🚀 Move Zeroes (In-Place & Efficient)

The Move Zeroes problem is a classic array manipulation question where the goal is to rearrange elements efficiently while preserving order.

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📌 Problem Statement

Given an array nums, move all 0s to the end while maintaining the relative order of non-zero elements.

The operation must be done in-place without using extra space.

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🔍 Examples

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

Example 2:
Input: [0]
Output: [0]

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🧠 Intuition

We need to:

• Keep all non-zero elements in the same order
• Move all zeros to the end

👉 This can be done efficiently using a two-pointer approach

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🔄 Approach (Optimal Two-Pointer Method)

Step-by-Step:

1. Initialize a pointer j = 0
   (this will track position for next non-zero element)

2. Traverse the array:

• If nums[i] != 0:

  • Swap nums[i] with nums[j]
  • Increment j

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💻 Python Code

def move_zeroes(nums):
j = 0

for i in range(len(nums)):
    if nums[i] != 0:
        nums[i], nums[j] = nums[j], nums[i]
        j += 1

return nums

print(move_zeroes([0, 1, 0, 3, 12]))

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🧾 Dry Run

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

Step-by-step:

• i = 0 → 0 → skip
• i = 1 → 1 → swap with index 0 → [1, 0, 0, 3, 12]
• i = 2 → 0 → skip
• i = 3 → 3 → swap with index 1 → [1, 3, 0, 0, 12]
• i = 4 → 12 → swap with index 2 → [1, 3, 12, 0, 0]

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

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⚡ Complexity

Time Complexity: O(n)
Space Complexity: O(1)

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🔥 Why This Works

• Maintains order of non-zero elements
• Minimizes unnecessary swaps
• Uses constant extra space

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⚠️ Follow-Up Optimization

To minimize operations:

👉 Only swap when i != j
This avoids swapping an element with itself.

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🏁 Conclusion

Move Zeroes is a simple yet powerful problem that teaches:

• Two-pointer technique
• In-place array manipulation
• Efficient traversal

Mastering this helps in many real-world and interview scenarios.