给定大小为N的数组arr [] ，任务是找到需要从数组中删除的数组元素的最小数量，以便将给定的数组转换为双音数组。
例子：
Input: arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 }
Output: 3
Explanation:
Removing arr[0], arr[1] and arr[5] modifies arr[] to { 1, 5, 6, 3, 1 }
Since the array elements follow an increasing order followed by a decreasing order, the required output is 3.
Input: arr[] = { 1, 3, 1 }
Output: 0
Explanation:
The given array is already a bitonic array. Therefore, the required output is 3.
方法：可以根据最长增长子序列问题的概念来解决该问题。请按照以下步骤解决问题：
- 初始化一个变量，例如left [] ，以使left [i]存储最长增长的子序列的长度，直到第i个索引。
- 初始化一个变量，例如right [] ，使right [i]在范围[i，N]上存储最长递减子序列的长度。
- 使用变量i遍历left []和right []数组，找到(left [i] + right [i] – 1)的最大值，并将其存储在变量maxLen中。
- 最后，打印N – maxLen的值。
下面是上述方法的实现：
C++14
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to coutnt minimum array elements
// required to be removed to make an array bitonic
void min_element_removal(int arr[], int N)
{
// left[i]: Stores the length
// of LIS up to i-th index
vector left(N, 1);
// right[i]: Stores the length
// of decreasing subsequence
// over the range [i, N]
vector right(N, 1);
// Calculate the length of LIS
// up to i-th index
for (int i = 1; i < N; i++) {
// Traverse the array
// upto i-th index
for (int j = 0; j < i; j++) {
// If arr[j] is less than arr[i]
if (arr[j] < arr[i]) {
// Update left[i]
left[i] = max(left[i],
left[j] + 1);
}
}
}
// Calculate the length of decreasing
// subsequence over the range [i, N]
for (int i = N - 2; i >= 0;
i--) {
// Traverse right[] array
for (int j = N - 1; j > i;
j--) {
// If arr[i] is greater
// than arr[j]
if (arr[i] > arr[j]) {
// Update right[i]
right[i] = max(right[i],
right[j] + 1);
}
}
}
// Stores length of the
// longest bitonic array
int maxLen = 0;
// Traverse left[] and right[] array
for (int i = 1; i < N - 1; i++) {
// Update maxLen
maxLen = max(maxLen, left[i] + right[i] - 1);
}
cout << (N - maxLen) << "\n";
}
// Function to print minimum removals
// required to make given array bitonic
void makeBitonic(int arr[], int N)
{
if (N == 1) {
cout << "0" << endl;
return;
}
if (N == 2) {
if (arr[0] != arr[1])
cout << "0" << endl;
else
cout << "1" << endl;
return;
}
min_element_removal(arr, N);
}
// Driver Code
int main()
{
int arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
makeBitonic(arr, N);
return 0;
} Java
// Java program to implement
// the above approach
class GFG {
// Function to coutnt minimum array elements
// required to be removed to make an array bitonic
static void min_element_removal(int arr[], int N)
{
// left[i]: Stores the length
// of LIS up to i-th index
int left[] = new int[N];
for(int i = 0; i < N; i++)
left[i] = 1;
// right[i]: Stores the length
// of decreasing subsequence
// over the range [i, N]
int right[] = new int[N];
for(int i = 0; i < N; i++)
right[i] = 1;
// Calculate the length of LIS
// up to i-th index
for (int i = 1; i < N; i++) {
// Traverse the array
// upto i-th index
for (int j = 0; j < i; j++) {
// If arr[j] is less than arr[i]
if (arr[j] < arr[i]) {
// Update left[i]
left[i] = Math.max(left[i],
left[j] + 1);
}
}
}
// Calculate the length of decreasing
// subsequence over the range [i, N]
for (int i = N - 2; i >= 0;
i--) {
// Traverse right[] array
for (int j = N - 1; j > i;
j--) {
// If arr[i] is greater
// than arr[j]
if (arr[i] > arr[j]) {
// Update right[i]
right[i] = Math.max(right[i],
right[j] + 1);
}
}
}
// Stores length of the
// longest bitonic array
int maxLen = 0;
// Traverse left[] and right[] array
for (int i = 1; i < N - 1; i++) {
// Update maxLen
maxLen = Math.max(maxLen, left[i] + right[i] - 1);
}
System.out.println(N - maxLen);
}
// Function to print minimum removals
