I am stuck in one of the problem related string in Java. My logic has worked well for some test cases, but not for all test cases. Please suggest me the actual logic of the following question::
I am given a string s of n character, comprising only of A's and B's . I can choose any index i and change s(i) to either A or B. Find the minimum no. Of changes that you must make to string S such that the resultant string is of format : AAAAA.....BBBBB. In other words, my task is to determine minimum no. of changes such that string s has x no. of A's in the beginning, followed by the remaining (n-x) no. of B's.
Sample Input is as below
4
3
AAB
5
AABAA
1
B
4
BABA
First line: A single integer denoting the number of test cases
For each test case:
First line contains a single integer denoting the size of the string
Second line contains string
My code is as below
import java.util.*;
class TestClass {
public static void main(String args[] ) throws Exception {
Scanner s = new Scanner(System.in);
TestClass t = new TestClass();
int test_case = s.nextInt();
for(int i = 0; i < test_case; i++){
int len = s.nextInt();
String none = s.nextLine();
String str = s.nextLine();
int cnta = t.count_ab(str,'A');
int cntb = t.count_ab(str,'B');
char c1 = '1';
if(cnta > cntb){
c1 = 'A';
}
else{
c1 = 'B';
}
int count = 0;
int c1_count = 0;
int c2_count = 0;
if(str.length() > 1){
String rev = "";
c1_count = t.cnt_init_a(str, 'A');
StringBuilder sb = new StringBuilder(str);
rev = sb.reverse().toString();
c2_count = t.cnt_init_a(rev, 'B');
int rem_len = str.length() - c2_count;
for(int h = c1_count; h < rem_len; h++){
if(Character.compare(str.charAt(h), c1) != 0){
count = count + 1;
}
}
}
System.out.println(count);
}
}
public int cnt_init_a(String str, char c){
int cnt = 0;
for(int l = 0; l < str.length(); l++){
if(Character.compare(str.charAt(l), c) == 0){
cnt = cnt + 1;
}
else{
break;
}
}
return cnt;
}
public int count_ab(String str, char c){
int cnt = 0;
for(int g = 0; g < str.length(); g++){
if(Character.compare(str.charAt(g), c) == 0){
cnt = cnt + 1;
}
}
return cnt;
}
Your logic fails for e.g. "BAAAAAAAAAABBBBBBBBBB" and "AAAAAAAAAABBBBBBBBBBA".
You should start by ignoring all leading A's and all trailing B's, since they should never be changed.
"BAAAAAAAAAABBBBBBBBBB" -> "BAAAAAAAAAA" (removed trailing B's)
"AAAAAAAAAABBBBBBBBBBA" -> "BBBBBBBBBBA" (removed leading A's)
Then change either the leading B's to A's, or the trailing A's to B's, whichever is shorter.
Then repeat the process.
So I have an integer z that is holding the values of zip codes and i need to store each digit of the zip code in a separate part of the array. How would i be able to do this?
private int [] zipDigits = new int [5];
private int digitCheck;
private int zipCode;
public ZipCode(int z)
{
for(int i = 0; i < 5; i++)
zipDigits[i] =
this.zipCode = z;
Something like this should work:
public int[] intToDigits(int z) {
int n = String.valueOf(z).length();
int[] res = new int[n];
int i = n - 1;
while (i >= 0) {
res[i--] = z % 10;
z = z / 10;
}
return res;
}
This first gets the length in a quite obvious matter, you could also do this using logarithms but I don't know if their precision would lead to issues. Then starts at the back of the array, takes the last digit with mod and then deletes this digit by dividing the number by 10.
