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Iterate over two strings checking to see if it matches with its pair

I am still new to Java, and I am currently working on a program that will take two strings as arguments and return the number of mismatched pairs. For my program I am working with ATGC because in science, A's always match up with T's and G's always match up with C's. I cant quite figure out how to iterate over the strings and see that the first character in string one (T for example) matches up with its intended pair (A), and if it doesn't it is a mismatched pair and it should be added to a counter to be totaled at the end. I believe I can use something called charAt(), but I am unsure of how that works.
I also need to figure out how to be able to take the absolute value of counter before it is added to the finalCounter. The main reason for this is because I just want to worry about getting the length difference between the two rather than making sure that the longer string is subracted from the smaller string.
Any help would be greatly appreciated!
''''
public class CountMismatches {
public static void main(String[] args) {
{
String seq1 = "TTCGATGGAGCTGTA";
String seq2 = "TAGCTAGCTCGGCATGA";
System.out.println(count_mismatches(seq1, seq2))
//*expected to print out 5 because there are 3 mismatched pairs and 2 that do not have a pair*
}
}
public static int count_mismatches(String seq1, String seq2) {
int mismatchCount = 0;
int counter = seq1.length() - seq2.length();
int finalCounter = mismatchCount + counter;
for(int i = 0; i < seq1.length(); i++) if (seq1.charAt(i) == seq2.charAt(i)) {
break; //checks to see if the length of seq1 and seq2 are the same
}
for(int i = 0; i < seq1.length(); i++) if (seq1.charAt(i) != seq2.charAt(i)) {
return counter; //figure out how to do absolute value for negative numbers
}
return finalCounter;
}
}
'''

Since you want to count only the places where there are differences, you can iterate through the minimum length present in both the strings and find out the places where they are different.
In the end, you can add absolute difference of length between seq1 and seq2 and return that value to the main function.
For the logic, all you have to do is apply 4 if conditions to check if character is A,G,C,T and if suitable pair is present in the other string.
public class CountMismatches {
public static void main(String[] args) {
{
String seq1 = "TTCGATGGAGCTGTA";
String seq2 = "TAGCTAGCTCGGCATGA";
System.out.println(count_mismatches(seq1, seq2));
}
}
public static int count_mismatches(String seq1, String seq2) {
int finalCounter = 0;
for (int i = 0; i < Math.min(seq1.length(), seq2.length()); i++) {
char c1 = seq1.charAt(i);
char c2 = seq2.charAt(i);
if (c1 == 'A') {
if (c2 == 'T')
continue;
else
finalCounter++;
} else if (c1 == 'T') {
if (c2 == 'A')
continue;
else
finalCounter++;
} else if (c1 == 'G') {
if (c2 == 'C')
continue;
else
finalCounter++;
} else if (c1 == 'C') {
if (c2 == 'G')
continue;
else
finalCounter++;
}
}
return finalCounter + (Math.abs(seq1.length() - seq2.length()));
}
}
and the output is as follows :
5

Make these refactorings:
To make the comparisons easy to code and understand, create a Map whose entires are each pair (both directions)
Iterate over the Strings up to the length of the shortest one, adding up the number of matching pairs as you go
The result is the length of the longest String minus the number of pairs
Like this:
public static int count_mismatches(String seq1, String seq2) {
Map<Character, Character> pairs = Map.of('A', 'T', 'T', 'A', 'G', 'C', 'C', 'G');
int count = 0;
for (int i = 0; i < Math.min(seq1.length(), seq2.length()); i++) {
if (pairs.get(seq1.charAt(i)) == seq2.charAt(i)) {
count++;
}
}
return Math.max(seq1.length(), seq2.length()) - count;
}
See live demo, which returns 5 for your sample input.

