What is the brute force approach and complexity for Move Zeroes?

Copy non-zeros into a new array, pad with zeros, copy back with time complexity O(N) and space complexity O(N).

What is the optimal approach and complexity for Move Zeroes?

Swap each non-zero into nums[write]; the zeros move back on their own with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Move Zeroes?

Look for key signals in the prompt: "move all 0s to the end" indicates Keep the non-zeros at the front → filter with read / write, `val = 0`.; "keep the relative order of the non-zero elements" indicates Write them forward one by one, never swap two non-zeros with each other.; "in place, without a copy" indicates Reuse the front of `nums`; swap so the zeros are not lost..

Move Zeroes

Easy·LeetCode #283 Pro OnlySlight Variation

LeetCode

Problem Statement

You are given an integer array nums.

Move every 0 to the end of the array, while the non-zero elements stay in the same relative order they started in.

Do it in place — change nums directly, without making a copy of the array. The function returns nothing.

Follow-up: can you keep the number of operations small?

Input

An integer array nums.

Output

Nothing is returned. nums itself must end with all non-zeros first (in original order) followed by all the zeros.

Example 1

Input: nums = [0, 1, 0, 3, 12]

Output: [1, 3, 12, 0, 0]

The non-zeros 1, 3, 12 keep their order and move to the front. The two zeros go to the end.

Example 2

Input: nums = [0]

Output: [0]

Only a zero — nothing to move.

Constraints

• 1 ≤ nums.length ≤ 10^4
• −2^31 ≤ nums[i] ≤ 2^31 − 1

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“move all 0s to the end”

→ Keep the non-zeros at the front → filter with read / write, `val = 0`.

“keep the relative order of the non-zero elements”

→ Write them forward one by one, never swap two non-zeros with each other.

“in place, without a copy”

→ Reuse the front of `nums`; swap so the zeros are not lost.

Approach Complexity Ladder

Brute Force — build a new arrayO(N)O(N)

Copy non-zeros into a new array, pad with zeros, copy back

Better — read / write, then fill zerosO(N)O(1)

Pack non-zeros to the front with read/write, then set the tail to 0

Optimal — read / write with swap (one pass)O(N)O(1)

Swap each non-zero into nums[write]; the zeros move back on their own

Brute Force Intuition

Make a new array. Append every non-zero in order, then append zeros until it has the same length as nums. Copy it back.

Recognition rule for this sub-pattern:

One pointer reads every element, the other marks where the next "kept" element should be written.

Remember this · Two Pointers

Trigger

Sorted array, pairs/triplets with a target, comparing from both ends, compacting an array in place.

Think

If I compare the elements at two positions, can I safely discard one side?

Key question

Which pointer moves, and why can the skipped elements never be the answer?

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