Find if string is K-Palindrome or not | Set 1
Last Updated :
23 Jul, 2025
Given a string S, find out if the string is K-Palindrome or not. A K-palindrome string transforms into a palindrome on removing at most K characters from it.
Examples :
Input: S = "abcdecba", k = 1
Output: Yes
Explanation: String can become palindrome by removing 1 character i.e. either d or e.
Input: S = "abcdeca", K = 2
Output: Yes
Explanation: Can become palindrome by removing 2 characters b and e.
Input: S= "acdcb", K = 1
Output: No
Explanation: String can not become palindrome by removing only one character.
If we carefully analyze the problem, the task is to transform the given string into its reverse by removing at most K characters from it. The problem is basically a variation of Edit Distance. We can modify the Edit Distance problem to consider the given string and its reverse as input and the only operation allowed is deletion. Since the given string is compared with its reverse, we will do at most N deletions from the first string and N deletions from the second string to make them equal. Therefore, for a string to be k-palindrome, 2*N <= 2*K should hold true.
Below are the detailed steps of the algorithm:
Process all characters one by one starting from either the left or right sides of both strings. Let us traverse from the right corner, there are two possibilities for every pair of characters being traversed.
- If the last characters of the two strings are same, we ignore last characters and get count for remaining strings. So we recur for lengths m-1 and n-1 where m is length of str1 and n is length of str2.
- If the last characters are not same, we consider removing operation on last character of first string and last character of the second string, recursively compute minimum cost for the operations and take minimum of two values.
- Remove last char from str1: Recur for m-1 and n.
- Remove last char from str2: Recur for m and n-1.
Below is the Naive recursive implementation of the above approach:
C++
// A Naive recursive C++ program to find
// if given string is K-Palindrome or not
#include<bits/stdc++.h>
using namespace std;
// find if given string is K-Palindrome or not
int isKPalRec(string str1, string str2, int m, int n)
{
// If first string is empty, the only option is to
// remove all characters of second string
if (m == 0) return n;
// If second string is empty, the only option is to
// remove all characters of first string
if (n == 0) return m;
// If last characters of two strings are same, ignore
// last characters and get count for remaining strings.
if (str1[m-1] == str2[n-1])
return isKPalRec(str1, str2, m-1, n-1);
// If last characters are not same,
// 1. Remove last char from str1 and recur for m-1 and n
// 2. Remove last char from str2 and recur for m and n-1
// Take minimum of above two operations
return 1 + min(isKPalRec(str1, str2, m-1, n), // Remove from str1
isKPalRec(str1, str2, m, n-1)); // Remove from str2
}
// Returns true if str is k palindrome.
bool isKPal(string str, int k)
{
string revStr = str;
reverse(revStr.begin(), revStr.end());
int len = str.length();
return (isKPalRec(str, revStr, len, len) <= k*2);
}
// Driver program
int main()
{
string str = "acdcb";
int k = 2;
isKPal(str, k)? cout << "Yes" : cout << "No";
return 0;
}
// A Naive recursive Java program
// to find if given string is
// K-Palindrome or not
import java.util.*;
import java.io.*;
class GFG
{
// find if given string is
// K-Palindrome or not
static int isKPalRec(String str1,
String str2, int m, int n)
{
// If first string is empty,
// the only option is to remove
// all characters of second string
if (m == 0)
{
return n;
}
// If second string is empty,
// the only option is to remove
// all characters of first string
if (n == 0)
{
return m;
}
// If last characters of two
// strings are same, ignore
// last characters and get
// count for remaining strings.
if (str1.charAt(m - 1) ==
str2.charAt(n - 1))
{
return isKPalRec(str1, str2,
m - 1, n - 1);
}
// If last characters are not same,
// 1. Remove last char from str1 and
// recur for m-1 and n
// 2. Remove last char from str2 and
// recur for m and n-1
// Take minimum of above two operations
return 1 + Math.min(isKPalRec(str1, str2, m - 1, n), // Remove from str1
isKPalRec(str1, str2, m, n - 1)); // Remove from str2
}
// Returns true if str is k palindrome.
