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.

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

Output:

2

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