最小回文子数组删除以使数组为空

给定一个由N个元素组成的数组arr [] ，任务是找到从数组中删除所有元素所需的最小回文子数组删除量。

例子：

Input: arr[] = {1, 3, 4, 1, 5}, N = 5
Output: 3
Explanation:
Removal of 4 from the array leaves {1, 3, 1, 5}.
Removal of {1, 3, 1} leaves {5}.
Removal of 5 makes the array empty.

Input: arr[] = {1, 2, 3, 5, 3, 1}, N = 5
Output: 2
Explanation:
Removal of {3, 5, 3} leaves {1, 2, 1}.
Removal of {1, 2, 1} makes the array empty.

方法：
我们可以使用间隔动态规划来解决此问题。请按照以下步骤解决问题：

初始化dp [] [] ，使每个dp [i] [j]代表从第i^个位置到第j^个位置的最小清除次数。

对于从i到j的间隔，答案可能是从i到k和k +1到j的两个间隔之和，即：

dp [i][j]= minimum (dp [i][j], dp [i][k] + dp [k + 1][j]) where i ≤ k

除了这种可能性之外，我们还需要检查arr [i] = arr [j] ，然后dp [i] [j] = dp [i + 1] [j – 1] 。

下面是上述方法的实现：

C++

// C++ Program for the above approach
#include 
using namespace std;
  
int dp[550][550];
int minSubarrayRemoval(vector& arr)
{
    int i, j, k, l;
    int n = arr.size();
  
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            dp[i][j] = n;
        }
    }
  
    for (i = 0; i < n; i++) {
        dp[i][i] = 1;
    }
  
    for (i = 0; i < n - 1; i++) {
        if (arr[i] == arr[i + 1]) {
            dp[i][i + 1] = 1;
        }
        else {
            dp[i][i + 1] = 2;
        }
    }
    for (l = 2; l < n; l++) {
        for (i = 0; i + l < n; i++) {
            j = i + l;
            if (arr[i] == arr[j]) {
                dp[i][j] = dp[i + 1][j - 1];
            }
            for (k = i; k < j; k++) {
                dp[i][j]
                    = min(
                        dp[i][j],
                        dp[i][k]
                            + dp[k + 1][j]);
            }
        }
    }
    return dp[0][n - 1];
}
// Driver Program
int main()
{
    vector arr
        = { 2, 3, 1, 2, 2, 1, 2 };
    int ans = minSubarrayRemoval(arr);
    cout << ans << endl;
}

Java

// Java program for the above approach
class GFG{ 
      
static int dp[][] = new int[550][550];
  
static int minSubarrayRemoval(int arr[])
{
    int i, j, k, l;
    int n = arr.length;
  
    for(i = 0; i < n; i++) 
    {
       for(j = 0; j < n; j++)
       {
          dp[i][j] = n;
       }
    }
  
    for(i = 0; i < n; i++)
    {
       dp[i][i] = 1;
    }
  
    for(i = 0; i < n - 1; i++)
    {
       if (arr[i] == arr[i + 1])
       {
           dp[i][i + 1] = 1;
       }
       else 
       {
           dp[i][i + 1] = 2;
       }
    }
      
    for(l = 2; l < n; l++)
    {
       for(i = 0; i + l < n; i++)
       {
          j = i + l;
          if (arr[i] == arr[j])
          {
              dp[i][j] = dp[i + 1][j - 1];
          }
          for(k = i; k < j; k++)
          {
             dp[i][j] = Math.min(dp[i][j], 
                                 dp[i][k] +
                                 dp[k + 1][j]);
          }
       }
    }
    return dp[0][n - 1];
}
      
// Driver code 
public static void main (String[] args) 
{ 
    int arr [] = new int[]{ 2, 3, 1, 2, 2, 1, 2 };
    int ans = minSubarrayRemoval(arr);
      
    System.out.println(ans);
} 
} 
  
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach
def minSubarrayRemoval(arr):
      
    n = len(arr)
    dp = []
      
    for i in range(n):
        l = [0] * n
        for j in range(n):
            l[j] = n
        dp.append(l)
      
    for i in range(n):
        dp[i][i] = 1
          
    for i in range(n - 1):
        if (arr[i] == arr[i + 1]):
            dp[i][i + 1] = 1
        else:
            dp[i][i + 1] = 2
              
    for l in range(2, n):
        for i in range(n - l):
            j = i + l
            if (arr[i] == arr[j]):
                dp[i][j] = dp[i + 1][j - 1]
              
            for k in range(i, j):
                dp[i][j] = min(dp[i][j], 
                               dp[i][k] + 
                               dp[k + 1][j])
      
    return dp[0][n - 1]
  
# Driver code
arr = [ 2, 3, 1, 2, 2, 1, 2 ]
ans = minSubarrayRemoval(arr)
  
print(ans)
  
# This code is contributed by shubhamsingh10

C#

// C# program for the above approach
using System;
class GFG{ 
      
static int [,]dp = new int[550, 550];
  
static int minSubarrayRemoval(int []arr)
{
    int i, j, k, l;
    int n = arr.Length;
  
    for(i = 0; i < n; i++) 
    {
       for(j = 0; j < n; j++)
       {
          dp[i, j] = n;
       }
    }
  
    for(i = 0; i < n; i++)
    {
       dp[i, i] = 1;
    }
  
    for(i = 0; i < n - 1; i++)
    {
       if (arr[i] == arr[i + 1])
       {
           dp[i, i + 1] = 1;
       }
       else
       {
           dp[i, i + 1] = 2;
       }
    }
      
    for(l = 2; l < n; l++)
    {
       for(i = 0; i + l < n; i++)
       {
          j = i + l;
          if (arr[i] == arr[j])
          {
              dp[i, j] = dp[i + 1, j - 1];
          }
          for(k = i; k < j; k++)
          {
             dp[i, j] = Math.Min(dp[i, j], 
                                 dp[i, k] +
                                 dp[k + 1, j]);
          }
       }
    }
    return dp[0, n - 1];
}
      
// Driver code 
public static void Main() 
{ 
    int []arr = new int[]{ 2, 3, 1, 2, 2, 1, 2 };
    int ans = minSubarrayRemoval(arr);
      
    Console.Write(ans);
} 
} 
  
// This code is contributed by Code_Mech

输出：

2