从 Array 中删除最大值和最小值所需的从前面或后面的最小删除

给定一个由整数组成的数组arr[] 。任务是找到从arr[]中删除初始最小和最大元素所需的最小删除。
注意：可以从正面或背面执行删除 的数组。

例子：

Input: arr[] = {5, 7, 2, 4, 3}
Output: 3
Explanation: Initial minimum = 2, Initial maximum = 7
Deleting first 3 from arr[] updates arr[] to {2, 4, 3} which is free from Initial maximum and minimum elements.
Therefore 3 operations are required which is minimum possible.

Input: arr[] = {2, -1, 3, 5, 8, -7}
Output: 2

方法：给定的问题可以通过使用贪心方法来解决。请按照以下步骤解决给定的问题。

• 初始化两个变量mn ， mx以分别存储最小和最大元素。
• 使用两个变量minIndex、maxIndex来存储最后出现的最小和最大元素。
• 用i迭代arr[]
  • 更新mn = min(mn, arr[i])
  • 更新mx = max(mx, arr[i])
• 使用两个变量 minIndex、maxIndex 来存储最后出现的最小和最大元素。
• 用 i 迭代 arr[]
  • 如果arr[i] = mn ，则更新minIndex = i
  • 如果arr[i] = mx ，则更新maxIndex = i
• 计算所有删除的情况并存储在变量x, y, z中。
• 返回x, y, z中的最小值作为最终答案。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate minimum deletions to
// remove initial minimum and maximum
int minDeletions(int *arr, int N)
{
    // To store initial minimum and maximum
    int mn = INT_MAX, mx = INT_MIN;
 
    // Iterate and find min and max in arr[]
    for(int i = 0; i < N; i++) {
        mn = min(mn, arr[i]);
        mx = max(mx, arr[i]);
    }
 
    // To store indices of last min and max
    int minIndex, maxIndex;
    for(int i = 0; i < N; i++) {
        if(arr[i] == mn) minIndex = i;
        if(arr[i] == mx) maxIndex = i;
    }
 
 
    int temp = max(minIndex, maxIndex);
    minIndex = min(minIndex, maxIndex);
    maxIndex = temp;
 
    // Calculating all possible case of
    // deletion operations
    int x = N - maxIndex + minIndex + 1;
    int y = N - minIndex;
    int z = maxIndex + 1;
 
    // Return minimum among all the three cases
    return min({x, y, z});
}
 
// Driver Code
int main()
{
    int N = 6;
    int arr[] = {2, -1, 9, 7, -2, 3};
 
    cout << minDeletions(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
   
    // Function to calculate minimum deletions to
    // remove initial minimum and maximum
    static int minDeletions(int arr[], int N)
    {
       
        // To store initial minimum and maximum
        int mn = Integer.MAX_VALUE, mx = Integer.MIN_VALUE;
 
        // Iterate and find min and max in arr[]
        for (int i = 0; i < N; i++) {
            mn = Math.min(mn, arr[i]);
            mx = Math.max(mx, arr[i]);
        }
 
        // To store indices of last min and max
        int minIndx = 0, maxIndx = 0;
        for (int i = 0; i < N; i++) {
            if (arr[i] == mn)
                minIndx = i;
            if (arr[i] == mx)
                maxIndx = i;
        }
 
        int temp = Math.max(minIndx, maxIndx);
        minIndx = Math.min(minIndx, maxIndx);
        maxIndx = temp;
 
        // Calculating all possible case of
        // deletion operations
        int x = N - maxIndx + minIndx + 1;
        int y = N - minIndx;
        int z = maxIndx + 1;
 
        // Return minimum among all the three cases
        return Math.min(x, Math.min(y, z));
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 2, -1, 9, 7, -2, 3 };
        int N = 6;
        System.out.println(minDeletions(arr, N));
    }
}
 
