Related
I was asked this question in an interview:
Given a string (1<=|s|<=10^5), check if it is possible to partition it into three palindromes. If there are multiple answers possible, output the one where the cuts are made the earliest. If no answer is possible, print "Impossible".
**Input:**
radarnoonlevel
aabab
abcdefg
**Output:**
radar noon level
a a bab (Notice how a, aba, b is also an answer, but we will output the one with the earliest cuts)
Impossible
I was able to give a brute force solution, running two loops and checking palindrome property for every 3 substrings ( 0-i, i-j, j-end). This was obviously not optimal, but I have not been able to find a better solution since then.
I need a way of checking that if I know the palindrome property of a string, then how removing a character from the start or adding one at the end can give me the property of the new string without having to do the check for the whole string again. I am thinking of using three maps where each character key is mapped to number of occurences but that too doesn't lead me down anything.
Still O(n^2) solution, but you can store the result of palindrome substrings in a table and use that to get to the answer.
vector<string> threePalindromicSubstrings(string word) {
int n = word.size();
vector<vector<bool>> dp (n,vector<bool>(n,false));
for(int i = 0 ; i < n ; ++i)
dp[i][i] = 1;
for(int l = 2 ; l <= n ; ++l){
for(int i = 0 ; i < n - l +1 ; ++i){
int j = i + l - 1;
if(l == 2)
dp[i][j] = (word[i] == word[j]);
else
dp[i][j] = (word[i] == word[j]) && (dp[i+1][j-1]);
}
}
vector<string> ans;
for(int i = 0 ; i < n - 2 ; ++i){
if(dp[0][i]) {
for(int j = i+1 ; j < n - 1 ; ++j){
if(dp[i+1][j] && dp[j+1][n-1]){
ans.push_back(word.substr(0,i + 1));
ans.push_back(word.substr(i+1,j-i));
ans.push_back(word.substr(j+1,n-j));
return ans;
}
}
}
}
if(ans.empty())
ans.push_back("Impossible");
return ans;
}
We are given a string and we have to find out the minimum number of swaps to convert it into a palindrome.
Ex-
Given string: ntiin
Palindrome: nitin
Minimum number of swaps: 1
If it is not possible to convert it into a palindrome, return -1.
I am unable to think of any approach except brute force. We can check on the first and last characters, if they are equal, we check for the smaller substring, and then apply brute force on it. But this will be of a very high complexity, and I feel this question can be solved in another way. Maybe dynamic programming. How to approach it?
First you could check if the string can be converted to a palindrome.
Just have an array of letters (26 chars if all letters are latin lowercase), and count the number of each letter in the input string.
If string length is even, all letters counts should be even.
If string length is odd, all letters counts should be even except one.
This first pass in O(n) will already treat all -1 cases.
If the string length is odd, start by moving the element with odd count to the middle.
Then you can apply following procedure:
Build a weighted graph with the following logic for an input string S of length N:
For every element from index 0 to N/2-1:
- If symmetric element S[N-index-1] is same continue
- If different, create edge between the 2 characters (alphabetic order), or increment weight of an existing one
The idea is that when a weight is even you can do a 'good swap' by forming two pairs in one swap.
When weight is odd, you cannot place two pairs in one swap, your swaps need to form a cycle
1. For instance "a b a b"
One edge between a,b of weight 2:
a - b (2)
Return 1
2. For instance: "a b c b a c"
a - c (1)
b - a (1)
c - b (1)
See the cycle: a - b, b - c, c - a
After a swap of a,c you get:
a - a (1)
b - c (1)
c - b (1)
Which is after ignoring first one and merge 2 & 3:
c - b (2)
Which is even, you get to the result in one swap
Return 2
3. For instance: "a b c a b c"
a - c (2)
One swap and you are good
So basically after your graph is generated, add to the result the weight/2 (integer division e.g. 7/3 = 3) of each edge
Plus find the cycles and add to the result length-1 of each cycle
there is the same question as asked!