EDIT:
I now see that you've edited your post to state that the length is known to be 5 digits, then we don't need to calculate the length ourselves and we can use the following algorithm:
public int[] intToDigits(int z) {
int[] res = new int[5];
int i = 4;
while (i >= 0) {
res[i--] = z % 10;
z = z / 10;
}
return res;
}
Since you know the number of digits, you can make a loop getting the rest of division for 10 and saving it into the array and then dividing for 10.
int zipCode = 1000;
int [] myArray = new int[5];
for(int i = 0; i < 5; i++){
if(zipCode > 0){
myArray[i] = zipCode%10;
zipCode /= 10;
}
else
myArray[i] = 0;
}
So you are asking this;
you have a
int zip = 90210
and you want to put this into a array like this
int [] zipArrary = new int [5];
String changeZip = Integer.toString(zip);
for (int i = 0; i < changeZip.length(); i++){
char c = s.charAt(i);
int place = Character.getNumericValue(c);
zipArray[i] = place;
}
Something like this?
int zip = 12345;
String strZip = Integer.toString(zip);
char[] charArrayZip = strZip.toCharArray();
Now your zip is in an array of char.
package pls;
import java.util.Scanner;
public class pls {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner keyboard = new Scanner(System.in);
int count = 0;
int amount;
int amount2;
int[] Array = new int[100];
int[] Array1 = new int[100];
int total;
char b;
String bad = "";
String finalA = "";
System.out.println("Input your first word/letter (must be upper case): ");
String first = keyboard.next();
amount = first.length();
System.out.println("Input your second word/letter (must be upper case): ");
String second = keyboard.next();
amount2 = second.length();
while (amount < amount2)//to add on 'A' to first word
{
first = 'A' + first;
amount++;
}
while (amount2 < amount)//to add on 'A' to second word
{
second = 'A' + second;
amount2++;
}
for (int i = 0; i < amount; i++) {
Array[i] = (int)first.charAt(i) - 65;
}
for (int i = 0; i < amount2; i++) {
Array1[i] = (int)second.charAt(i) - 65;
}
for (int i = amount; i >= 0; i--) {
total = Array[amount] + Array1[amount2];
if (total > 25) {
int pls = total % 26;
b = (char)(pls + 65);
finalA = b + finalA;
total = total / 26;
b = (char)(total + 65);
finalA = b + finalA;
total = pls + total;
b = (char)(total + 65);
finalA = b + finalA;
}
else {
b = (char)(total + 65);
finalA = b + finalA;
}
amount--;
amount2--;
}
finalA = finalA.substring(0, finalA.length()-1);
while (finalA.charAt(0) == 'A') {
finalA = finalA.substring(0, 0) + finalA.substring(0 + 1);
}
System.out.println(finalA);
}
}
I need to make a program that lets users add words going from right to left.. They're given numerical values and through that, you can add them. Here are a few examples: CAT & DOG make FOZ, MOM & DAD make POP, ABA & A make BA, BCD & BC make BDF.
Now up to this point, I've managed to make it work, however this next one screws me up... ZZZ + ZZZ = BZZY.. This is how this part of the program is explained: "For example, Z + Z = BY (since Z + Z = 50, and 50 modulo 26 is 24, which is Y, and the carry is the quotient you get when you divide 50 by 26, i.e. 1, which represents B)." I appreciate any help! Thanks in advance!(Code at the top because errors.)
In the loop, you only need to create one character per pass, the one calculated with the mod operator.
The other calculation should store the division as the carry.
The total has to include both letters and the carry from the previous calculation.
You do NOT have to remove the last character of finalA when you are done.
I'm not sure, when I run this in eclipse it says . What's happening is, the input Y is a year that's less than 10000, and I have to find the next year after that one that has all different digits. For example, 2010 would print 2013 because 2013 is the next year after 2010 with all different digits.
package from1987To2013;
public class from1987To2013 {
public static void main(String[] args) {
//Y is year input
int Y = 2011;
//number of digits in Y
int length = String.valueOf(Y).length();
//Turns the Y into an int array arrayY
String temp = Integer.toString(Y);
int[] arrayY = new int[temp.length()];
for (int i = 0; i < temp.length(); i++)
{
arrayY[i] = temp.charAt(i) - '0';
}
//first for loop
for (int i = 0; i < 10000; i++) {
//find every value from Y to 10000
int a = Y + i;
//changes if its true or not
boolean bool = true;
//this loop goes through once if y = 2, twice if y = 33, thrice if y = 456, four times if y == 4666, etc
int d = 0;
for (int b = 0; b < length; b++)
//d is the b - 1 which will be compared with the position at b
d = b-1;
int b = 0;
//checks if they're all equal
if (arrayY[d] != (arrayY[b])) {
} else {
bool = false;
break;
}
if (bool = true){
System.out.print(a);
}
}
}
}
As well as changing the code as per #ZouZou's comment, I'm not quite sure why, but i had to change this line:
System.out.print(a);
to
System.out.println(a);
in order to get any output from eclipse.
p.s. your logic doesn't work, but at least this gives you output so you can debug it now.
