DSA interview question

Left Rotate an Array by K Places

Easy · Algorithms · Topics: array

The problem

The advert carousel from the previous problem gets a "skip ahead K adverts" button: the first K adverts move to the end in one go. Rotating by one, K times, is O(n·K), which is too slow when both are large. Copying into a new array is O(n) but uses O(n) memory the controller doesn't have. The elegant answer is the reversal algorithm: reverse the first K elements, reverse the rest, then reverse the whole array. That's O(n) time and O(1) space, and it's worth being able to explain why it works.

Example: [1, 2, 3, 4, 5, 6, 7], K = 2 → [3, 4, 5, 6, 7, 1, 2].

Function signature

void leftRotate(vector<int>& a, int k);   // in place; k may exceed n
void rightRotate(vector<int>& a, int k);  // in place (the LeetCode 189 direction)

Test cases

Input `(a, k)`	Output	Explanation
`([1, 2, 3, 4, 5, 6, 7], 2)` left	`[3, 4, 5, 6, 7, 1, 2]`
`([1, 2, 3, 4, 5, 6, 7], 0)` left	`[1, 2, 3, 4, 5, 6, 7]`	Nothing to do
`([1, 2, 3, 4, 5, 6, 7], 7)` left	`[1, 2, 3, 4, 5, 6, 7]`	Full cycle
`([1, 2, 3, 4, 5, 6, 7], 9)` left	`[3, 4, 5, 6, 7, 1, 2]`	9 mod 7 = 2
`([1, 2], 3)` left	`[2, 1]`	3 mod 2 = 1
`([], 3)` left	`[]`	Guard against dividing by zero
`([1, 2, 3, 4, 5, 6, 7], 3)` right	`[5, 6, 7, 1, 2, 3, 4]`	LeetCode 189 example

(Confirm: k ≥ 0 and may be larger than n; O(1) extra space.)

How interviewers phrase it

"Rotate an array by K steps" (LeetCode 189 rotates to the right).
"Left rotate an array by D places."
"Do it in O(n) time and O(1) extra space."

Clarifying questions to ask

Left or right rotation?
Can k be larger than n, or negative?
Is O(n) extra space allowed?

Follow-ups to expect

"Why does the reversal trick work?"
"Why take k mod n?"
"How do you rotate right instead?"
"What other O(n) approaches exist?"
"Complexity?"

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