制作给定数组 Bitonic 所需的最小移除量

给定一个大小为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：该问题可以基于最长递增子序列问题的概念来解决。请按照以下步骤解决问题：

• 初始化一个变量，比如left[] ，这样left[i]存储最长递增子序列的长度，直到第i^个索引。
• 初始化一个变量，比如right[] ，这样right[i]存储了[i, N]范围内最长递减子序列的长度。
• 使用变量i遍历left[]和right[]数组并找到(left[i] + right[i] – 1)的最大值并将其存储在一个变量中，例如maxLen 。
• 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 AnkitRai01

Python3

# 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 29

C#

// 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

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to coutnt minimum array elements
// required to be removed to make an array bitonic
function min_element_removal(arr, N)
{
    // left[i]: Stores the length
    // of LIS up to i-th index
    var left = Array(N).fill(1);
 
    // right[i]: Stores the length
    // of decreasing subsequence
    // over the range [i, N]
    var right = Array(N).fill(1);
 
    // Calculate the length of LIS
    // up to i-th index
    for (var i = 1; i < N; i++) {
 
        // Traverse the array
        // upto i-th index
        for (var 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 (var i = N - 2; i >= 0;
         i--) {
 
        // Traverse right[] array
        for (var 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
    var maxLen = 0;
 
    // Traverse left[] and right[] array
    for (var i = 1; i < N - 1; i++) {
 
        // Update maxLen
        maxLen = Math.max(maxLen, left[i] + right[i] - 1);
    }
 
    document.write((N - maxLen) + "<br>");
}
 
// Function to print minimum removals
// required to make given array bitonic
function makeBitonic(arr, N)
{
    if (N == 1) {
        document.write( "0" + "<br>");
        return;
    }
 
    if (N == 2) {
        if (arr[0] != arr[1])
            document.write( "0" + "<br>");
        else
            document.write( "1" + "<br>");
 
        return;
    }
 
    min_element_removal(arr, N);
}
 
// Driver Code
var arr = [2, 1, 1, 5, 6, 2, 3, 1];
var N = arr.length;
makeBitonic(arr, N);
 
// This code is contributed by rutvik_56.

输出：

3

时间复杂度： O(N ² )
辅助空间： O(N)

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