static boolean isKPal(String str, int k)
{
String revStr = str;
revStr = reverse(revStr);
int len = str.length();
return (isKPalRec(str, revStr, len, len) <= k * 2);
}
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.valueOf(temparray);
}
// Driver code
public static void main(String[] args)
{
String str = "acdcb";
int k = 2;
if (isKPal(str, k))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by Rajput-Ji
# A Naive recursive Python3
# code to find if given string
# is K-Palindrome or not
# Find if given string
# is K-Palindrome or not
def isKPalRec(str1, str2, m, n):
# If first string is empty,
# the only option is to remove
# all characters of second string
if not m: return n
# If second string is empty,
# the only option is to remove
# all characters of first string
if not n: return m
# If last characters of two strings
# are same, ignore last characters
# and get count for remaining strings.
if str1[m-1] == str2[n-1]:
return isKPalRec(str1, str2, m-1, n-1)
# If last characters are not same,
# 1. Remove last char from str1 and recur for m-1 and n
# 2. Remove last char from str2 and recur for m and n-1
# Take minimum of above two operations
res = 1 + min(isKPalRec(str1, str2, m-1, n), # Remove from str1
(isKPalRec(str1, str2, m, n-1))) # Remove from str2
return res
# Returns true if str is k palindrome.
def isKPal(string, k):
revStr = string[::-1]
l = len(string)
return (isKPalRec(string, revStr, l, l) <= k * 2)
# Driver program
string = "acdcb"
k = 2
print("Yes" if isKPal(string, k) else "No")
# This code is contributed by Ansu Kumari.
// A Naive recursive C# program
// to find if given string is
// K-Palindrome or not
using System;
class GFG
{
// find if given string is
// K-Palindrome or not
static int isKPalRec(String str1,
String str2, int m, int n)
{
// If first string is empty,
// the only option is to remove
// all characters of second string
if (m == 0)
{
return n;
}
// If second string is empty,
// the only option is to remove
// all characters of first string
if (n == 0)
{
return m;
}
// If last characters of two
// strings are same, ignore
// last characters and get
// count for remaining strings.
if (str1[m - 1] ==
str2[n - 1])
{
return isKPalRec(str1, str2,
m - 1, n - 1);
}
// If last characters are not same,
// 1. Remove last char from str1 and
// recur for m-1 and n
// 2. Remove last char from str2 and
// recur for m and n-1
// Take minimum of above two operations
return 1 + Math.Min(isKPalRec(str1, str2, m - 1, n), // Remove from str1
isKPalRec(str1, str2, m, n - 1)); // Remove from str2
}
// Returns true if str is k palindrome.
static bool isKPal(String str, int k)
{
String revStr = str;
revStr = reverse(revStr);
int len = str.Length;
return (isKPalRec(str, revStr, len, len) <= k * 2);
}
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 = "acdcb";
int k = 2;
if (isKPal(str, k))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by 29AjayKumar
<script>
// A Naive recursive javascript program
// to find if given string is
// K-Palindrome or not // find if given string is
// K-Palindrome or not
function isKPalRec( str1, str2 , m , n)
{
// If first string is empty,
// the only option is to remove
// all characters of second string
if (m == 0) {
return n;
}
// If second string is empty,
// the only option is to remove
// all characters of first string
if (n == 0) {
return m;
}
// If last characters of two
// strings are same, ignore
// last characters and get
// count for remaining strings.
if (str1.charAt(m - 1) == str2[n - 1]) {
return isKPalRec(str1, str2, m - 1, n - 1);
}
// If last characters are not same,
// 1. Remove last char from str1 and
// recur for m-1 and n
// 2. Remove last char from str2 and
// recur for m and n-1
// Take minimum of above two operations
return 1 + Math.min(isKPalRec(str1, str2, m - 1, n), // Remove from str1
isKPalRec(str1, str2, m, n - 1)); // Remove from str2
}
// Returns true if str is k palindrome.
