✅✅✅Simple || Faster than 100%
IntuitionWe want to check if after shifting the cycle is it as original or not.Approach1.Get one pointer at the low at 0 and another at the high at k .
2.To che
# Intuition
We want to check if after shifting the cycle is it as original or not.

# Approach
1.Get one pointer at the low at 0 and another at the high at k .
2.To check after cycling it is as original or not .
3.Keep Incrementing the high and round it off in the columns.
4.check each time is it equal or not.

# Complexity
- Time complexity:
O(n*m);

- Space complexity:
O(1)

# Code
```c []
bool areSimilar(int** mat, int matSize, int* matColSize, int k) 
{
    int i,low,high;

    for(i=0;i<matSize;i++)
    {
        high = k;
        for(low=0;low<matColSize[0];low++)
        {
            if(mat[i][low]!=mat[i][high%matColSize[0]])
            {
                return false;
            }

            high++;
        }
    }    

    return true;
}
```
['Array', 'C', 'Matrix']