Given an array, cyclically rotate the array clockwise by one.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}
Output: arr[] = {5, 1, 2, 3, 4}
Following are steps.
1) Store last element in a variable say x.
2) Shift all elements one position ahead.
3) Replace first element of array with x.
Javascript
<script>// JavaScript code for program // to cyclically rotate// an array by onefunction rotate(arr, n){ var x = arr[n-1], i; for(i = n-1; i > 0; i--) arr[i] = arr[i-1]; arr[0] = x; }var arr = [1, 2, 3, 4, 5];var n = arr.length;document.write("Given array is <br>");for(var i = 0; i< n; i++) document.write(arr[i] + " "); rotate(arr, n);document.write("<br>Rotated array is <br>");for(var i = 0; i < n; i++) document.write(arr[i] + " "); </script> |
Given array is 1 2 3 4 5 Rotated array is 5 1 2 3 4
Time Complexity: O(n) As we need to iterate through all the elements
Auxiliary Space: O(1)
The above question can also be solved by using reversal algorithm.
Another approach:
We can use two pointers, say i and j which point to first and last element of array respectively. As we know in cyclic rotation we will bring last element to first and shift rest in forward direction, so start swapping arr[i] and arr[j] and keep j fixed and i moving towards j. Repeat till i is not equal to j.
Javascript
<script>// JavaScript code for program// to cyclically rotate// an array by one using pointers i,jfunction rotate(arr, n){ var i = 0 var j = n-1 while(i != j){ let temp; temp = arr[i]; arr[i] = arr[j]; arr[j]= temp; i =i+1 }}var arr = [1, 2, 3, 4, 5];var n = arr.length;document.write("Given array is <br>");for(var i = 0; i< n; i++) document.write(arr[i] + " "); rotate(arr, n);document.write("<br>Rotated array is <br>");for(var i = 0; i < n; i++) document.write(arr[i] + " ");</script> |
Given array is 1 2 3 4 5 Rotated array is 5 1 2 3 4
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Program to cyclically rotate an array by one for more details!
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