function isKPal( str , k) {
var revStr = str;
revStr = reverse(revStr);
var len = str.length;
return (isKPalRec(str, revStr, len, len) <= k * 2);
}
function reverse( input) {
var temparray = input;
var left, right = 0;
right = temparray.length - 1;
for (left = 0; left < right; left++, right--)
{
// Swap values of left and right
var temp = temparray[left];
temparray[left] = temparray[right];
temparray[right] = temp;
}
return temparray;
}
// Driver code
var str = "acdcb";
var k = 2;
if (isKPal(str, k)) {
document.write("Yes");
} else {
document.write("No");
}
// This code is contributed by umadevi9616
</script>
Time complexity: O(2n), It is exponential. In the worst case, we may end up doing O(2n) operations and the worst case happens string contains all distinct characters.
Auxiliary Space: O(1), since no extra space has been taken.
This problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, re-computations of the same subproblems can be avoided by constructing a temporary array that stores the results of subproblems .
Below is a Bottom-up implementation of the above recursive approach :
C++
// C++ program to find if given string is K-Palindrome or not
#include <bits/stdc++.h>
using namespace std;
// find if given string is K-Palindrome or not
int isKPalDP(string str1, string str2, int m, int n)
{
// Create a table to store results of subproblems
int dp[m + 1][n + 1];
// Fill dp[][] in bottom up manner
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
// If first string is empty, only option is to
// remove all characters of second string
if (i == 0)
dp[i][j] = j; // Min. operations = j
// If second string is empty, only option is to
// remove all characters of first string
else if (j == 0)
dp[i][j] = i; // Min. operations = i
// If last characters are same, ignore last character
// and recur for remaining string
else if (str1[i - 1] == str2[j - 1])
dp[i][j] = dp[i - 1][j - 1];
// If last character are different, remove it
// and find minimum
else
dp[i][j] = 1 + min(dp[i - 1][j], // Remove from str1
dp[i][j - 1]); // Remove from str2
}
}
return dp[m][n];
}
// Returns true if str is k palindrome.
bool isKPal(string str, int k)
{
string revStr = str;
reverse(revStr.begin(), revStr.end());
int len = str.length();
return (isKPalDP(str, revStr, len, len) <= k*2);
}
// Driver program
int main()
{
string str = "acdcb";
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
{
// find if given string is
// K-Palindrome or not
static int isKPalDP(String str1,
String str2, int m, int n)
{
// Create a table to store
// results of subproblems
int dp[][] = new int[m + 1][n + 1];
// Fill dp[][] in bottom up manner
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
// If first string is empty,
// only option is to remove all
// characters of second string
if (i == 0)
{
// Min. operations = j
dp[i][j] = j;
}
// If second string is empty,
// only option is to remove all
// characters of first string
else if (j == 0)
{
// Min. operations = i
dp[i][j] = i;
}
// If last characters are same,
// ignore last character and
// recur for remaining string
else if (str1.charAt(i - 1) ==
str2.charAt(j - 1))
{
dp[i][j] = dp[i - 1][j - 1];
}
// If last character are different,
// remove it and find minimum
else
{
// Remove from str1
// Remove from str2
dp[i][j] = 1 + Math.min(dp[i - 1][j],
dp[i][j - 1]);
}
}
}
return dp[m][n];
}
// Returns true if str is k palindrome.
static boolean isKPal(String str, int k)
{
String revStr = str;
revStr = reverse(revStr);
int len = str.length();
return (isKPalDP(str, revStr,
len, len) <= k * 2);
}
static String reverse(String str)
{
StringBuilder sb = new StringBuilder(str);
sb.reverse();
return sb.toString();
}
// Driver program
public static void main(String[] args)
{
String str = "acdcb";
int k = 2;
if (isKPal(str, k))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by
// PrinciRaj1992
# Python program to find if given
# string is K-Palindrome or not
# Find if given string is
# K-Palindrome or not
def isKPalDP(str1, str2, m, n):
# Create a table to store
# results of subproblems
dp = [[0] * (n + 1) for _ in range(m + 1)]
# Fill dp[][] in bottom up manner
for i in range(m + 1):
for j in range(n + 1):
# If first string is empty,
# only option is to remove
# all characters of second string
if not i :
dp[i][j] = j # Min. operations = j
# If second string is empty,
# only option is to remove
# all characters of first string
elif not j :
dp[i][j] = i # Min. operations = i
# If last characters are same,
# ignore last character and
# recur for remaining string
elif (str1[i - 1] == str2[j - 1]):
dp[i][j] = dp[i - 1][j - 1]
# If last character are different,
# remove it and find minimum
else:
dp[i][j] = 1 + min(dp[i - 1][j], # Remove from str1
(dp[i][j - 1])) # Remove from str2
return dp[m][n]
