Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4]. How many different ways do you know to solve this problem?
Using intermediate array
- First find out correct number of rotations using mod
- Create temporary array
- Push array elements into that temporary array
- Copy that temporary array into original array
function rotateArray(nums,k) { if(isNaN(k) || k < 0) { throw "provide valid order"; } if(nums.length == 0) { throw "provide valid array"; } if(k > nums.length) k=k%nums.length; var result = []; for(var i=0; i < k; i++){ result[i] = nums[nums.length-k+i]; } var j=0; for(var i=k; i<nums.length; i++){ result[i] = nums[j]; j++; } nums = result.slice(); console.log(nums); }
Complexity
Observe above demo. It will take O(n) space and O(n) time
Using bubble rotate
By using bubble swap, we can solve this problem. Move array elements from right side to left side of the array using bubble swap
function rotateArray2(arr, order) { if(isNaN(order) || order < 0) { throw "provide valid order"; } if(arr.length == 0) { throw "provide valid array"; } for (var i = 0; i < order; i++) { for (var j = arr.length - 1; j > 0; j--) { var temp = arr[j]; arr[j] = arr[j - 1]; arr[j - 1] = temp; } } console.log(arr); }
Complexity
Observe above demo. It will take O(1) space and O(n*k) time
Using Reversal
Divide array into 2 parts based on given order. Reverse first part and then reverse second, then reverse whole array
function rotate3(arr, order) { if(isNaN(order) || order < 0) { throw "provide valid order"; } if(nums.length == 0) { throw "provide valid array"; } order = order % arr.length; //length of first part var a = arr.length - order; reverse(arr, 0, a-1); reverse(arr, a, arr.length-1); reverse(arr, 0, arr.length-1); console.log(arr); } function reverse(arr, left, right){ while(left < right){ var temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; left++; right--; } } rotate3([1,2,3,4,5,6],3);
Complexity
Observe above demo. It will take O(1) space and O(n) time
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