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.

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.

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++ 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;
                // 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],
    /* 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;



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