if __name__ == "__main__" :
arr = [ 8, 7, 3, 5, 2, 9 ];
n = len(arr);
# Function calling
print(maxLength(arr, n));
# This code is contributed by AnkitRai01C#
// C# program to find maximum length
// of decreasing sequence by removing
// at most one element
using System;
class GFG {
// Function to find the maximum length
static int maxLength(int []a, int n)
{
// Intialise maximum length to 1
int maximum = 1;
// Intialise left[] to find the
// length of decreasing sequence
// from left to right
int []left = new int[n];
// Intialise right[] to find the
// length of decreasing sequence
// from right to left
int []right = new int[n];
// Initially store 1 at each index of
// left and right array
for (int i = 0; i < n; i++) {
left[i] = 1;
right[i] = 1;
}
// Iterate over the array arr[] to
// store length of decreasing
// sequence that can be obtained
// at every index in the right array
for (int i = n - 2; i >= 0; i--) {
if (a[i] > a[i + 1]) {
right[i] = right[i + 1] + 1;
}
// Store the length of longest
// continuous decreasing
// sequence in maximum
maximum = Math.Max(maximum, right[i]);
}
// Iterate over the array arr[] to
// store length of decreasing
// sequence that can be obtained
// at every index in the left array
for (int i = 1; i < n; i++) {
if (a[i] < a[i - 1]) {
left[i] = left[i - 1] + 1;
}
}
if (n > 2) {
// Check if we can obtain a
// longer decreasing sequence
// after removal of any element
// from the array arr[] with
// the help of left[] & right[]
for (int i = 1; i < n - 1; i++) {
if (a[i - 1] > a[i + 1]) {
maximum = Math.Max(maximum, left[i - 1] + right[i + 1]);
}
}
}
// Return maximum length of sequence
return maximum;
}
// Driver code
public static void Main (String[] args)
{
int []arr = { 8, 7, 3, 5, 2, 9 };
int n = arr.Length;
// Function calling
Console.WriteLine(maxLength(arr, n));
}
}
// This code is contributed by AnkitRai01输出：
4
时间复杂度： O(n)
辅助空间： O(n)