# Returns true if str
# is k palindrome.
def isKPal(string, k):
revStr = string[::-1]
l = len(string)
return (isKPalDP(string, revStr, l, l) <= k * 2)
# Driver program
string = "acdcb"
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 {
// find if given string is
// K-Palindrome or not
static int isKPalDP(string str1,
string str2, int m, int n)
{
// Create a table to store
// results of subproblems
int[,] dp = new int[m + 1, n + 1];
// Fill dp[][] in bottom up manner
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
// If first string is empty,
// only option is to remove all
// characters of second string
if (i == 0)
{
// Min. operations = j
dp[i, j] = j;
}
// If second string is empty,
// only option is to remove all
// characters of first string
else if (j == 0)
{
// Min. operations = i
dp[i, j] = i;
}
// If last characters are same,
// ignore last character and
// recur for remaining string
else if (str1[i - 1] == str2[j - 1])
{
dp[i, j] = dp[i - 1, j - 1];
}
// If last character are different,
// remove it and find minimum
else
{
// Remove from str1
// Remove from str2
dp[i, j] = 1 + Math.Min(dp[i - 1, j], dp[i, j - 1]);
}
}
}
return dp[m, n];
}
// Returns true if str is k palindrome.
static bool isKPal(string str, int k)
{
string revStr = str;
revStr = reverse(revStr);
int len = str.Length;
return (isKPalDP(str, revStr, len, len) <= k * 2);
}
static string reverse(string str)
{
char[] sb = str.ToCharArray();
Array.Reverse(sb);
return new string(sb);
}
static void Main() {
string str = "acdcb";
int k = 2;
if (isKPal(str, k))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by divyesh072019
<script>
// javascript program to find if given
// string is K-Palindrome or not // find if given string is
// K-Palindrome or not
function isKPalDP( str1, str2 , m , n) {
// Create a table to store
// results of subproblems
var dp = Array(m + 1).fill().map(()=>Array(n + 1).fill(0));
// Fill dp in bottom up manner
for (i = 0; i <= m; i++) {
for (j = 0; j <= n; j++) {
// If first string is empty,
// only option is to remove all
// characters of second string
if (i == 0) {
// Min. operations = j
dp[i][j] = j;
}
// If second string is empty,
// only option is to remove all
// characters of first string
else if (j == 0) {
// Min. operations = i
dp[i][j] = i;
}
// If last characters are same,
// ignore last character and
// recur for remaining string
else if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
}
// If last character are different,
// remove it and find minimum
else {
// Remove from str1
// Remove from str2
dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[m][n];
}
// Returns true if str is k palindrome.
function isKPal( str , k) {
var revStr = str;
revStr = reverse(revStr);
var len = str.length;
return (isKPalDP(str, revStr, len, len) <= k * 2);
}
function reverse( str) {
return str.split('').reverse().join('');
}
// Driver program
var str = "acdcb";
var k = 2;
if (isKPal(str, k)) {
document.write("Yes");
} else {
document.write("No");
}
// This code is contributed by Rajput-Ji
</script>
Time complexity: O(n2), Where n is the length of the given string
Auxiliary Space: O(n2), For creating 2D dp array
Alternate Approach :
1) Find Length of the Longest Palindromic Subsequence
2) If the difference between lengths of the original string and the longest palindromic subsequence is less than or equal k, return true. Else return false.
Find if string is K-Palindrome or not | Set 2 (Using LCS)
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