Good Evening,
Something seems off here, this snippet of code:
for(int i = 0; i < seq1.length(); i++)
if (seq1.charAt(i) == seq2.charAt(i)) {
break; //checks to see if the length of seq1 and seq2 are the same
}
Does not do what you think it does. This cycle will loop through all characters in sequence1 using i < seq1.length() and for each character that exists in seq1, it will check if said character is equal to the character with the same index in seq2.
This means that a correction is in order:
int countMismatches = 0;
for(int i = 0; i < seq1.length();i++){
switch(seq1.charAt(i)){
case 'A':
if(seq2.charAt(i) != 'T') countMismatches++;
break;
}
}
Repeat this process for the other letters, and voilá, you should be able to count your mismatches this way.
Do be careful with sequences having different lengths, as if that happens, as soon as you step out of a bound, you will receive an IndexOutOfBoundsException, indicating you've tried to check a character that does not exist.

First you must find out which string is the shortest in length. Also you need to get the length difference when calculating the shortest string. After that, use that length as a terminating condition in your for loop. You can use booleans to check whether the values are present before incrementing the counter with an if statement.
The absolute value of any number can be obtained by calling the static method abs() from the Math class. Last, just add the mismatchCounts to the absolute value of the length difference in order to obtain the result.
Here is my solution.
public class App {
public static void main(String[] args) throws Exception {
String seq1 = "TTCGATGGAGCTGTA";
String seq2 = "TAGCTAGCTCGGCATGA";
System.out.println(compareStrings(seq1, seq2));
}
public static int compareStrings(String stringOne, String stringTwo) {
Character A = 'A', T = 'T', G = 'G', C = 'C';
int mismatchCount = 0;
int lowestStringLenght = 0;
int length_one = stringOne.length();
int length_two = stringTwo.length();
int lenght_difference = 0;
if (length_one < length_two) {// string one lenght is greater
lowestStringLenght = length_one;
lenght_difference = length_one - length_two;
} else if (length_one > length_two) {// string two lenght is greater
lowestStringLenght = length_two;
lenght_difference = length_two - length_one;
} else { // lenghts must be equal, use either
lowestStringLenght = length_one;
lenght_difference = 0; // there is no difference because they are equal
}
for (int i = 0; i < lowestStringLenght; i++) {
// A matches with T
// G matches with C
// evaluate if the values A, T, G, C are present
boolean A_T_PRESENT = stringOne.charAt(i) == A && stringTwo.charAt(i) == T;
boolean G_C_PRESENT = stringOne.charAt(i) == G && stringTwo.charAt(i) == C;
boolean T_A_PRESENT = stringOne.charAt(i) == T && stringTwo.charAt(i) == A;
boolean C_G_PRESENT = stringOne.charAt(i) == C && stringTwo.charAt(i) == G;
boolean TWO_EQUAL = stringOne.charAt(i) == stringTwo.charAt(i);
// characters are equal, increase mismatch counter
if (TWO_EQUAL) {
mismatchCount++;
continue;
}
// all booleans evaluated to false, it means that the characters are not proper
// matches. Increment mismatchCount
else if (!A_T_PRESENT && !G_C_PRESENT && !T_A_PRESENT && !C_G_PRESENT) {
mismatchCount++;
continue;
} else {
continue;
}
}
// calculate the sum of the mismatches plus the abs of the lenght difference
lenght_difference = Math.abs(lenght_difference);
return mismatchCount + lenght_difference;
}
}

