Minimum palindromic subarray removals to make array Empty

0 0 5 minutes read

Given an array arr[] consisting of N elements, the task is to find the minimum palindromic subarray removals required to remove all elements from the array.
Examples:

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.

Approach:
We can use Interval Dynamic Programming to solve this problem. Follow the steps below to solve the problem:

Initialize dp[][] such that every dp[i][j] represents the minimum number of removals required from the i^th position to the j^th position.
For the interval from i to j, the answer may be the sum of the two intervals from i to k and k + 1 to j, that is:

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

In addition to this possibility, we need to check if arr[i] = arr[j], then dp[i][j] = dp[i + 1][j – 1].

Below is the implementation of the above approach:

C++

// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
int dp[550][550];
int minSubarrayRemoval(vector<int>& 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<int> 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

Javascript

<script>
 
// JavaScript program for the above approach
 
let dp = new Array(550);
// Loop to create 2D array using 1D array
for (var i = 0; i < dp.length; i++) {
    dp[i] = new Array(2);
}
   
function minSubarrayRemoval(arr)
{
    let i, j, k, l;
    let 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
  
    let arr = [ 2, 3, 1, 2, 2, 1, 2 ];
    let ans = minSubarrayRemoval(arr);
       
    document.write(ans);
               
</script>

Output

Time Complexity: O(n³), where n is the length of the array.

Auxiliary Space: O(n²)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!