Firstly here is the problem:
A positive integer is called a palindrome if its representation in the decimal system is the same when read from left to right and from right to left. For a given positive integer K of not more than 1000000 digits, write the value of the smallest palindrome larger than K to output. Numbers are always displayed without leading zeros.
Input: The first line contains integer t, the number of test cases. Integers K are given in the next t lines.
Output: For each K, output the smallest palindrome larger than K.
Example
Input:
2
808
2133
Output:
818
2222
Secondly here is my code:
// I know it is bad practice to not cater for erroneous input,
// however for the purpose of the execise it is omitted
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.lang.Exception;
import java.math.BigInteger;
public class Main
{
public static void main(String [] args){
try{
Main instance = new Main(); // create an instance to access non-static
// variables
// Use java.util.Scanner to scan the get the input and initialise the
// variable
Scanner sc=null;
BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
String input = "";
int numberOfTests = 0;
String k; // declare any other variables here
if((input = r.readLine()) != null){
sc = new Scanner(input);
numberOfTests = sc.nextInt();
}
for (int i = 0; i < numberOfTests; i++){
if((input = r.readLine()) != null){
sc = new Scanner(input);
k=sc.next(); // initialise the remainder of the variables sc.next()
instance.palindrome(k);
} //if
}// for
}// try
catch (Exception e)
{
e.printStackTrace();
}
}// main
public void palindrome(String number){
StringBuffer theNumber = new StringBuffer(number);
int length = theNumber.length();
int left, right, leftPos, rightPos;
// if incresing a value to more than 9 the value to left (offset) need incrementing
int offset, offsetPos;
boolean offsetUpdated;
// To update the string with new values
String insert;
boolean hasAltered = false;
for(int i = 0; i < length/2; i++){
leftPos = i;
rightPos = (length-1) - i;
offsetPos = rightPos -1; offsetUpdated = false;
// set values at opposite indices and offset
left = Integer.parseInt(String.valueOf(theNumber.charAt(leftPos)));
right = Integer.parseInt(String.valueOf(theNumber.charAt(rightPos)));
offset = Integer.parseInt(String.valueOf(theNumber.charAt(offsetPos)));
if(left != right){
// if r > l then offest needs updating
if(right > left){
// update and replace
right = left;
insert = Integer.toString(right);
theNumber.replace(rightPos, rightPos + 1, insert);
offset++; if (offset == 10) offset = 0;
insert = Integer.toString(offset);
theNumber.replace(offsetPos, offsetPos + 1, insert);
offsetUpdated = true;
// then we need to update the value to left again
while (offset == 0 && offsetUpdated){
offsetPos--;
offset =
Integer.parseInt(String.valueOf(theNumber.charAt(offsetPos)));
offset++; if (offset == 10) offset = 0;
// replace
insert = Integer.toString(offset);
theNumber.replace(offsetPos, offsetPos + 1, insert);
}
// finally incase right and offset are the two middle values
left = Integer.parseInt(String.valueOf(theNumber.charAt(leftPos)));
if (right != left){
right = left;
insert = Integer.toString(right);
theNumber.replace(rightPos, rightPos + 1, insert);
}
}// if r > l
else
// update and replace
right = left;
insert = Integer.toString(right);
theNumber.replace(rightPos, rightPos + 1, insert);
}// if l != r
}// for i
System.out.println(theNumber.toString());
}// palindrome
}
Finally my explaination and question.
My code compares either end and then moves in
if left and right are not equal
if right is greater than left
(increasing right past 9 should increase the digit
to its left i.e 09 ---- > 10) and continue to do
so if require as for 89999, increasing the right
most 9 makes the value 90000
before updating my string we check that the right
and left are equal, because in the middle e.g 78849887
we set the 9 --> 4 and increase 4 --> 5, so we must cater for this.