// required to make given array bitonic
static void makeBitonic(int arr[], int N)
{
if (N == 1) {
System.out.println("0");
return;
}
if (N == 2) {
if (arr[0] != arr[1])
System.out.println("0");
else
System.out.println("1");
return;
}
min_element_removal(arr, N);
}
// Driver Code
public static void main (String[] args) {
int arr[] = { 2, 1, 1, 5, 6, 2, 3, 1 };
int N = arr.length;
makeBitonic(arr, N);
}
}
// This code is contributed by AnkitRai01Python3
# Python3 program to implement
# the above approach
# Function to coutnt minimum array elements
# required to be removed to make an array bitonic
def min_element_removal(arr, N):
# left[i]: Stores the length
# of LIS up to i-th index
left = [1] * N
# right[i]: Stores the length
# of decreasing subsequence
# over the range [i, N]
right = [1] * (N)
#Calculate the length of LIS
#up to i-th index
for i in range(1, N):
#Traverse the array
#upto i-th index
for j in range(i):
#If arr[j] is less than arr[i]
if (arr[j] < arr[i]):
#Update left[i]
left[i] = max(left[i], left[j] + 1)
# Calculate the length of decreasing
# subsequence over the range [i, N]
for i in range(N - 2, -1, -1):
# Traverse right[] array
for j in range(N - 1, i, -1):
# If arr[i] is greater
# than arr[j]
if (arr[i] > arr[j]):
# Update right[i]
right[i] = max(right[i], right[j] + 1)
# Stores length of the
# longest bitonic array
maxLen = 0
# Traverse left[] and right[] array
for i in range(1, N - 1):
# Update maxLen
maxLen = max(maxLen, left[i] + right[i] - 1)
print((N - maxLen))
# Function to prminimum removals
# required to make given array bitonic
def makeBitonic(arr, N):
if (N == 1):
print("0")
return
if (N == 2):
if (arr[0] != arr[1]):
print("0")
else:
print("1")
return
min_element_removal(arr, N)
# Driver Code
if __name__ == '__main__':
arr=[2, 1, 1, 5, 6, 2, 3, 1]
N = len(arr)
makeBitonic(arr, N)
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to coutnt minimum array elements
// required to be removed to make an array bitonic
static void min_element_removal(int []arr, int N)
{
// left[i]: Stores the length
// of LIS up to i-th index
int []left = new int[N];
for(int i = 0; i < N; i++)
left[i] = 1;
// right[i]: Stores the length
// of decreasing subsequence
// over the range [i, N]
int []right = new int[N];
for(int i = 0; i < N; i++)
right[i] = 1;
// Calculate the length of LIS
// up to i-th index
for(int i = 1; i < N; i++)
{
// Traverse the array
// upto i-th index
for(int j = 0; j < i; j++)
{
// If arr[j] is less than arr[i]
if (arr[j] < arr[i])
{
// Update left[i]
left[i] = Math.Max(left[i],
left[j] + 1);
}
}
}
// Calculate the length of decreasing
// subsequence over the range [i, N]
for(int i = N - 2; i >= 0; i--)
{
// Traverse right[] array
for(int j = N - 1; j > i; j--)
{
// If arr[i] is greater
// than arr[j]
if (arr[i] > arr[j])
{
// Update right[i]
right[i] = Math.Max(right[i],
right[j] + 1);
}
}
}
// Stores length of the
// longest bitonic array
int maxLen = 0;
// Traverse left[] and right[] array
for(int i = 1; i < N - 1; i++)
{
// Update maxLen
maxLen = Math.Max(maxLen, left[i] +
right[i] - 1);
}
Console.WriteLine(N - maxLen);
}
// Function to print minimum removals
// required to make given array bitonic
static void makeBitonic(int []arr, int N)
{
if (N == 1)
{
Console.WriteLine("0");
return;
}
if (N == 2)
{
if (arr[0] != arr[1])
Console.WriteLine("0");
else
Console.WriteLine("1");
return;
}
min_element_removal(arr, N);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 2, 1, 1, 5, 6, 2, 3, 1 };
int N = arr.Length;
makeBitonic(arr, N);
}
}
// This code is contributed by AnkThon输出：
3时间复杂度： O(N 2 )
辅助空间： O(N)