Avoid char
The char type is legacy, essentially broken. As a 16-bit value, char is physically incapable of representing most characters. The char type in your particular case would work. But using char is a bad habit generally, as such code may break when encountering any of about 75,000 characters defined in Unicode.
Code point
Use code point integer numbers instead. A code point is the number assigned to each of the over 140,000 characters defined by the Unicode Consortium.
Here we get an IntStream, a series of int values, one for each character in the input string. Then we collect these integer numbers into an array of int values.
int[] codePoints1 = seq1.codePoints().toArray() ;
int[] codePoints2 = seq2.codePoints().toArray() ;
You said the input strings may be of unequal length. So our two arrays may be jagged, of different lengths. Figure out the size of the shorter array.
int smallerSize = Math.min( codePoints1.length , codePoints2.length ) ;
Keep track of the index number of mismatched rows.
List<Integer> mismatchIndices = new ArrayList <>();
Loop the arrays based on that smaller size.
for( int i = 0 ; i < smallerSize ; i ++ )
{
if ( isBasePairValid( codePoint first , codePoint second ) )
{
…
} else
{
mismatchIndices.add( i ) ;
}
}
Write an isBasePairValid method
Write the isBasePairValid method, taking two arguments, the code points of the two nucleobase letters.
static int A = "A".codePointAt( 0 ) ; // Annoying zero-based index counting. So first character is number zero.
static int C = "C".codePointAt( 0 ) ;
static int G = "G".codePointAt( 0 ) ;
static int T = "T".codePointAt( 0 ) ;
if( first == A ) return ( second == T )
else if( first == T ) return ( second == A )
else if( first == C ) return ( second == G )
else if( first == G ) return ( second == C )
else { throw new IllegalStateException( … ) ; }
Count the mismatches.
int countMismatches = mismatchIndices.size() ;

The numerical sum of chars T & A and G & C is fixed and unique for legal nucleobase pairs. So you just need to ensure that the corresponding bases have one of those sums.
String seq1 = "TTCGATGGAGCTGTA";
String seq2 = "TAGCTAGCTCGGCATGA";
System.out.println(count_mismatches(seq1, seq2));
prints
5
find max length to iterate
establish fixed sums for comparison
iterate and compare to expected pairing and update count appropriately
public static int count_mismatches(String seq1, String seq2) {
int len1 = seq1.length();
int len2 = seq2.length();
int len = len1;
if (len1 > len2) {
len = len2;
}
int sumTA = 'T'+'A';
int sumGC = 'G'+'C';
int misMatchCount = Math.abs(len1-len2);
for (int i = 0; i < len; i++) {
int pair = seq1.charAt(i) + seq2.charAt(i);
if (pair != sumTA && pair != sumGC) {
misMatchCount++;
}
}
return misMatchCount;
}

How do I manually find a substring in a string? (Java)

public int lookFor(String s) {
final int EXIST = 1;
final int NOT_EXIST = -1;
int thisIndex = 0;
int otherIndex = 0;
char thisNext;
char otherNext;
if (s == null || s.length() == 0)
return NOT_EXIST;
for(; thisIndex < this.mainString.length() ; ) {
thisNext = this.mainString.charAt(thisIndex);
otherNext = s.charAt(otherIndex);
if (thisNext == otherNext) {
thisIndex++;
otherIndex++;
}
else if (thisNext != otherNext)
thisIndex++;
if (otherIndex == s.length()-1)
return EXIST;
}
return NOT_EXIST;
}
This is my attempt so far.
mainString = the main string I want to find the substring in.
s = the substring.
So my idea was to get the first chars of both strings, see if they equal. if they don't, i'll get the second char of mainString, see if they equal (mainString second char to s first char). If they're not equal, i'll get the third char of mainString and so forth. Once they're equal, i'll get the next char of both strings and see if they both equal.
Basically the loops knows that mainString contains s when index of s equals to s length minus one (that means the loop looped all the way to the last char inc, of s, so s index == s length -1).
Is the logic I'm trying to work with incorrect? or I just executed it not good? i'll happy to get answers!

