How do you solve Move Zeroes on LeetCode?

Move Zeroes (LeetCode #283) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using in-place operations minimizes space complexity.

What is the brute force solution for Move Zeroes?

The brute force approach for Move Zeroes is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves creating a new array to hold all non-zero elements while counting the zeros. After that, we fill the remaining positions in the original array with zeros. This is straightforward but not efficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Move Zeroes?

Move Zeroes uses the following concepts: Array, Two Pointers. The recommended patterns to study are: Two Pointers, In-place Array Manipulation.

What are the interview tips for Move Zeroes?

Clarify the constraints and requirements (in-place, order). Discuss edge cases (all zeros, no zeros). Explain your thought process clearly while coding.

What are common mistakes when solving Move Zeroes?

Not maintaining the order of non-zero elements. Using extra space unnecessarily when in-place is required.

#283

Move Zeroes

Easy

ArrayTwo PointersTwo PointersIn-place Array Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a two-pointer technique to efficiently rearrange the elements in-place. One pointer tracks the position for non-zero elements, while the other iterates through the array.

⚙️

Algorithm

4 steps

1Step 1: Initialize a pointer 'lastNonZeroFoundAt' to 0.
2Step 2: Iterate through the array with another pointer 'current'.
3Step 3: If the current element is non-zero, place it at 'lastNonZeroFoundAt' and increment 'lastNonZeroFoundAt'.
4Step 4: After the loop, fill the remaining positions in the array with zeros.

solution.py8 lines

1def moveZeroes(nums):
2    lastNonZeroFoundAt = 0
3    for current in range(len(nums)):
4        if nums[current] != 0:
5            nums[lastNonZeroFoundAt] = nums[current]
6            lastNonZeroFoundAt += 1
7    for i in range(lastNonZeroFoundAt, len(nums)):
8        nums[i] = 0

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once. The space complexity is O(1) since we are rearranging the elements in-place without using extra space.

1Using in-place operations minimizes space complexity.
2Two-pointer technique is efficient for rearranging elements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.