# Another O(1) space O(N) time solution, 24MS, idea is swapping block by block

• Idea is swapping the k-element block to the left, each time 1 block is swapped.
If some elements are remained at the left rotate these element to the right and vice - versa.

``````  class Solution {
public:
void rotate(vector<int>& nums, int k) {
k = k % nums.size();
if( k==0) return;
rotateRight(nums, k, 0, nums.size()-1);
}
private:
void rotateLeft(vector<int>& nums, int k, int start, int end)
{
int t = start;
int remaining = end-start+1-k;
while(remaining>=k)
{
for(int i=t; i< t+k; i++)
swap( nums[i], nums[i+k] );
remaining -= k;
t+=k;
}
if(remaining > 0)
rotateRight(nums, remaining, end-remaining-k+1, end);
}

void rotateRight(vector<int>& nums, int k, int start, int end)
{
int t = end - k +1;
int remaining = t-start;
while(remaining>=k)
{
for(int i=t; i< t+k; i++)
swap( nums[i], nums[i-k] );
remaining -= k;
t -= k;
}
if(remaining > 0)
rotateLeft(nums, remaining, start, start+remaining+k-1);
}
};``````

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