The problem is from spoj.pl an online judge system. My code works for all the test can provide but when I submit it, I get a time limit exceeded error and my answer is not accepted.
Does anyone have any suggestions as to how I can improve my algorithm. While writing this question i thought that instead of my while (offset == 0 && offsetUpdated) loop i could use a boolean to to make sure i increment the offset on my next [i] iteration. Confirmation of my chang or any suggestion would be appreciated, also let me know if i need to make my question clearer.
This seems like a lot of code. Have you tried a very naive approach yet? Checking whether something is a palindrome is actually very simple.
private boolean isPalindrome(int possiblePalindrome) {
String stringRepresentation = String.valueOf(possiblePalindrome);
if ( stringRepresentation.equals(stringRepresentation.reverse()) ) {
return true;
}
}
Now that might not be the most performant code, but it gives you a really simple starting point:
private int nextLargestPalindrome(int fromNumber) {
for ( int i = fromNumber + 1; ; i++ ) {
if ( isPalindrome( i ) ) {
return i;
}
}
}
Now if that isn't fast enough you can use it as a reference implementation and work on decreasing the algorithmic complexity.
There should actually be a constant-time (well it is linear on the number of digits of the input) way to find the next largest palindrome. I will give an algorithm that assumes the number is an even number of digits long (but can be extended to an odd number of digits).
Find the decimal representation of the input number ("2133").
Split it into the left half and right half ("21", "33");
Compare the last digit in the left half and the first digit in the right half.
a. If the right is greater than the left, increment the left and stop. ("22")
b. If the right is less than the left, stop.
c. If the right is equal to the left, repeat step 3 with the second-last digit in the left and the second digit in the right (and so on).
Take the left half and append the left half reversed. That's your next largest palindrome. ("2222")
Applied to a more complicated number:
1. 1234567887654322
2. 12345678 87654322
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ equal
3. 12345678 87654322
^ ^ greater than, so increment the left
3. 12345679
4. 1234567997654321 answer
This seems a bit similar to the algorithm you described, but it starts at the inner digits and moves to the outer.
There is no reason to fiddle with individual digits when the only needed operation is one simple addition. The following code is based on Raks' answer.
The code stresses simplicity over execution speed, intentionally.
import static org.junit.Assert.assertEquals;
import java.math.BigInteger;
import org.junit.Test;
public class NextPalindromeTest {
public static String nextPalindrome(String num) {
int len = num.length();
String left = num.substring(0, len / 2);
String middle = num.substring(len / 2, len - len / 2);
String right = num.substring(len - len / 2);
if (right.compareTo(reverse(left)) < 0)
return left + middle + reverse(left);
String next = new BigInteger(left + middle).add(BigInteger.ONE).toString();
return next.substring(0, left.length() + middle.length())
+ reverse(next).substring(middle.length());
}
private static String reverse(String s) {
return new StringBuilder(s).reverse().toString();
}
#Test
public void testNextPalindrome() {
assertEquals("5", nextPalindrome("4"));
assertEquals("11", nextPalindrome("9"));
assertEquals("22", nextPalindrome("15"));
assertEquals("101", nextPalindrome("99"));
assertEquals("151", nextPalindrome("149"));
assertEquals("123454321", nextPalindrome("123450000"));
assertEquals("123464321", nextPalindrome("123454322"));
}
}
Well I have constant order solution(atleast of order k, where k is number of digits in the number)
Lets take some examples
suppose n=17208
divide the number into two parts from middle
and reversibly write the most significant part onto the less significant one.
