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February 22, 2017 00:47
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import java.io.*; | |
import java.util.*; | |
public class Solution { | |
public static void main(String[] args) { | |
//initializing regular code... | |
Scanner s = new Scanner(System.in); | |
int n = s.nextInt(); | |
int d = s.nextInt(); | |
int[] arr = new int[n]; | |
for(int i = 0; i < n; i++) | |
arr[i] = s.nextInt(); | |
//for optimal solution, if n==d or d is multiple of n then we do nothing. | |
//example: move array of 5 elements, 5 place to the left. We do nothing. | |
//Therefor we remove from d the multiple of n. | |
//if d is not multiple of n | |
if(d%n != 0) { | |
if(d > n) { | |
while(d>n) { | |
//here we remove n from d. (But we need to check when to stop.) | |
if(d < n) | |
break; | |
d -= n; | |
} | |
} | |
/* | |
after all the work we got d that is 100% less than n. which is good, so we don't need to do mutliple | |
rotations to get to the same result. | |
My optimal solution: | |
Create new list of size n. Add all elements from arr[] to new list, starting from index d (included) | |
up to index arr.length (which is n) . That way only the "rotating elements from left to right" will | |
be added later, and the "not rotating elements from left to right" are added immidietly, without checking. | |
*/ | |
int[] newArr = new int[n]; | |
for(int i = 0; i < (n - d); i++) | |
newArr[i] = arr[d+i]; | |
//we done adding all "non rotating elements" | |
//now append all "moving elements" | |
//they are in arr[] from index (include) 0 to index (include) d-1 . | |
for(int i = 0; i < d; i++) | |
newArr[n-d+i] = arr[i]; | |
//and we done! | |
//print result | |
for(int i = 0; i < n; i++) | |
System.out.print(newArr[i] + " "); | |
} | |
else | |
//print result | |
for(int i = 0; i < n; i++) | |
System.out.print(arr[i] + " "); | |
} | |
} |
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