Minimum steps to delete a string after repeated deletion of palindrome substrings

Given a string containing characters as integers only. We need to delete all character of this string in a minimum number of steps where in one step we can delete the substring which is a palindrome. After deleting a substring remaining parts are concatenated.
Examples:

Input : s = “2553432”
Output : 2
We can delete all character of above string in
2 steps, first deleting the substring s[3, 5] “343”  
and then remaining string completely  s[0, 3] “2552”

Input : s = “1234”
Output : 4
We can delete all character of above string in 4 
steps only because each character need to be deleted 
separately. No substring of length 2 is a palindrome 
in above string.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

We can solve this problem using Dynamic programming. Let dp[i][j] denotes the number of steps it takes to delete the substring s[i, j]. Each character will be deleted alone or as part of some substring so in the first case we will delete the character itself and call subproblem (i+1, j). In the second case we will iterate over all occurrence of the current character in right side, if K is the index of one such occurrence then the problem will reduce to two subproblems (i+1, K – 1) and (K+1, j). We can reach to this subproblem (i+1, K-1) because we can just delete the same character and call for mid substring. We need to take care of a case when first two characters are same in that case we can directly reduce to the subproblem (i+2, j)

So after above discussion of relation among subproblems, we can write dp relation as follows,

dp[i][j] = min(1 + dp[i+1][j],
          dp[i+1][K-1] + dp[K+1][j],  where s[i] == s[K]
          1 + dp[i+2][j] )

Total time complexity of above solution is O(n^3)

C++

//  C++ program to find minimum step to delete a string 
#include <bits/stdc++.h> 
using namespace std; 
  
/* method returns minimum step for deleting the string, 
   where in one step a palindrome is removed */
int minStepToDeleteString(string str) 
{ 
    int N = str.length(); 
  
    //  declare dp array and initialize it with 0s 
    int dp[N + 1][N + 1]; 
    for (int i = 0; i <= N; i++) 
        for (int j = 0; j <= N; j++) 
            dp[i][j] = 0; 
  
    // loop for substring length we are considering 
    for (int len = 1; len <= N; len++) 
    { 
        // loop with two variables i and j, denoting 
        // starting and ending of substrings 
        for (int i = 0, j = len - 1; j < N; i++, j++) 
        { 
            // If substring length is 1, then 1 step 
            // will be needed 
            if (len == 1) 
                dp[i][j] = 1; 
            else
            { 
                // delete the ith char individually 
                // and assign result for subproblem (i+1,j) 
                dp[i][j] = 1 + dp[i + 1][j]; 
  
                // if current and next char are same, 
                // choose min from current and subproblem 
                // (i+2,j) 
                if (str[i] == str[i + 1]) 
                    dp[i][j] = min(1 + dp[i + 2][j], dp[i][j]); 
  
                /*  loop over all right characters and suppose 
                    Kth char is same as ith character then 
                    choose minimum from current and two 
                    substring after ignoring ith and Kth char */
                for (int K = i + 2; K <= j; K++) 
                    if (str[i] == str[K]) 
                        dp[i][j] = min(dp[i+1][K-1] + dp[K+1][j], 
                                                       dp[i][j]); 
            } 
        } 
    } 
  
    /* Uncomment below snippet to print actual dp tablex 
    for (int i = 0; i < N; i++, cout << endl) 
        for (int j = 0; j < N; j++) 
            cout << dp[i][j] << " ";    */
  
    return dp[0][N - 1]; 
} 
  
//  Driver code to test above methods 
int main() 
{ 
    string str = "2553432"; 
    cout << minStepToDeleteString(str) << endl; 
    return 0; 
} 

Java

// Java program to find minimum step to  
// delete a string 
public class GFG  
{                             
    /* method returns minimum step for deleting 
       the string, where in one step a 
       palindrome is removed 
     */
    static int minStepToDeleteString(String str) { 
        int N = str.length(); 
  
        // declare dp array and initialize it with 0s 
        int[][] dp = new int[N + 1][N + 1]; 
        for (int i = 0; i <= N; i++) 
            for (int j = 0; j <= N; j++) 
                dp[i][j] = 0; 
  
