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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
#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
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