Here's my naïve approach:
private final int EXIST = 1;
private final int NOT_EXIST = -1;
private int lookFor(String a, String b, int index) {
for (int i = 0; i < b.length(); i++) {
if ((index + i) >= a.length()) return NOT_EXIST;
if (a.charAt(index + i) != b.charAt(i)) return NOT_EXIST;
}
return EXIST;
}
public int lookFor(String a, String b) {
char start = b.charAt(0);
for (int i=0; i < a.length(); i++) {
if (a.charAt(i) == start) {
if (lookFor(a, b, i) == EXIST) return EXIST;
}
}
return NOT_EXIST;
}
Though, I'm not sure why you would do this when you could just do:
int ret = a.contains(b) ? EXIST : NOT_EXIST
However I wanted to actually answer your question.
Here's a slightly improved version that satisfies your "all in one method" requirement.
public static int lookFor(String a, String b) {
// Fancy way of preventing errors when one of the strings is empty
boolean az = a.length() == 0;
boolean bz = b.length() == 0;
if (az ^ bz) return NOT_EXIST;
// Need this next line if you want to interpret two empty strings as containing eachother
if (az && bz) return EXIST;
char start = b.charAt(0);
// This is known as a "label". Some say it's bad practice.
outer:
for (int i=0; i < a.length(); i++) {
if (a.charAt(i) == start) {
// Instead of using two methods, we can condense it like so
for (int q = 0; q < b.length(); q++) {
if ((i + q) >= a.length()) continue outer;
if (a.charAt(i + q) != b.charAt(q)) continue outer;
}
return EXIST;
}
}
return NOT_EXIST;
}

To find a substring "by hand", you need a nested loop; i.e. a loop inside a loop.
The outer loop tries all of the possible start positions for the substring in the string.
For a given start position, the inner loop tests all of the characters of the string that you are looking for against the string you are searching in.
In the naive substring search algorithm, the outer loop steps starts at index zero, and increments the index by one until it gets to the end of the string being searched. This can be improved on:
Every non-null string "contains" the empty string. It may be worth treating this as a special case.
It is easy to see that the outer loop can usually stop before the final. If you are searching for a string of length (say) 3, then the outer loop can stop at 3 from the end. (Think about it ....)
There are some clever algorithms which allow the outer loop to skip over some indexes. If you are interested, start by Googling for "Boyer-Moore string search".
(Note: the looping could be replaced with / written using recursion, but it is still there.)
Your code doesn't have a nested loops. By my reading, it is only going to find a match if the string you are searching for is at the start of the string you are searching. That is not correct.

Java: How to implement Word Break?

I'm trying to understanding the following Java + Dynamic Programming implementation (https://pingzhblog.wordpress.com/2015/09/17/word-break/):
public class Solution {
public boolean wordBreak(String s, Set<String> wordDict) {
if(s == null) {
return false;
}
boolean[] wordBreakDp = new boolean[s.length() + 1];
wordBreakDp[0] = true;
for(int i = 1; i <= s.length(); i++) {
for(int j = 0; j < i; j++) {
String word = s.substring(j, i);
if(wordBreakDp[j] && wordDict.contains(word)) {
wordBreakDp[i] = true;
break;
}
}
}//end for i
return wordBreakDp[s.length()];
}
}
But need some clarification with String s = "abcxyz" and Set <String> wordDict = ["z", "xy", "ab", "c"].
I'm still unclear as to what wordBreakDp[] represents, and setting one to true means.
So I made the attempt and got wordBreakDP[2,3,5,6]=true, but what do those indexes tell? Couldn't I have just checked for i=6 since all we are checking is if last index of wordBreakDp[] is true, wordBreakDp[s.length()];?
And say for example I got ab for s.substring(0, 2);, but then how can we just assume that next loop, s.substring(1, 2);, is not useful and just break; out of the loop?
Thank you

This isn't really an answer, but it might be helpful in understanding the loops. It prints the value of substring(i,j) and also wordBreakDp[j] at each iteration. It also prints the final solution (segmentation) at the end of the method.
public boolean wordBreak(String s, Set<String> wordDict) {
if(s == null) {
return false;
}
boolean[] wordBreakDp = new boolean[s.length() + 1];
wordBreakDp[0] = true;
for(int i = 1; i <= s.length(); i++) {
for(int j = 0; j < i; j++) {
String word = s.substring(j, i);
System.out.println("["+j+","+i+"]="+s.substring(j,i)+", wordBreakDP["+j+"]="+wordBreakDp[j]);
if(wordBreakDp[j] && wordDict.contains(word)) {
wordBreakDp[i] = true;
break;
}
}
}//end for i
for (int i = 1, start=0; i <= s.length(); i++) {
if (wordBreakDp[i]) {
System.out.println(s.substring(start,i));
start = i;
}
}
return wordBreakDp[s.length()];
}