// This code is contributed by dwivediyash

Python3

# python program for the above approach
INT_MIN = -2147483648
INT_MAX = 2147483647
 
# Function to calculate minimum deletions to
# remove initial minimum and maximum
def minDeletions(arr, N):
 
        # To store initial minimum and maximum
    mn = INT_MAX
    mx = INT_MIN
 
    # Iterate and find min and max in arr[]
    for i in range(0, N):
        mn = min(mn, arr[i])
        mx = max(mx, arr[i])
 
        # To store indices of last min and max
    minIndex = 0
    maxIndex = 0
    for i in range(0, N):
        if(arr[i] == mn):
            minIndex = i
        if(arr[i] == mx):
            maxIndex = i
 
    temp = max(minIndex, maxIndex)
    minIndex = min(minIndex, maxIndex)
    maxIndex = temp
 
    # Calculating all possible case of
    # deletion operations
    x = N - maxIndex + minIndex + 1
    y = N - minIndex
    z = maxIndex + 1
 
    # Return minimum among all the three cases
    return min({x, y, z})
 
# Driver Code
if __name__ == "__main__":
 
    N = 6
    arr = [2, -1, 9, 7, -2, 3]
 
    print(minDeletions(arr, N))
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
class GFG
{
   
    // Function to calculate minimum deletions to
    // remove initial minimum and maximum
    static int minDeletions(int[] arr, int N)
    {
       
        // To store initial minimum and maximum
        int mn = int.MaxValue, mx = int.MinValue;
 
        // Iterate and find min and max in arr[]
        for (int i = 0; i < N; i++) {
            mn = Math.Min(mn, arr[i]);
            mx = Math.Max(mx, arr[i]);
        }
 
        // To store indices of last min and max
        int minIndx = 0, maxIndx = 0;
        for (int i = 0; i < N; i++) {
            if (arr[i] == mn)
                minIndx = i;
            if (arr[i] == mx)
                maxIndx = i;
        }
 
        int temp = Math.Max(minIndx, maxIndx);
        minIndx = Math.Min(minIndx, maxIndx);
        maxIndx = temp;
 
        // Calculating all possible case of
        // deletion operations
        int x = N - maxIndx + minIndx + 1;
        int y = N - minIndx;
        int z = maxIndx + 1;
 
        // Return minimum among all the three cases
        return Math.Min(x, Math.Min(y, z));
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 2, -1, 9, 7, -2, 3 };
        int N = 6;
        Console.Write(minDeletions(arr, N));
    }
}
 
// This code is contributed by gfgking

Javascript

// Javascript program for the above approach
 
// Function to calculate minimum deletions to
// remove initial minimum and maximum
function minDeletions(arr, N)
{
 
  // To store initial minimum and maximum
  let mn = Number.MAX_SAFE_INTEGER,
    mx = Number.MIN_SAFE_INTEGER;
 
  // Iterate and find min and max in arr[]
  for (let i = 0; i < N; i++) {
    mn = Math.min(mn, arr[i]);
    mx = Math.max(mx, arr[i]);
  }
 
  // To store indices of last min and max
  let minIndex, maxIndex;
  for (let i = 0; i < N; i++) {
    if (arr[i] == mn) minIndex = i;
    if (arr[i] == mx) maxIndex = i;
  }
 
  let temp = Math.max(minIndex, maxIndex);
  minIndex = Math.min(minIndex, maxIndex);
  maxIndex = temp;
 
  // Calculating all possible case of
  // deletion operations
  let x = N - maxIndex + minIndex + 1;
  let y = N - minIndex;
  let z = maxIndex + 1;
 
  // Return minimum among all the three cases
  return Math.min(x, Math.min(y, z));
}
 
// Driver Code
 
let N = 6;
let arr = [2, -1, 9, 7, -2, 3];
 
document.write(minDeletions(arr, N));
 
// This code is contributed by gfgking.

输出

4

时间复杂度： 在）

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