ie, 17271
if the so generated number is greater than your n it is your palindrome, if not just increase the center number(pivot) ie, you get 17371
other examples
n=17286
palidrome-attempt=17271(since it is less than n increment the pivot, 2 in this case)
so palidrome=17371
n=5684
palidrome1=5665
palidrome=5775
n=458322
palindrome=458854
now suppose n = 1219901
palidrome1=1219121
incrementing the pivot makes my number smaller here
so increment the number adjacent pivot too
1220221
and this logic could be extended
public class NextPalindrome
{
int rev, temp;
int printNextPalindrome(int n)
{
int num = n;
for (int i = num+1; i >= num; i++)
{
temp = i;
rev = 0;
while (temp != 0)
{
int remainder = temp % 10;
rev = rev * 10 + remainder;
temp = temp / 10;
}
if (rev == i)
{
break;
}
}
return rev;
}
public static void main(String args[])
{
NextPalindrome np = new NextPalindrome();
int nxtpalin = np.printNextPalindrome(11);
System.out.println(nxtpalin);
}
}
Here is my code in java. Whole idea is from here.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter number of tests: ");
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
System.out.println("Enter number: ");
String numberToProcess = sc.next(); // ne proveravam dal su brojevi
nextSmallestPalindrom(numberToProcess);
}
}
private static void nextSmallestPalindrom(String numberToProcess) {
int i, j;
int length = numberToProcess.length();
int[] numberAsIntArray = new int[length];
for (int k = 0; k < length; k++)
numberAsIntArray[k] = Integer.parseInt(String
.valueOf(numberToProcess.charAt(k)));
numberToProcess = null;
boolean all9 = true;
for (int k = 0; k < length; k++) {
if (numberAsIntArray[k] != 9) {
all9 = false;
break;
}
}
// case 1, sve 9ke
if (all9) {
whenAll9(length);
return;
}
int mid = length / 2;
if (length % 2 == 0) {
i = mid - 1;
j = mid;
} else {
i = mid - 1;
j = mid + 1;
}
while (i >= 0 && numberAsIntArray[i] == numberAsIntArray[j]) {
i--;
j++;
}
// case 2 already polindrom
if (i == -1) {
if (length % 2 == 0) {
i = mid - 1;
j = mid;
} else {
i = mid;
j = i;
}
addOneToMiddleWithCarry(numberAsIntArray, i, j, true);
} else {
// case 3 not polindrom
if (numberAsIntArray[i] > numberAsIntArray[j]) { // 3.1)
while (i >= 0) {
numberAsIntArray[j] = numberAsIntArray[i];
i--;
j++;
}
for (int k = 0; k < numberAsIntArray.length; k++)
System.out.print(numberAsIntArray[k]);
System.out.println();
} else { // 3.2 like case 2
if (length % 2 == 0) {
i = mid - 1;
j = mid;
} else {
i = mid;
j = i;
}
addOneToMiddleWithCarry(numberAsIntArray, i, j, false);
}
}
}
private static void whenAll9(int length) {
for (int i = 0; i <= length; i++) {
if (i == 0 || i == length)
System.out.print('1');
else
System.out.print('0');
}
}
private static void addOneToMiddleWithCarry(int[] numberAsIntArray, int i,
int j, boolean palindrom) {
numberAsIntArray[i]++;
numberAsIntArray[j] = numberAsIntArray[i];
while (numberAsIntArray[i] == 10) {
numberAsIntArray[i] = 0;
numberAsIntArray[j] = numberAsIntArray[i];
i--;
j++;
numberAsIntArray[i]++;
numberAsIntArray[j] = numberAsIntArray[i];
}
if (!palindrom)
while (i >= 0) {
numberAsIntArray[j] = numberAsIntArray[i];
i--;
j++;
}
for (int k = 0; k < numberAsIntArray.length; k++)
System.out.print(numberAsIntArray[k]);
System.out.println();
}
}
Try this
public static String genNextPalin(String base){
//check if it is 1 digit
if(base.length()==1){
if(Integer.parseInt(base)==9)
return "11";
else
return (Integer.parseInt(base)+1)+"";
}
boolean check = true;
//check if it is all 9s
for(char a: base.toCharArray()){
if(a!='9')
check = false;
}
if(check){
String num = "1";
for(int i=0; i<base.length()-1; i++)
