RoatateNumberInRightKTimes.java

packageXtraQuestions.StringsArrayMixed;
// You are given two numbers n and k. You are required to rotate n k times to the right. If kis positive, rotate to the right l.e. remove rightmost digit and make it leftmost. Do the reverse for negative value of k. Also k can have an absolute value larger than number of digits in n. 2. Take as input n and k. 3. Print the rotated number. 4. 
// Note - Assume that the number of rotations will not cause leading O's in the result, e.g. such an input will not be given
// n = 12340056 k= 3 r = 05612340
// Constraints 1 <= n < 10^9 - 10^9 < k < 10^9
// Format
// Input
// "n" where n is any integer. "K" where k is any integer.
// Output
// "r", the rotated number
// SAMPLE INPUT
// 562984
// 2
// OUTPUT
// 845629

importjava.util.*;

publicclassRoatateNumberInRightKTimes {
publicstaticvoidmain(String[] args) {
// int n = 12340056; // This type of number is not allowed
intn = 124452;
System.out.println(n);

intk = 2;

intfrontnum = 0;
for(inti=0; i<k; i++){
intrem = n%10;
intremaining = n/10;

// for 0's in between we have to use this
// if(rem==0){
// String padded = String.format("%08d" , remaining);
// // System.out.println("padded: "+ padded);
// n = Integer.parseInt(padded);
// // n = padded;
// System.out.println("the n is : " + n);
// continue;
// }
intcount = 0;
inttemp = remaining;
while(temp!=0){
count++;
temp = temp/10;
 }
// System.out.println("count : " + count);
frontnum = rem*(int)(Math.pow(10,count)) + remaining;
n = frontnum;
// System.out.println("frontnum : " + frontnum);
 }

System.out.println("The number after " + k + " rotations: " + frontnum);
 }
}