// C++ program to find if given string is K-Palindrome // or not #include    using namespace std; /* Returns length of LCS for X[0..m-1] Y[0..n-1] */ int lcs( string X string Y int m int n ) {  int L[m + 1][n + 1];  /* Following steps build L[m+1][n+1] in bottom up  fashion. Note that L[i][j] contains length of  LCS of X[0..i-1] and Y[0..j-1] */  for (int i = 0; i <= m; i++)  {  for (int j = 0; j <= n; j++)  {  if (i == 0 || j == 0)  L[i][j] = 0;  else if (X[i - 1] == Y[j - 1])  L[i][j] = L[i - 1][j - 1] + 1;  else  L[i][j] = max(L[i - 1][j] L[i][j - 1]);  }  }  // L[m][n] contains length of LCS for X and Y  return L[m][n]; } // find if given string is K-Palindrome or not bool isKPal(string str int k) {  int n = str.length();  // Find reverse of string  string revStr = str;  reverse(revStr.begin() revStr.end());  // find longest palindromic subsequence of  // given string  int lps = lcs(str revStr n n);  // If the difference between longest palindromic  // subsequence and the original string is less  // than equal to k then the string is k-palindrome  return (n - lps <= k); } // Driver program int main() {  string str = 'abcdeca';  int k = 2;  isKPal(str k) ? cout << 'Yes' : cout << 'No';  return 0; } 

// Java program to find if given  // String is K-Palindrome or not import java.util.*; import java.io.*; class GFG  {  /* Returns length of LCS for  X[0..m-1] Y[0..n-1] */  static int lcs(String X String Y  int m int n)   {  int L[][] = new int[m + 1][n + 1];  /* Following steps build L[m+1][n+1]  in bottom up fashion. Note that L[i][j]   contains length of LCS of X[0..i-1]  and Y[0..j-1] */  for (int i = 0; i <= m; i++)  {  for (int j = 0; j <= n; j++)   {  if (i == 0 || j == 0)   {  L[i][j] = 0;  }   else if (X.charAt(i - 1) == Y.charAt(j - 1))  {  L[i][j] = L[i - 1][j - 1] + 1;  }   else  {  L[i][j] = Math.max(L[i - 1][j] L[i][j - 1]);  }  }  }  // L[m][n] contains length   // of LCS for X and Y   return L[m][n];  }  // find if given String is  // K-Palindrome or not   static boolean isKPal(String str int k)   {  int n = str.length();  // Find reverse of String   StringBuilder revStr = new StringBuilder(str);  revStr = revStr.reverse();  // find longest palindromic   // subsequence of given String   int lps = lcs(str revStr.toString() n n);  // If the difference between longest   // palindromic subsequence and the   // original String is less than equal   // to k then the String is k-palindrome   return (n - lps <= k);  }  // Driver code   public static void main(String[] args)   {  String str = 'abcdeca';  int k = 2;  if (isKPal(str k))  {  System.out.println('Yes');  }  else  System.out.println('No');  } } // This code is contributed by Rajput-JI 

# Python program to find # if given string is K-Palindrome # or not # Returns length of LCS # for X[0..m-1] Y[0..n-1]  def lcs(X Y m n ): L = [[0]*(n+1) for _ in range(m+1)] # Following steps build # L[m+1][n+1] in bottom up # fashion. Note that L[i][j] # contains length of # LCS of X[0..i-1] and Y[0..j-1]  for i in range(m+1): for j in range(n+1): if not i or not j: L[i][j] = 0 elif X[i - 1] == Y[j - 1]: L[i][j] = L[i - 1][j - 1] + 1 else: L[i][j] = max(L[i - 1][j] L[i][j - 1]) # L[m][n] contains length # of LCS for X and Y return L[m][n] # find if given string is # K-Palindrome or not def isKPal(string k): n = len(string) # Find reverse of string revStr = string[::-1] # find longest palindromic # subsequence of # given string lps = lcs(string revStr n n) # If the difference between # longest palindromic # subsequence and the original # string is less # than equal to k then # the string is k-palindrome return (n - lps <= k) # Driver program string = 'abcdeca' k = 2 print('Yes' if isKPal(string k) else 'No') # This code is contributed # by Ansu Kumari. 

// C# program to find if given  // String is K-Palindrome or not  using System; class GFG  {   /* Returns length of LCS for   X[0..m-1] Y[0..n-1] */  static int lcs(String X String Y   int m int n)   {   int []L = new int[m + 1n + 1];   /* Following steps build L[m+1n+1]   in bottom up fashion. Note that L[ij]   contains length of LCS of X[0..i-1]   and Y[0..j-1] */  for (int i = 0; i <= m; i++)   {   for (int j = 0; j <= n; j++)   {   if (i == 0 || j == 0)   {   L[i j] = 0;   }   else if (X[i - 1] == Y[j - 1])   {   L[i j] = L[i - 1 j - 1] + 1;   }   else  {   L[i j] = Math.Max(L[i - 1 j]  L[i j - 1]);   }   }   }     // L[mn] contains length   // of LCS for X and Y   return L[m n];   }   // find if given String is   // K-Palindrome or not   static bool isKPal(String str int k)   {   int n = str.Length;   // Find reverse of String   str = reverse(str);   // find longest palindromic   // subsequence of given String   int lps = lcs(str str n n);   // If the difference between longest   // palindromic subsequence and the   // original String is less than equal   // to k then the String is k-palindrome   return (n - lps <= k);   }   static String reverse(String input)  {  char[] temparray = input.ToCharArray();  int left right = 0;  right = temparray.Length - 1;  for (left = 0; left < right; left++ right--)   {    // Swap values of left and right   char temp = temparray[left];  temparray[left] = temparray[right];  temparray[right] = temp;  }  return String.Join(''temparray);  }    // Driver code   public static void Main(String[] args)   {   String str = 'abcdeca';   int k = 2;   if (isKPal(str k))   {   Console.WriteLine('Yes');   }   else  Console.WriteLine('No');   }  }  // This code is contributed by PrinciRaj1992 

<script> // JavaScript program to find // if given string is K-Palindrome // or not // Returns length of LCS // for X[0..m-1] Y[0..n-1]  function lcs(X Y m n ){  let L = new Array(m+1);  for(let i=0;i<m+1;i++){  L[i] = new Array(n+1).fill(0);  }  // Following steps build  // L[m+1][n+1] in bottom up  // fashion. Note that L[i][j]  // contains length of  // LCS of X[0..i-1] and Y[0..j-1]   for(let i = 0; i < m + 1; i++)  {  for(let j = 0; j < n + 1; j++)  {  if(!i || !j)  L[i][j] = 0  else if(X[i - 1] == Y[j - 1])  L[i][j] = L[i - 1][j - 1] + 1  else  L[i][j] = Math.max(L[i - 1][j] L[i][j - 1])  }  }  // L[m][n] contains length  // of LCS for X and Y  return L[m][n] } // find if given string is // K-Palindrome or not function isKPal(string k){  let n = string.length  // Find reverse of string  let revStr = string.split('').reverse().join('')  // find longest palindromic  // subsequence of  // given string  let lps = lcs(string revStr n n)  // If the difference between  // longest palindromic  // subsequence and the original  // string is less  // than equal to k then  // the string is k-palindrome  return (n - lps <= k) } // Driver program let string = 'abcdeca' let k = 2 document.write(isKPal(string k)?'Yes' : 'No') // This code is contributed by shinjanpatra </script>