num+="0";
num+="1";
return num;
}
boolean isBasePalin = isPalindrome(base);
int mid = base.length()/2;
if(isBasePalin){
//if base is palin and it is odd increase mid and return
if(base.length()%2==1){
BigInteger leftHalf = new BigInteger(base.substring(0,mid+1));
String newLeftHalf = leftHalf.add(BigInteger.ONE).toString();
String newPalin = genPalin2(newLeftHalf.substring(0,mid),newLeftHalf.charAt(mid));
return newPalin;
}
else{
BigInteger leftHalf = new BigInteger(base.substring(0,mid));
String newLeftHalf = leftHalf.add(BigInteger.ONE).toString();
String newPalin = genPalin(newLeftHalf.substring(0,mid));
return newPalin;
}
}
else{
if(base.length()%2==1){
BigInteger leftHalf = new BigInteger(base.substring(0,mid));
BigInteger rightHalf = new BigInteger(reverse(base.substring(mid+1,base.length())));
//check if leftHalf is greater than right half
if(leftHalf.compareTo(rightHalf)==1){
String newPalin = genPalin2(base.substring(0,mid),base.charAt(mid));
return newPalin;
}
else{
BigInteger leftHalfMid = new BigInteger(base.substring(0,mid+1));
String newLeftHalfMid = leftHalfMid.add(BigInteger.ONE).toString();
String newPalin = genPalin2(newLeftHalfMid.substring(0,mid),newLeftHalfMid.charAt(mid));
return newPalin;
}
}
else{
BigInteger leftHalf = new BigInteger(base.substring(0,mid));
BigInteger rightHalf = new BigInteger(reverse(base.substring(mid,base.length())));
//check if leftHalf is greater than right half
if(leftHalf.compareTo(rightHalf)==1){
return genPalin(base.substring(0,mid));
}
else{
BigInteger leftHalfMid = new BigInteger(base.substring(0,mid));
String newLeftHalfMid = leftHalfMid.add(BigInteger.ONE).toString();
return genPalin(newLeftHalfMid);
}
}
}
}
public static String genPalin(String base){
return base + new StringBuffer(base).reverse().toString();
}
public static String genPalin2(String base, char middle){
return base + middle +new StringBuffer(base).reverse().toString();
}
public static String reverse(String in){
return new StringBuffer(in).reverse().toString();
}
static boolean isPalindrome(String str) {
int n = str.length();
for( int i = 0; i < n/2; i++ )
if (str.charAt(i) != str.charAt(n-i-1))
return false;
return true;
}
HI Here is another simple algorithm using python,
def is_palindrome(n):
if len(n) <= 1:
return False
else:
m = len(n)/2
for i in range(m):
j = i + 1
if n[i] != n[-j]:
return False
return True
def next_palindrome(n):
if not n:
return False
else:
if is_palindrome(n) is True:
return n
else:
return next_palindrome(str(int(n)+1))
print next_palindrome('1000010')
I have written comments to clarify what each step is doing in this python code.
One thing that need to be taken into consideration is that input can be very large that we can not simply perform integer operations on it. So taking input as string and then manipulating it would be much easier.
tests = int(input())
results = []
for i in range(0, tests):
pal = input().strip()
palen = len(pal)
mid = int(palen/2)
if palen % 2 != 0:
if mid == 0: # if the number is of single digit e.g. next palindrome for 5 is 6
ipal = int(pal)
if ipal < 9:
results.append(int(pal) + 1)
else:
results.append(11) # for 9 next palindrome will be 11
else:
pal = list(pal)
pl = l = mid - 1
pr = r = mid + 1
flag = 'n' # represents left and right half of input string are same
while pl >= 0:
if pal[pl] > pal[pr]:
flag = 'r' # 123483489 in this case pal[pl] = 4 and pal[pr] = 3 so we just need to copy left half in right half
break # 123484321 will be the answer
elif pal[pl] < pal[pr]:
flag = 'm' # 123487489 in this case pal[pl] = 4 and pal[pr] = 9 so copying left half in right half will make number smaller
