Javascript Program for Block swap algorithm for array rotation
Write a function rotate(ar[], d, n) that rotates arr[] of size n by d elements.
Rotation of the above array by 2 will make array
Algorithm :
Initialize A = arr[0..d-1] and B = arr[d..n-1] 1) Do following until size of A is equal to size of B a) If A is shorter, divide B into Bl and Br such that Br is of same length as A. Swap A and Br to change ABlBr into BrBlA. Now A is at its final place, so recur on pieces of B. b) If A is longer, divide A into Al and Ar such that Al is of same length as B Swap Al and B to change AlArB into BArAl. Now B is at its final place, so recur on pieces of A. 2) Finally when A and B are of equal size, block swap them.
Recursive Implementation:
<script>
let leftRotate = (arr, d, n) =>{
/* Return If number of elements to be rotated
is zero or equal to array size */
if(d == 0 || d == n)
return;
/*If number of elements to be rotated
is exactly half of array size */
if(n - d == d) {
arr = swap(arr, 0, n - d, d);
return;
}
/* If A is shorter*/
if(d < n - d) {
arr = swap(arr, 0, n - d, d);
leftRotate(arr, d, n - d);
}
else{
/* If B is shorter*/
arr = swap(arr, 0, d, n - d);
/*This is tricky*/
leftRotate(arr + n - d, 2 * d - n, d);
}
}
/*UTILITY FUNCTIONS*/
/* function to print an array */
let printArray = (arr, size) =>{
ans = ''
for(let i = 0; i < size; i++)
ans += arr[i]+" ";
document.write(ans)
}
/*This function swaps d elements
starting at index fi
with d elements starting at index si */
let swap = (arr, fi, si, d) =>{
for(let i = 0; i < d; i++) {
let temp = arr[fi + i];
arr[fi + i] = arr[si + i];
arr[si + i] = temp;
}
return arr
}
// Driver Code
arr = [1, 2, 3, 4, 5, 6, 7];
leftRotate(arr, 2, 7);
printArray(arr, 7);
</script>
Output:
3 5 4 6 7 1 2
Time Complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(N), due to recursive stack space.
Iterative Implementation:
Here is iterative implementation of the same algorithm. Same utility function swap() is used here.
<script>
// JavaScript code for above implementation
function leftRotate(arr, d, n)
{
if(d == 0 || d == n)
return;
let i = d;
let j = n - d;
while (i != j)
{
if(i < j)
{
// A is shorter
arr = swap(arr, d - i, d + j - i, i);
j -= i;
}
else{ // B is shorter
arr = swap(arr, d - i, d, j);
i -= j;
}
}
arr = swap(arr, d - i, d, i);
// This code is contributed by rohitsingh04052.
}
</script>
Time Complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please see following posts for other methods of array rotation:
https://www.geeksforgeeks.org/array-rotation/
https://www.geeksforgeeks.org/program-for-array-rotation-continued-reversal-algorithm/
Please write comments if you find any bug in the above programs/algorithms or want to share any additional information about the block swap algorithm.
Please refer complete article on Block swap algorithm for array rotation for more details!