Rotate Array

Problem

Rotate an array of n elements to the right by k steps.

Example

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].

Solution

If you rotate one step to the right each time and rotate k times (code below), you will get Time Limit Exceeded. Time complexity O(k*n), Space complexity O(1).

// !!! This will time out
k %= nums.size();
if (k == 0) return;
int i, j, tmp;
for (i = 0; i < k; i++) {
    tmp = nums[nums.size()-1];
    for (j = nums.size()-1; j > 0; j--) {
        nums[j] = nums[j-1];
    }
    nums[0] = tmp;
}

A better solution is to create a new array and fill each element right into the position. Time complexity O(n), Space complexity O(n).

k %= nums.size();
if (k == 0) return;
int i;
vector<int> tmp;
for (i = 0; i < nums.size(); i++) tmp.push_back(nums[i]);
for (i = 0; i < tmp.size(); i++) {
    nums[(i+k)%nums.size()] = tmp[i];
}

To further reduce space complexity, first reverse the whole array in place. Then reverse the first k elements, and also the rest n-k elements. Time complexity O(n), Space complexity O(1).

k %= nums.size();
if (k == 0) return;
int i, j, tmp;
for (i = 0, j = nums.size()-1; i < j; i++, j--) {
    tmp = nums[i];
    nums[i] = nums[j];
    nums[j] = tmp;
}
for (i = 0, j = k-1; i < j; i++, j--) {
    tmp = nums[i];
    nums[i] = nums[j];
    nums[j] = tmp;
}
for (i = k, j = nums.size()-1; i < j; i++, j--) {
    tmp = nums[i];
    nums[i] = nums[j];
    nums[j] = tmp;
}