https://www.codechef.com/problems/ENCD12
I got ac for this solution
https://www.ideone.com/8wF9DT
//minimum adjacent swaps to make a string to its palindrome
#include<bits/stdc++.h>
using namespace std;
bool check(string s)
{
int n=s.length();
map<char,int> m;
for(auto i:s)
{
m[i]++;
}
int cnt=0;
for(auto i=m.begin();i!=m.end();i++)
{
if(i->second%2)
{
cnt++;
}
}
if(n%2&&cnt==1){return true;}
if(!(n%2)&&cnt==0){return true;}
return false;
}
int main()
{
string a;
while(cin>>a)
{
if(a[0]=='0')
{
break;
}
string s;s=a;
int n=s.length();
//first check if
int cnt=0;
bool ini=false;
if(n%2){ini=true;}
if(check(s))
{
for(int i=0;i<n/2;i++)
{
bool fl=false;
int j=0;
for(j=n-1-i;j>i;j--)
{
if(s[j]==s[i])
{
fl=true;
for(int k=j;k<n-1-i;k++)
{
swap(s[k],s[k+1]);
cnt++;
// cout<<cnt<<endl<<flush;
}
// cout<<" "<<i<<" "<<cnt<<endl<<flush;
break;
}
}
if(!fl&&ini)
{
for(int k=i;k<n/2;k++)
{
swap(s[k],s[k+1]);
cnt++;
}
// cout<<cnt<<" "<<i<<" "<<endl<<flush;
}
}
cout<<cnt<<endl;
}
else{
cout<<"Impossible"<<endl;
}
}
}
Hope it helps!
Technique behind my code is Greedy
first check if palindrome string can exist for the the string and if it can
there would be two cases one is when the string length would be odd then only count of one char has be odd
and if even then no count should be odd
then
from index 0 to n/2-1 do the following
fix this character and search for this char from n-i-1 to i+1
if found then swap from that position (lets say j) to its new position n-i-1
if the string length is odd then every time you encounter a char with no other occurence shift it to n/2th position..
My solution revolves around the palindrome property that first element and last element should match and if their adjacent elements also do not match then its not a palindrome. Keep comparing and swapping till both reach the same element or adjacent elements.
Written solution in java as below:
public static void main(String args[]){
String input = "natinat";
char[] arr = input.toCharArray();
int swap = 0;
int i = 0;
int j = arr.length-1;
char temp;
while(i<j){
if(arr[i] != arr[j]){
if(arr[i+1] == arr[j]){
//swap i and i+1 and increment i, decrement j, swap++
temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
i++;j--;
swap++;
} else if(arr[i] == arr[j-1]){
//swap j and j-1 and increment i, decrement j, swap++
temp = arr[j];
arr[j] = arr[j-1];
arr[j-1] = temp;
i++;j--;
swap++;
} else if(arr[i+1] == arr[j-1] && i+1 != j-1){
//swap i and i+1, swap j and j-1 and increment i, decrement j, swap+2
temp = arr[j];
arr[j] = arr[j-1];
arr[j-1] = temp;
temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
i++;j--;
swap = swap+2;
}else{
swap = -1;break;
}
} else{
//increment i, decrement j
i++;j--;
}
}
System.out.println("No Of Swaps: "+swap);
}
My solution in java for any type of string i.e Binary String, Numbers
public int countSwapInPalindrome(String s){
int length = s.length();
if (length == 0 || length == 1) return -1;
char[] str = s.toCharArray();
int start = 0, end = length - 1;
int count = 0;
while (start < end) {
if (str[start] != str[end]){
boolean isSwapped = false;
for (int i = start + 1; i < end; i++){
if (str[start] == str[i]){
char temp = str[i];
str[i] = str[end];
str[end] = temp;
count++;
isSwapped = true;
break;
}else if (str[end] == str[i]){
char temp = str[i];
str[i] = str[start];
str[start] = temp;
count++;
isSwapped = true;
break;
}
}
if (!isSwapped) return -1;
}
start++;
end--;
}
return (s.equals(String.valueOf(str))) ? -1 : count;
}
I hope it helps
string s;
cin>>s;
int n = s.size(),odd=0;
vi cnt(26,0);
unordered_map<int,set<int>>mp;
for(int i=0;i<n;i++){
cnt[s[i]-'a']++;
mp[s[i]-'a'].insert(i);
}
for(int i=0;i<26;i++){
if(cnt[i]&1) odd++;
}
int ans=0;
if((n&1 && odd == 1)|| ((n&1) == 0 && odd == 0)){
int left=0,right=n-1;
while(left < right){
if(s[left] == s[right]){
cnt[left]--;
cnt[right]--;
mp[s[left]-'a'].erase(left);
mp[s[right]-'a'].erase(right);
left++;
right--;
}else{
if(cnt[left]&1 == 0){
ans++;
int index = *mp[s[left]-'a'].rbegin();
mp[s[left]-'a'].erase(index);
mp[s[right]-'a'].erase(right);
mp[s[right]-'a'].insert(index);
swap(s[right],s[index]);
cnt[left]-=2;
}else{
ans++;
int index = *mp[s[right]-'a'].begin();
mp[s[right]-'a'].erase(index);
mp[s[left]-'a'].erase(left);
mp[s[left]-'a'].insert(index);
swap(s[left],s[index]);
cnt[right]-=2;
}
left++;
right--;
}
}
}else{
// cout<<odd<<" ";
cout<<"-1\n";
return;
}
cout<<ans<<"\n";
I have an exercise to solve in MIPS assembly (where I have some doubts but other things are clear) but I have some problem to write it's code. The exercise ask me:
Write a programm that, obtained a string from keyboard, count the occurrences of the character with the higher number of occurrences and show it.