I think I have the answer to your last question, which was:
And say for example I got ab for s.substring(0, 2);, but then how can
we just assume that next loop, s.substring(1, 2);, is not useful and
just break; out of the loop?
You are right that if you got "ab" for s.substring(0,2) and you break out of the loop, this doesn't necessarily mean that s.substring(1,2) is not useful. Let's say that s.substring(1,2) is useful. This means that "a" is also a valid word, and so is "b". The solution found by this algorithm would start with "ab" whereas the solution found by not breaking would be "a" followed by "b". Both these solutions are correct (assuming that the rest of the string can be broken into valid words as well). The algorithm doesn't find all valid solutions. It just returns true if the string can be broken into valid words, i.e., if there is one solution that satisfies the condition. There may be more than one solution, of course, but that is not the purpose of this algorithm.

Note, f[i] means whether the first i characters of s is valid. An example,
String = catsand
Dict = [cat, cats, sand, and]
s = c a t s a n d
dp = T F F T T F F ?
i = 0 1 2 3 4 5 6 7
*
Note dp[0] is True, because empty string is valid. Say we have dp[0] through dp[6].
Let's calculate dp[7], which denotes whether catsand is valid.
For catsand to be valid, there must be a non-empty suffix that is in dict, AND the remaining prefix is valid.
"catsand" is a valid, if:
If "" is valid, and "catsand" is inDict, or
If "c" is valid, and "atsand" is inDict, or
If "ca" is valid, and "tsand" is inDict, or
If "cat" is valid, and "sand" is inDict, or
If "cats" is valid, and "and" is inDict, or
If "catsa" is valid, and "nd" is inDict, or
If "catsan" is valid, and "d" is inDict
Translated to pesudo code:
dp[7] is True, if:
j j i
If dp[0] && inDict( s[0..7) ) or
If dp[1] && inDict( s[1..7) ) or
If dp[2] && inDict( s[2..7) ) or
If dp[3] && inDict( s[3..7) ) or
If dp[4] && inDict( s[4..7) ) or
If dp[5] && inDict( s[5..7) ) or
If dp[6] && inDict( s[6..7) )

public class Solution {
public int wordBreak(String a, ArrayList<String> b) {
int n = a.length();
boolean dp[] = new boolean[n+1];
dp[0] = true;
for(int i = 0;i<n;i++){
if(!dp[i])continue;
for(String s : b){
int len = s.length();
int end = len + i;
if(end > n || dp[end])continue;
if(a.substring(i,end).equals(s)){
dp[end] = true;
}
}
}
if(dp[n])return 1;
else return 0;
}
}

Determine if a given string is a k-palindrome

I'm trying to solve the following interview practice question:
A k-palindrome is a string which transforms into a palindrome on removing at most
k characters.
Given a string S, and an integer K, print "YES" if S is a k-palindrome;
otherwise print "NO".
Constraints:
S has at most 20,000 characters.
0 <= k <= 30
Sample Test Cases:
Input - abxa 1
Output - YES
Input - abdxa 1
Output - NO
My approach I've decided is going to be taking all possible String combinations of length s.length - k or greater, i.e. "abc" and k = 1 -> "ab" "bc" "ac" "abc" and checking if they are palindromes. I have the following code so far, but can't seem to figure out a proper way to generate all these string combinations in the general case:
public static void isKPalindrome(String s, int k) {
// Generate all string combinations and call isPalindrome on them,
// printing "YES" at first true
}
private static boolean isPalindrome(String s) {
char[] c = s.toCharArray()
int slow = 0;
int fast = 0;
Stack<Character> stack = new Stack<>();
while (fast < c.length) {
stack.push(c[slow]);
slow += 1;
fast += 2;
}
if (c.length % 2 == 1) {
stack.pop();
}
while (!stack.isEmpty()) {
if (stack.pop() != c[slow++]) {
return false;
}
}
return true;
}
Can anyone figure out a way to implement this, or perhaps demonstrate a better way?

