Imagine you are handed the array $\text{nums}=[1,2,3,4,5,6,7]$ and told to rotate it to the right by $k=3$ steps. If you shift elements one step at a time, the last element moves to the front, and everything else shuffles over. Doing this once turns our array into $[7,1,2,3,4,5,6]$. Doing it three times successfully yields $[5,6,7,1,2,3,4]$. But what happens when we scale up to $k=10$? Shifting element-by-element 10 separate times requires moving 70 individual items, even though the array length $n=7$ means rotating by 10 is identical to rotating by $10(\mathrm{mod}7)=3$ steps. As $k$ grows larger than the array length, naive shifting wastes redundant operations, highlighting the need for a smarter way to handle array shifts without looping $k$ times.

Instead of shifting individual elements of $[1,2,3,4,5,6,7]$ one position to the right $k$ times—which forces redundant movements—we can achieve the exact same final state by manipulating contiguous blocks of memory. The three-step reversal algorithm treats the array as two joined segments, $A$ and $B$, where $A$ contains the first $n-k$ elements and $B$ contains the last $k$ elements. To swap their positions from $AB$ to $BA$, we reverse the entire array first, then independently reverse the first $k$ elements, and finally reverse the remaining $n-k$ elements. When $k=3$, $A=[1,2,3,4]$ and $B=[5,6,7]$, and this geometric choreography reorganizes the elements in $O(n)$ time without allocating a secondary array.

Now that we know the three-step reversal sequence, let us apply it directly to our locked working example: Given $\text{nums}=[1,2,3,4,5,6,7]$ and $k=3$.

Now that we have successfully traced our 7-element array with $k=3$ and $k=10$, let us test the limits of our 3-step reversal pattern on a completely different array size and rotation count. Suppose we are given a new array of length $n=5$, specifically $nums=[10,20,30,40,50]$, and we need to rotate it by $k=12$ steps to the right. We first normalize the large shift count using modulo arithmetic: $k=12(\mathrm{mod}5)=2$. Next, we apply our proven three-step recipe: reverse the entire array to get $[50,40,30,20,10]$, reverse the first $k=2$ elements to get $[40,50,30,20,10]$, and finally reverse the remaining $n-k=3$ elements. This structural transfer proves that the triple-reversal algorithm is independent of specific array values and scales cleanly to any positive integer length.

def rotate_transfer_check():
    nums = [10, 20, 30, 40, 50]
    k = 12
    # Apply the 3-step reversal model here
    # Expected final state: [40, 50, 10, 20, 30]