How I can check all the 26 characters and find who has the higher occurences?
Example:
Give me a string: Hello world!
The character with the higher occurrences is: l
Thanks alot for the future answer.
P.s.
This is my first part of the programm:
#First message
li $v0, 4
la $a0, mess
syscall
#Stack space allocated
addi $sp, $sp, -257
#Read the string
move $a0, $sp
li $a1, 257
li $v0, 8
syscall
Since this is your assignment I'll leave the MIPS assembly implementation to you. I'll just show you the logic for the code in a higher-level language:
// You'd keep these variables in some MIPS registers of your choice
int c, i, count, max_count=0;
char max_char;
// Iterate over all ASCII character codes
for (c = 0; c < 128; c+=1) {
count = 0;
// Count the number of occurences of this character in the string
for (i = 0; string[i]!=0; i+=1) {
if (string[i] == c) count++;
}
// Was is greater than the current max?
if (count > max_count) {
max_count = count;
max_char = c;
}
}
// max_char now hold the ASCII code of the character with the highest number
// of occurences, and max_count hold the number of times that character was
// found in the string.
#Michael, I saw you answered before I posted, I just want to repeat that with a more detailed answer. If you edit your own to add some more explanations, then I will delete mine. I did not edit yours directly, because I was already half-way there when you posted. Anyway:
#Marco:
You can create a temporary array of 26 counters (initialized to 0).
Each counter corresponds to each letter (i.e. the number each letter occurs). For example counter[0] corresponds to the number of occurences of letter 'a', counter[1] for letter 'b', etc...
Then iterate over each character in the input character-sequence and for each character do:
a) Obtain the index of the character in the counter array.
b) Increase counter["obtained index"] by 1.
To obtain the index of the character you can do the following:
a) First make sure the character is not capital, i.e. only 'a' to 'z' allowed and not 'A' to 'Z'. If it is not, convert it.
b) Substract the letter 'a' from the character. This way 'a'-'a' gives 0, 'b'-'a' gives 1, 'c'-'a' gives 2, etc...
I will demonstrate in C language, because it's your exercise on MIPS (I mean the goal is to learn MIPS Assembly language):
#include <stdio.h>
int main()
{
//Maximum length of string:
int stringMaxLength = 100;
//Create string in stack. Size of string is length+1 to
//allow the '\0' character to mark the end of the string.
char str[stringMaxLength + 1];
//Read a string of maximum stringMaxLength characters:
puts("Enter string:");
scanf("%*s", stringMaxLength, str);
fflush(stdin);
//Create array of counters in stack:
int counter[26];
//Initialize the counters to 0:
int i;
for (i=0; i<26; ++i)
counter[i] = 0;
//Main counting loop:
for (i=0; str[i] != '\0'; ++i)
{
char tmp = str[i]; //Storing of str[i] in tmp, to write tmp if needed,
//instead of writing str[i] itself. Optional operation in this particular case.
if (tmp >= 'A' && tmp <= 'Z') //If the current character is upper:
tmp = tmp + 32; //Convert the character to lower.
if (tmp >= 'a' && tmp <='z') //If the character is a lower letter:
{
//Obtain the index of the letter in the array:
int index = tmp - 'a';
//Increment its counter by 1:
counter[index] = counter[index] + 1;
}
//Else if the chacacter is not a lower letter by now, we ignore it,
//or we could inform the user, for example, or we could ignore the
//whole string itself as invalid..