break # in this case we need to take left half increment by 1 and the copy in right half 123494321 will be the anwere
else:
pl = pl -1
pr = pr + 1
if flag == 'm' or flag == 'n': # increment left half by one and copy in right half
if pal[mid] != '9': # if mid element is < 9 the we can simply increment the mid number only and copy left in right half
pal[mid] = str(int(pal[mid]) + 1)
while r < palen:
pal[r] = pal[l]
r = r + 1
l = l - 1
results.append(''.join(pal))
else: # if mid element is 9 this will effect entire left half because of carry
pal[mid] = '0' # we need to take care of large inputs so we can not just directly add 1 in left half
pl = l
while pal[l] == '9':
pal[l] = '0'
l = l - 1
if l >= 0:
pal[l] = str(int(pal[l]) + 1)
while r < palen:
pal[r] = pal[pl]
r = r + 1
pl = pl - 1
if l < 0:
pal[0] = '1'
pal[palen - 1] = '01'
results.append(''.join(pal))
else:
while r < palen: # when flag is 'r'
pal[r] = pal[l]
r = r + 1
l = l - 1
results.append(''.join(pal))
else: # even length almost similar concept here with flags having similar significance as in case of odd length input
pal = list(pal)
pr = r = mid
pl = l = mid - 1
flag = 'n'
while pl >= 0:
if pal[pl] > pal[pr]:
flag = 'r'
break
elif pal[pl] < pal[pr]:
flag = 'm'
break
else:
pl = pl -1
pr = pr + 1
if flag == 'r':
while r < palen:
pal[r] = pal[l]
r = r + 1
l = l - 1
results.append(''.join(pal))
else:
if pal[l] != '9':
pal[l] = str(int(pal[l]) + 1)
while r < palen:
pal[r] = pal[l]
r = r + 1
l = l - 1
results.append(''.join(pal))
else:
pal[mid] = '0'
pl = l
while pal[l] == '9':
pal[l] = '0'
l = l - 1
if l >= 0:
pal[l] = str(int(pal[l]) + 1)
while r < palen:
pal[r] = pal[pl]
r = r + 1
pl = pl - 1
if l < 0:
pal[0] = '1'
pal[palen - 1] = '01'
results.append(''.join(pal))
for xx in results:
print(xx)
We can find next palindrome easily like below.
private void findNextPalindrom(int i) {
i++;
while (!checkPalindrom(i)) {
i++;
}
Log.e(TAG, "findNextPalindrom:next palindrom is===" + i);
}
private boolean checkPalindrom(int num) {
int temp = num;
int rev = 0;
while (num > 0) {
int rem = num % 10;
rev = rev * 10 + rem;
num = num / 10;
}
return temp == rev;
}
Simple codes and test output:
class NextPalin
{
public static void main( String[] args )
{
try {
int[] a = {2, 23, 88, 234, 432, 464, 7887, 7657, 34567, 99874, 7779222, 2569981, 3346990, 229999, 2299999 };
for( int i=0; i<a.length; i++)
{
int add = findNextPalin(a[i]);
System.out.println( a[i] + " + " + add + " = " + (a[i]+add) );
}
}
catch( Exception e ){}
}
static int findNextPalin( int a ) throws Exception
{
if( a < 0 ) throw new Exception();
if( a < 10 ) return a;
int count = 0, reverse = 0, temp = a;
while( temp > 0 ){
reverse = reverse*10 + temp%10;
count++;
temp /= 10;
}
//compare 'half' value
int halfcount = count/2;
int base = (int)Math.pow(10, halfcount );
int reverseHalfValue = reverse % base;
int currentHalfValue = a % base;
if( reverseHalfValue == currentHalfValue ) return 0;
if( reverseHalfValue > currentHalfValue ) return (reverseHalfValue - currentHalfValue);
if( (((a-currentHalfValue)/base)%10) == 9 ){
//cases like 12945 or 1995
int newValue = a-currentHalfValue + base*10;
int diff = findNextPalin(newValue);
return base*10 - currentHalfValue + diff;
}
else{
return (base - currentHalfValue + reverseHalfValue );
}
}
}
$ java NextPalin
2 + 2 = 4
23 + 9 = 32
88 + 0 = 88
234 + 8 = 242
432 + 2 = 434
464 + 0 = 464
7887 + 0 = 7887
7657 + 10 = 7667
34567 + 76 = 34643
99874 + 25 = 99899
7779222 + 555 = 7779777
2569981 + 9771 = 2579752
3346990 + 443 = 3347433
229999 + 9933 = 239932
2299999 + 9033 = 2309032