I think there is a better way
package se.wederbrand.stackoverflow;
public class KPalindrome {
public static void main(String[] args) {
KPalindrome kPalindrome = new KPalindrome();
String s = args[0];
int k = Integer.parseInt(args[1]);
if (kPalindrome.testIt(s, k)) {
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
boolean testIt(String s, int k) {
if (s.length() <= 1) {
return true;
}
while (s.charAt(0) == s.charAt(s.length()-1)) {
s = s.substring(1, s.length()-1);
if (s.length() <= 1) {
return true;
}
}
if (k == 0) {
return false;
}
// Try to remove the first or last character
return testIt(s.substring(0, s.length() - 1), k - 1) || testIt(s.substring(1, s.length()), k - 1);
}
}
Since K is max 30 it's likely the string can be invalidated pretty quick and without even examining the middle of the string.
I've tested this with the two provided test cases as well as a 20k characters long string with just "ab" 10k times and k = 30;
All tests are fast and returns the correct results.

This can be solved using Edit distance dynamic programming algorithm. Edit distance DP algorithm is used to find the minimum operations required to convert a source string to destination string. The operations can be either addition or deletion of characters.
The K-palindrome problem can be solved using Edit distance algorithm by checking the minimum operation required to convert the input string to its reverse.
Let editDistance(source,destination) be the function which takes source string and destination string and returns the minimum operations required to convert the source string to destination string.
A string S is K-palindrome if editDistance(S,reverse(S))<=2*K
This is because we can transform the given string S into its reverse by deleting atmost K letters and then inserting the same K letters in different position.
This will be more clear with an example.
Let S=madtam and K=1.
To convert S into reverse of S (i.e matdam) first we have to remove the character 't' at index 3 ( 0 based index) in S.
Now the intermediate string is madam. Then we have to insert the character 't' at index 2 in the intermediate string to get "matdam" which is the reverse of string s.
If you look carefully you will know that the intermediate string "madam" is the palindrome that is obtained by removing k=1 characters.

I found the length of a longest string such that after removing characters >= k, we will be having a palindrome. I have used dynamic programming here. The palindrome I have considered need not be consecutive. Its like abscba has a longest palindromic length of 4.
So now this can be used further, such that whenever k >= (len - len of longest palindrome), it results to true else false.
public static int longestPalindrome(String s){
int len = s.length();
int[][] cal = new int[len][len];
for(int i=0;i<len;i++){
cal[i][i] = 1; //considering strings of length = 1
}
for(int i=0;i<len-1;i++){
//considering strings of length = 2
if (s.charAt(i) == s.charAt(i+1)){
cal[i][i+1] = 2;
}else{
cal[i][i+1] = 0;
}
}
for(int p = len-1; p>=0; p--){
for(int q=p+2; q<len; q++){
if (s.charAt(p)==s.charAt(q)){
cal[p][q] = 2 + cal[p+1][q-1];
}else{
cal[p][q] = max(cal[p+1][q], cal[p][q-1]);
}
}
}
return cal[0][len-1];
}

This is a common interview question, and I'm little surprised that no one has mentioned dynamic programming yet. This problem exhibits optimal substructure (if a string is a k-palindrome, some substrings are also k-palindromes), and overlapping subproblems (the solution requires comparing the same substrings more than once).
This is a special case of the edit distance problem, where we check if a string s can be converted to string p by only deleting characters from either or both strings.
Let the string be s and its reverse rev. Let dp[i][j] be the number of deletions required to convert the first i characters of s to the first j characters of rev. Since deletions have to be done in both strings, if dp[n][n] <= 2 * k, then the string is a k-palindrome.
Base case: When one of the strings is empty, all characters from the other string need to be deleted in order to make them equal.
Time complexity: O(n^2).
Scala code:
def kPalindrome(s: String, k: Int): Boolean = {
val rev = s.reverse
val n = s.length
val dp = Array.ofDim[Int](n + 1, n + 1)
for (i <- 0 to n; j <- 0 to n) {
dp(i)(j) = if (i == 0 || j == 0) i + j
else if (s(i - 1) == rev(j - 1)) dp(i - 1)(j - 1)
else 1 + math.min(dp(i - 1)(j), dp(i)(j - 1))
}
dp(n)(n) <= 2 * k
}
Since we are doing bottom-up DP, an optimization is to return false if at any time i == j && dp[i][j] > 2 * k, since all subsequent i == j must be greater.