}
//Now find the maximum occurences of a letter:
int indexOfMaxCount = 0;
int maxCount = counter[0];
for (i=1; i<26; ++i)
if (counter[i] > maxCount)
{
maxCount = counter[i];
indexOfMaxCount = i;
}
//Convert the indexOfMaxCount back to the character it corresponds to:
char maxChar = 'a' + indexOfMaxCount;
//Inform the user of the letter with maximum occurences:
printf("Maximum %d occurences for letter '%c'.\n", maxCount, maxChar);
return 0;
}
If you don't understand why I convert the upper letter to lower by adding 32, then read on:
Each character corresponds to an integer value in memory, and when you make arithmetic operations on characters, it's like you are making them to their corresponding number in the encoding table.
An encoding is just a table which matches those letters with numbers.
For example 'a' corresponds to number 97 in ASCII encoding/decoding/table.
For example 'b' corresponds to number 98 in ASCII encoding/decoding/table.
So 'a'+1 gives 97+1=98 which is the character 'b'. They are all numbers in memory, and the difference is how you represent (decode) them. The same table of the encoding, is also used for decoding of course.
Examples:
printf("%c", 'a'); //Prints 'a'.
printf("%d", (int) 'a'); //Prints '97'.
printf("%c", (char) 97); //Prints 'a'.
printf("%d", 97); //Prints '97'.
printf("%d", (int) 'b'); //Prints '98'.
printf("%c", (char) (97 + 1)); //Prints 'b'.
printf("%c", (char) ( ((int) 'a') + 1 ) ); //Prints 'b'.
//Etc...
//All the casting in the above examples is just for demonstration,
//it would work without them also, in this case.
I am trying to count two binary numbers from string. The maximum number of counting digits have to be 253. Short numbers works, but when I add there some longer numbers, the output is wrong. The example of bad result is "10100101010000111111" with "000011010110000101100010010011101010001101011100000000111000000000001000100101101111101000111001000101011010010111000110".
#include <iostream>
#include <stdlib.h>
using namespace std;
bool isBinary(string b1,string b2);
int main()
{
string b1,b2;
long binary1,binary2;
int i = 0, remainder = 0, sum[254];
cout<<"Get two binary numbers:"<<endl;
cin>>b1>>b2;
binary1=atol(b1.c_str());
binary2=atol(b2.c_str());
if(isBinary(b1,b2)==true){
while (binary1 != 0 || binary2 != 0){
sum[i++] =(binary1 % 10 + binary2 % 10 + remainder) % 2;
remainder =(binary1 % 10 + binary2 % 10 + remainder) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (remainder != 0){
sum[i++] = remainder;
}
--i;
cout<<"Result: ";
while (i >= 0){
cout<<sum[i--];
}
cout<<endl;
}else cout<<"Wrong input"<<endl;
return 0;
}
bool isBinary(string b1,string b2){
bool rozhodnuti1,rozhodnuti2;
for (int i = 0; i < b1.length();i++) {
if (b1[i]!='0' && b1[i]!='1') {
rozhodnuti1=false;
break;
}else rozhodnuti1=true;
}
for (int k = 0; k < b2.length();k++) {
if (b2[k]!='0' && b2[k]!='1') {
rozhodnuti2=false;
break;
}else rozhodnuti2=true;
}
if(rozhodnuti1==false || rozhodnuti2==false){ return false;}
else{ return true;}
}
One of the problems might be here: sum[i++]
This expression, as it is, first returns the value of i and then increases it by one.
Did you do it on purporse?
Change it to ++i.
It'd help if you could also post the "bad" output, so that we can try to move backward through the code starting from it.
EDIT 2015-11-7_17:10
Just to be sure everything was correct, I've added a cout to check what binary1 and binary2 contain after you assing them the result of the atol function: they contain the integer numbers 547284487 and 18333230, which obviously dont represent the correct binary-to-integer transposition of the two 01 strings you presented in your post.
Probably they somehow exceed the capacity of atol.
Also, the result of your "math" operations bring to an even stranger result, which is 6011111101, which obviously doesnt make any sense.
What do you mean, exactly, when you say you want to count these two numbers? Maybe you want to make a sum? I guess that's it.
