Move Zeroes

Today, we’ll tackle a common problem that often arises when working with integer arrays: how to effectively move all 0’s to the end while preserving the order of the non-zero elements. While it may seem like a simple task at first, this problem requires careful consideration of array manipulation techniques and optimization strategies. By the end of this article, you’ll have a better understanding of array manipulation in general.

Problem Statement:

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

Note that you must do this in-place without making a copy of the array.

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Approach 1: Two Pointers

1. Iterate through the array using two pointers, one to track the last non-zero index seen and another to iterate through the array.

2. For each non-zero element encountered, swap it with the element at the last non-zero index.

3. This approach avoids the need for an additional array.

// ES6 Arrow Function
const moveZeroes = nums => {
    let slowPointer = 0;

    for(let i = 0; i < nums.length; i++) {
        if(nums[i] !== 0) {
            [nums[slowPointer], nums[i]] = [nums[i], nums[slowPointer]];
            slowPointer++;
        }
    }

    return nums;
}

Time Complexity: O(N)

Space Complexity: O(1)

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Approach 2: Optimized Two-Pointers

Similar to the two-pointer approach, instead of swapping non-zero elements with the last non-zero element, simply replace the last non-zero element with the current non-zero element. This avoids unnecessary swaps and can improve performance.

// ES6 Arrow Function
const moveZeroes = nums => {
    let counter = 0;

    for(let i = 0; i < nums.length; i++) {
        if(nums[i] !== 0) {
            nums[counter] = nums[i];
            counter++;
        }
    }

    for(let i = counter; i < nums.length; i++) {
        nums[i] = 0;
    }

    return nums;
}

Time Complexity: O(N)

Space Complexity: O(1)

Note: This approach has the same O(n) time complexity as the previous approach, but is faster in practice due to the reduced number of operations needed.

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And there you have it guys! I hope this article has provided you with valuable insights and helped you better understand the different approaches to solving this problem. Happy coding!

Problem - Leetcode 283