Move Zeroes Leetcode Problem 283 [Python Solution]

Welcome to another exciting problem-solving session, we’re going to tackle the LeetCode problem titled Move Zeroes.

This problem falls under the category of “Two Pointers” and is classified as an “Easy” difficulty level.

We’ll provide a Python solution to this problem, explaining it step by step, optimizing for SEO, and ensuring clarity for beginners.

But first, let’s set the stage with a clear problem statement.

Problem Overview

The task at hand is to take an integer array, nums, and move all the zeros in the array to the end while maintaining the relative order of the non-zero elements.

The catch is that we must do this operation in-place, without creating a separate copy of the array.

Let’s dive into the problem by examining an example:

Example 1:

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

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

In this example, we start with an array containing zeros and non-zero values.

After performing the operation, we move all the zeros to the end, and the relative order of non-zero elements is preserved.

To ensure clarity for beginners, let’s break down the problem further.

Understanding Move Zeroes Constraints

Before we jump into the solution, it’s important to understand the constraints of the problem:

• The length of the nums array will be between 1 and 10,000.
• The values in the nums array can range from -2,147,483,648 to 2,147,483,647. Now that we have a clear understanding of the problem, let’s proceed to the solution.

Move Zeroes LeetCode Problem Solution

Efficient Approach

To solve this problem efficiently, we’ll use a two-pointer approach.

We’ll maintain two pointers, left and right, initialized at the beginning of the array.

The right pointer will iterate through the entire array.

def moveZeroes(self, nums: List[int]) -> None:
    """
    Do not return anything, modify nums in-place instead.
    """
    left = 0  # Initialize the left pointer
    for right in range(len(nums)):
        if nums[right] != 0:
            nums[left], nums[right] = nums[right], nums[left]
            left += 1

The idea behind this approach is to keep the left pointer pointing to the position where the next non-zero value should be placed.

Whenever we encounter a non-zero value at the right pointer, we swap it with the value at the left pointer.

This effectively moves the non-zero value to the left side of the array while preserving the relative order of non-zero values.

The efficient approach ensures that we iterate through the array only once, resulting in a time complexity of O(n), where n is the length of the nums array.

Additionally, this approach uses constant space, making it an in-place solution with a space complexity of O(1).

Time and Space Complexity

To summarize, the efficient approach for the Move Zeroes problem has the following time and space complexities:

• Time Complexity: O(n) – We iterate through the entire array exactly once.
• Space Complexity: O(1) – We use a constant amount of extra space for the two pointers and a few variables.

Reasoning Behind Our Approach

The key to solving this problem efficiently lies in understanding that we can reverse the problem statement.

Instead of moving zeros to the end, we can think of it as moving non-zero values to the beginning.

This subtle shift in perspective allows us to maintain the relative order of non-zero elements while solving the problem in-place.

By using two pointers, we efficiently track the positions for swapping and achieve the desired result.

This approach is not only efficient but also a valuable technique in problem-solving, applicable to various scenarios beyond this specific problem.

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Conclusion

In this blog post, we tackled the Move Zeroes LeetCode problem with an efficient Python solution.

We explored the problem statement, understood its constraints, and provided a step-by-step explanation of our approach.

By reversing the problem statement and employing a two-pointer technique, we successfully solved the problem in-place while preserving the relative order of non-zero elements.

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If you have any questions, suggestions, or would like to discuss other problem-solving techniques, please don’t hesitate to leave a comment.

For more exciting coding challenges and in-depth explanations, stay tuned for more content on your daily dose of neat code.

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Remember, every small problem you solve brings you one step closer to becoming a confident coder.

Happy coding!