But then, again, what you got there is two signed integer numbers and not two binaries, which means those %10 and %2 operations are (probably) misused.
EDIT 2015-11-07_17:20
I've tried to use your program with small binary strings and it actually works; with small binary strings.
It's a fact(?), at this point, that atol cant handle numerical strings that long.
My suggestion: use char arrays instead of strings and replace 0 and 1 characters with numerical values (if (bin1[i]){bin1[i]=1;}else{bin1[i]=0}) with which you'll be able to perform all the math operations you want (you've already written a working sum function, after all).
Once done with the math, you can just convert the char array back to actual characters for 0 and 1 and cout it on the screen.
EDIT 2015-11-07_17:30
Tested atol on my own: it correctly converts only strings that are up to 10 characters long.
Anything beyond the 10th character makes the function go crazy.
There are 2 strings , how can we check if one is a rotated version of another ?
For Example : hello --- lohel
One simple solution is by concatenating first string with itself and checking if the other one is a substring of the concatenated version.
Is there any other solution to it ?
I was wondering if we could use circular linked list maybe ? But I am not able to arrive at the solution.
One simple solution is by concatenating them and checking if the other one is a substring of the concatenated version.
I assume you mean concatenate the first string with itself, then check if the other one is a substring of that concatenation.
That will work, and in fact can be done without any concatenation at all. Just use any string searching algorithm to search for the second string in the first, and when you reach the end, loop back to the beginning.
For instance, using Boyer-Moore the overall algorithm would be O(n).
There's no need to concatenate at all.
First, check the lengths. If they're different then return false.
Second, use an index that increments from the first character to the last of the source. Check if the destination starts with all the letters from the index to the end, and ends with all the letters before the index. If at any time this is true, return true.
Otherwise, return false.
EDIT:
An implementation in Python:
def isrot(src, dest):
# Make sure they have the same size
if len(src) != len(dest):
return False
# Rotate through the letters in src
for ix in range(len(src)):
# Compare the end of src with the beginning of dest
# and the beginning of src with the end of dest
if dest.startswith(src[ix:]) and dest.endswith(src[:ix]):
return True
return False
print isrot('hello', 'lohel')
print isrot('hello', 'lohell')
print isrot('hello', 'hello')
print isrot('hello', 'lohe')
You could compute the lexicographically minimal string rotation of each string and then test if they were equal.
Computing the minimal rotation is O(n).
This would be good if you had lots of strings to test as the minimal rotation could be applied as a preprocessing step and then you could use a standard hash table to store the rotated strings.
Trivial O(min(n,m)^2) algorithm: (n - length of S1, m - length of S2)
isRotated(S1 , S2):
if (S1.length != S2.length)
return false
for i : 0 to n-1
res = true
index = i
for j : 0 to n-1
if S1[j] != S2[index]
res = false
break
index = (index+1)%n
if res == true
return true
return false
EDIT:
Explanation -
Two strings S1 and S2 of lengths m and n respectively are cyclic identical if and only if m == n and exist index 0 <= j <= n-1 such S1 = S[j]S[j+1]...S[n-1]S[0]...S[j-1].
So in the above algorithm we check if the length is equal and if exist such an index.
A very straightforward solution is to rotate one of the words n times, where n is the length of the word. For each of those rotations, check to see if the result is the same as the other word.
You can do it in O(n) time and O(1) space:
def is_rot(u, v):
n, i, j = len(u), 0, 0
if n != len(v):
return False
while i < n and j < n:
k = 1
while k <= n and u[(i + k) % n] == v[(j + k) % n]:
k += 1
if k > n:
return True
if u[(i + k) % n] > v[(j + k) % n]:
i += k
else:
j += k
return False
See my answer here for more details.
Simple solution in Java. No need of iteration or concatenation.
private static boolean isSubString(String first, String second){
int firstIndex = second.indexOf(first.charAt(0));
if(first.length() == second.length() && firstIndex > -1){
if(first.equalsIgnoreCase(second))
return true;
int finalPos = second.length() - firstIndex ;
return second.charAt(0) == first.charAt(finalPos)
&& first.substring(finalPos).equals(second.subSequence(0, firstIndex));
}
return false;
}
Test case:
String first = "bottle";
String second = "tlebot";
Logic:
Take the first string's first character, find the index in the second string. Subtract the length of the second with the index found, check if first character of the second at 0 is same as character at the difference of length of the second and index found and substrings between those 2 characters are the same.