        // loop for substring length we are considering 
        for (int len = 1; len <= N; len++) { 
              
            // loop with two variables i and j, denoting 
            // starting and ending of substrings 
            for (int i = 0, j = len - 1; j < N; i++, j++) { 
      
                // If substring length is 1, then 1 step 
                // will be needed 
                if (len == 1) 
                    dp[i][j] = 1; 
                      
                else { 
                    // delete the ith char individually 
                    // and assign result for  
                    // subproblem (i+1,j) 
                    dp[i][j] = 1 + dp[i + 1][j]; 
  
                    // if current and next char are same, 
                    // choose min from current and  
                    // subproblem (i+2, j) 
                    if (str.charAt(i) == str.charAt(i + 1)) 
                        dp[i][j] = Math.min(1 + dp[i + 2][j],  
                                               dp[i][j]); 
  
                    /* loop over all right characters and  
                      suppose Kth char is same as ith  
                      character then choose minimum from  
                      current and two substring after  
                      ignoring ith and Kth char 
                     */
                    for (int K = i + 2; K <= j; K++) 
                        if (str.charAt(i) == str.charAt(K)) 
                            dp[i][j] = Math.min( 
                                         dp[i + 1][K - 1] + 
                                        dp[K + 1][j], dp[i][j]); 
                } 
            } 
        } 
  
        /* Uncomment below snippet to print actual dp tablex  
           
           for (int i = 0; i < N; i++){ 
           System.out.println();  
           for (int j = 0; j < N; j++)  
           System.out.print(dp[i][j] + " "); 
           } 
            */
              
        return dp[0][N - 1]; 
    } 
  
    // Driver code to test above methods 
    public static void main(String args[]) { 
        String str = "2553432"; 
        System.out.println(minStepToDeleteString(str)); 
    } 
} 
// This code is contributed by Sumit Ghosh 

C#

// C# program to find minimum step to  
// delete a string 
using System; 
  
class GFG {     
      
    /* method returns minimum step for deleting 
    the string, where in one step a 
    palindrome is removed */
    static int minStepToDeleteString(string str) 
    { 
        int N = str.Length; 
  
        // declare dp array and initialize it  
        // with 0s 
        int [,]dp = new int[N + 1,N + 1]; 
          
        for (int i = 0; i <= N; i++) 
            for (int j = 0; j <= N; j++) 
                dp[i,j] = 0; 
  
        // loop for substring length we are 
        // considering 
        for (int len = 1; len <= N; len++) { 
              
            // loop with two variables i and j, 
            // denoting starting and ending of  
            // substrings 
            for (int i = 0, j = len - 1; j < N; i++, j++) 
            { 
      
                // If substring length is 1, then 1 
                // step will be needed 
                if (len == 1) 
                    dp[i,j] = 1; 
                      
                else 
                { 
                    // delete the ith char individually 
                    // and assign result for  
                    // subproblem (i+1,j) 
                    dp[i,j] = 1 + dp[i + 1,j]; 
  
                    // if current and next char are same, 
                    // choose min from current and  
                    // subproblem (i+2, j) 
                    if (str[i] == str[i + 1]) 
                        dp[i,j] = Math.Min(1 + dp[i + 2,j],  
                                                  dp[i,j]); 
  
                    /* loop over all right characters and  
                    suppose Kth char is same as ith  
                    character then choose minimum from  
                    current and two substring after  
                    ignoring ith and Kth char 
                    */
                    for (int K = i + 2; K <= j; K++) 
                        if (str[i] == str[K]) 
                            dp[i,j] = Math.Min( 
                                        dp[i + 1,K - 1] + 
                                     dp[K + 1,j], dp[i,j]); 
                } 
            } 
        } 
  
        /* Uncomment below snippet to print actual dp tablex  
          
        for (int i = 0; i < N; i++){ 
        System.out.println();  
        for (int j = 0; j < N; j++)  
        System.out.print(dp[i][j] + " "); 
        } */
              
        return dp[0,N - 1]; 
    } 
  
    // Driver code to test above methods 
    public static void Main()  
    { 
        string str = "2553432"; 
        Console.Write(minStepToDeleteString(str)); 
    } 
} 
  
// This code is contributed by nitin mittal 

Output:

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