Valid Palindrome III in C++

Suppose we have a string s and another number k; we have to check whether the given string is a K-Palindrome or not.

A string is said to be K-Palindrome if it can be transformed into a palindrome by removing at most k characters from it.

So, if the input is like s = "abcdeca", k = 2, then the output will be true as by removing 'b' and 'e' characters, it will be palindrome.

To solve this, we will follow these steps −

• Define a function lcs(), this will take s, t,

• n := size of s

• add one blank space before s

• add one blank space before t

• Define one 2D array dp of size (n + 1) x (n + 1)

• for initialize i := 1, when i <= n, update (increase i by 1), do −

  • for initialize j := 1, when j <= n, update (increase j by 1), do −

    • dp[i, j] := maximum of dp[i - 1, j] and dp[i, j - 1]

    • if s[i] is same as t[j], then −

      • dp[i, j] := maximum of dp[i, j] and 1 + dp[i - 1, j - 1]

• return dp[n, n]

• From the main method do the following −

• if not size of s, then −

  • return true

• x := blank space

• for initialize i := size of s, when i >= 0, update (decrease i by 1), do −

  • x := x + s[i]

• return size of s

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; class Solution {    public:    int lcs(string s, string t){       int n = s.size();       s = " " + s;       t = " " + t;       vector<vector<int> > dp(n + 1, vector<int>(n + 1));       for (int i = 1; i <= n; i++) {          for (int j = 1; j <= n; j++) {             dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);             if (s[i] == t[j])             dp[i][j] = max(dp[i][j], 1 + dp[i - 1][j - 1]);          }       }       return dp[n][n];    }    bool isValidPalindrome(string s, int k) {       if (!s.size())       return true;       string x = "";       for (int i = s.size() - 1; i >= 0; i--)          x += s[i];       return s.size() - lcs(s, x) <= k;    } }; main(){    Solution ob;    cout << (ob.isValidPalindrome("abcdeca", 2)); }

Input

"abcdeca", 2

Output

1