Another python implementation (without concatenation) although not efficient but it's O(n), looking forward for comments if any.
Assume that there are two strings s1 and s2.
Obviously, if s1 and s2 are rotations, there exists two sub strings of s2 in s1, the sum of them will total to the length of the string.
The question is to find that partition for which I increment an index in s2 whenever a char of s2 matches with that of s1.
def is_rotation(s1, s2):
if len(s1) != len(s2):
return False
n = len(s1)
if n == 0: return True
j = 0
for i in range(n):
if s2[j] == s1[i]:
j += 1
return (j > 0 and s1[:n - j] == s2[j:] and s1[n - j:] == s2[:j])
The second and condition is just to ensure that the counter incremented for s2 are a sub string match.
input1= "hello" input2="llohe" input3="lohel"(input3 is special case)
if length's of input 1 & input2 are not same return 0.Let i and j be two indexes pointing to input1 and input2 respectively and initialize count to input1.length. Have a flag called isRotated which is set to false
while(count != 0){
When the character's of input1 matches input2
increment i & j
decrement count
If the character's donot match
if isRotated = true(it means even after rotation there's mismatch) so break;
else Reset j to 0 as there's a mismatch. Eg:
Please find the code below and let me know if it fails for some other combination I may not have considered.
public boolean isRotation(String input1, String input2) {
boolean isRotated = false;
int i = 0, j = 0, count = input1.length();
if (input1.length() != input2.length())
return false;
while (count != 0) {
if (i == input1.length() && !isRotated) {
isRotated = true;
i = 0;
}
if (input1.charAt(i) == input2.charAt(j)) {
i++;
j++;
count--;
}
else {
if (isRotated) {
break;
}
if (i == input1.length() - 1 && !isRotated) {
isRotated = true;
}
if (i < input1.length()) {
j = 0;
count = input1.length();
}
/* To handle the duplicates. This is the special case.
* This occurs when input1 contains two duplicate elements placed side-by-side as "ll" in "hello" while
* they may not be side-by-side in input2 such as "lohel" but are still valid rotations.
Eg: "hello" "lohel"
*/
if (input1.charAt(i) == input2.charAt(j)) {
i--;
}
i++;
}
}
if (count == 0)
return true;
return false;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println(new StringRotation().isRotation("harry potter",
"terharry pot"));
System.out.println(new StringRotation().isRotation("hello", "llohe"));
System.out.println(new StringRotation().isRotation("hello", "lohell"));
System.out.println(new StringRotation().isRotation("hello", "hello"));
System.out.println(new StringRotation().isRotation("hello", "lohe"));
}
Solving the problem in O(n)
void isSubstring(string& s1, string& s2)
{
if(s1.length() != s2.length())
cout<<"Not rotation string"<<endl;
else
{
int firstI=0, secondI=0;
int len = s1.length();
while( firstI < len )
{
if(s1[firstI%len] == s2[0] && s1[(firstI+1) %len] == s2[1])
break;
firstI = (firstI+1)%len;
}
int len2 = s2.length();
int i=0;
bool isSubString = true;
while(i < len2)
{
if(s1[firstI%len] != s2[i])
{
isSubString = false;
break;
}
i++;
}
if(isSubString)
cout<<"Is Rotation String"<<endl;
else
cout<<"Is not a rotation string"<<endl;
}
}
String source = "avaraavar";
String dest = "ravaraava";
System.out.println();
if(source.length()!=dest.length())
try {
throw (new IOException());
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
int i = 0;
int j = 0;
int totalcount=0;
while(true)
{
i=i%source.length();
if(source.charAt(i)==dest.charAt(j))
{
System.out.println("i="+i+" , j = "+j);
System.out.println(source.charAt(i)+"=="+dest.charAt(j));
i++;
j++;
totalcount++;
}
else
{
System.out.println("i="+i+" , j = "+j);
System.out.println(source.charAt(i)+"!="+dest.charAt(j));
i++;
totalcount++;
j=0;
}
if(j==source.length())
{
System.out.println("Yes its a rotation");
break;
}
if(totalcount >(2*source.length())-1)
{
System.out.println("No its a rotation");
break;
}
}