Thanks to Andreas, that algo worked like a charm. Here my implementation for anyone who's curious. Slightly different, but fundamentally your same logic:
public static boolean kPalindrome(String s, int k) {
if (s.length() <= 1) {
return true;
}
char[] c = s.toCharArray();
if (c[0] != c[c.length - 1]) {
if (k <= 0) {
return false;
} else {
char[] minusFirst = new char[c.length - 1];
System.arraycopy(c, 1, minusFirst, 0, c.length - 1);
char[] minusLast = new char[c.length - 1];
System.arraycopy(c, 0, minusLast, 0, c.length - 1);
return kPalindrome(String.valueOf(minusFirst), k - 1)
|| kPalindrome(String.valueOf(minusLast), k - 1);
}
} else {
char[] minusFirstLast = new char[c.length - 2];
System.arraycopy(c, 1, minusFirstLast, 0, c.length - 2);
return kPalindrome(String.valueOf(minusFirstLast), k);
}
}

This problem can be solved using the famous Longest Common Subsequence(LCS) method. When LCS is applied with the string and the reverse of the given string, then it gives us the longest palindromic subsequence present in the string.
Let the longest palindromic subsequence length of a given string of length string_length be palin_length. Then (string_length - palin_length) gives the number of characters required to be deleted to convert the string to a palindrome. Thus, the given string is k-palindrome if (string_length - palin_length) <= k.
Let me give some examples,
Initial String: madtam (string_length = 6)
Longest Palindromic Subsequence: madam (palin_length = 5)
Number of non-contributing characters: 1 ( string_length - palin_length)
Thus this string is k-palindromic where k>=1. This is because you need to delete atmost k characters ( k or less).
Here is the code snippet:
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define MAX 10000
int table[MAX+1][MAX+1];
int longest_common_subsequence(char *first_string, char *second_string){
int first_string_length = strlen(first_string), second_string_length = strlen(second_string);
int i, j;
memset( table, 0, sizeof(table));
for( i=1; i<=first_string_length; i++ ){
for( j=1; j<=second_string_length; j++){
if( first_string[i-1] == second_string[j-1] )
table[i][j] = table[i-1][j-1] + 1;
else
table[i][j] = max(table[i-1][j], table[i][j-1]);
}
}
return table[first_string_length][second_string_length];
}
char first_string[MAX], second_string[MAX];
int main(){
scanf("%s", first_string);
strcpy(second_string, first_string);
reverse(second_string, second_string+strlen(second_string));
int max_palindromic_length = longest_common_subsequence(first_string, second_string);
int non_contributing_chars = strlen(first_string) - max_palindromic_length;
if( k >= non_contributing_chars)
printf("K palindromic!\n");
else
printf("Not K palindromic!\n");
return 0;
}

I designed a solution purely based on recursion -
public static boolean isKPalindrome(String str, int k) {
if(str.length() < 2) {
return true;
}
if(str.charAt(0) == str.charAt(str.length()-1)) {
return isKPalindrome(str.substring(1, str.length()-1), k);
} else{
if(k == 0) {
return false;
} else {
if(isKPalindrome(str.substring(0, str.length() - 1), k-1)) {
return true;
} else{
return isKPalindrome(str.substring(1, str.length()), k-1);
}
}
}
}
There is no while loop in above implementation as in the accepted answer.
Hope it helps somebody looking for it.

public static boolean failK(String s, int l, int r, int k) {
if (k < 0)
return false;
if (l > r)
return true;
if (s.charAt(l) != s.charAt(r)) {
return failK(s, l + 1, r, k - 1) || failK(s, l, r - 1, k - 1);
} else {
return failK(s, l + 1, r - 1, k);
}
}

How to simplify this array method?

