DSA interview question

Move Zeros to the End

Easy · Algorithms · Topics: array, two pointer

The problem

A sensor log stores one reading per second, with 0 meaning "no reading". The analytics team wants all real readings at the front in their original order, followed by all the zeros, so a downstream job can process the prefix without skipping blanks. It must be done in place. Removing each zero and appending it is O(n²) because of all the shifting. A two-pointer pass does it in O(n): one pointer scans, the other marks where the next non-zero value should go.

Example: [0, 1, 0, 3, 12] → [1, 3, 12, 0, 0].

Function signature

void moveZeroes(vector<int>& a);  // in place; non-zero elements keep their relative order

Test cases

Input `a`	Output	Explanation
`[0, 1, 0, 3, 12]`	`[1, 3, 12, 0, 0]`	Order of 1, 3, 12 preserved
`[4, 0, 5, 0, 0, 6]`	`[4, 5, 6, 0, 0, 0]`
`[0, 0, 1]`	`[1, 0, 0]`
`[1, 2, 3]`	`[1, 2, 3]`	No zeros: unchanged
`[0]`	`[0]`
`[]`	`[]`

(Confirm: in place, O(1) extra space; the relative order of non-zero elements must not change.)

How interviewers phrase it

"Move all zeros to the end while keeping the order of the other elements" (LeetCode 283).
Variants: "move all negatives to the front," "partition by a condition while keeping order."

Clarifying questions to ask

Must the non-zero elements keep their order? (That rules out simple swapping from both ends.)
In place?
Should I minimise the number of writes?

Follow-ups to expect

"Why does swapping preserve order?"
"Can you reduce writes?"
"Why not two pointers from both ends?"
"How is this related to partitioning in quick sort?"
"Complexity?"

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