# 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.```

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;`
`            ``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;`
`}`

Output:

```2

```

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits:http://www.geeksforgeeks.org/minimum-steps-to-delete-a-string-after-repeated-deletion-of-palindrome-substrings/
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