Here is my code to an array method:
private int _a;
public static void main(String[] args) {}
public int[] countAll(String s) {
int[] xArray = new int[27];
int[] yArray = new int[27];
_a = (int)'a';
for (int i = 0; i < xArray.length; i++) {
xArray[i] = _a;
_a = _a++;
}
for (int j = 0; j < s.length(); j++) {
s = s.toLowerCase();
char c = s.charAt(j);
int g = (int) c;
int letterindex = g - yArray[0];
if (letterindex >= 0 && letterindex <= 25) {
xArray[letterindex]++;
} else if (letterindex < 0 || letterindex > 25) {
xArray[26]++;
}
}
return xArray;
}
This code works in java but I was told that there is a simpler way. I am having a lot of trouble figuring out a simplified version of my code. Please help me.

If all you want to do is count the upper and lower case, that's a very roundabout way of doing it, what's wrong with something like:
public static int countUpper(String str)
{
int upper = 0;
for(char c : str.toCharArray())
{
if(Character.isUpperCase(c))
{
upper++;
}
}
return upper;
}
Then just the same thing with Character.isLowerCase(c) for the opposite.

public static int[] countAll(String s) {
int[] xArray = new int[27];
for (char c : s.toLowerCase().toCharArray()){
if (Character.isLetter(c))
xArray[c -'a']++;
else
xArray[26]++;
}
return xArray;
}

It looks like your program is trying to find frequencies of different alphabets in a string, and you are counting the non letters in special index 26. In that case your code to initialize the count is wrong. It is getting pre-initialized with some values in following for loop:
for (int i = 0; i < xArray.length; i++) {
xArray[i] = _a;
_a = _a++;
}
I think the method can be simply something like:
s = s.toLowerCase();
int histogram[] = new int[27];
for (char c: s.toCharArray()) {
int index = c - 'a';
if (index < 0 || index > 25) {
index = 26;
}
histogram[index]++;
}

Here are two important improvements that should be made to your code:
Add a method javadoc for countAll, so that readers don't have to trawl through 20+ lines of turgid code to reverse engineer what the method is supposed to be.
Get rid of the _a abomination. According to the most widely accepted Java coding standard, the underscore character has no place in a variable name. Besides, a is about the most useless field name I've ever come across. If it is intended to convey some meaning to the reader ... you have totally lost me.
(Oh I get it. It shouldn't be a field at all. Bzzzt!!!)
Then there is the yArray array. As far as I can tell the only place it is used is here:
int letterindex = g - yArray[0];
which is actually the same as:
int letterindex = g;
since yArray[0] is never assigned to. In short yArray is completely redundant.
And this:
if (letterindex >= 0 && letterindex <= 25) {
xArray[letterindex]++;
} else if (letterindex < 0 || letterindex > 25) {
xArray[26]++;
}
The condition in the else part is redundant. Your code will be easier to read if you just write this:
if (letterindex >= 0 && letterindex <= 25) {
xArray[letterindex]++;
} else {
xArray[26]++;
}
The two are equivalent. Do you see why?
Finally the initialization of the xArray elements looks plain wrong to me. If xArray contains counts, the elements need to start at zero. (Didn't you wonder why your code was telling you that every string contained lots of "zees"?)
"This code works in java ..."
I don't think so. Maybe it compiles. Maybe it runs without crashing. But it doesn't give correct answers!

public static int[] countAll(String s) {
int[] count = new int[26];
for (char c : s.toLowerCase().toCharArray()) {
if ('a' <= c && c <= 'z') {
count[c - 'a']++;
}
}
return count;
}
First.. your arrays where to big.
Second.. why do you need two arrays at all?
Third.. your code didn't seemt to work.. the word "hello" returned an array with the number 97 (26 times) and the number 102.
Edit: Made it shorter.