1. Introduction
In this post, we will explore a solution to the "Move Zeroes" problem, a common task in array manipulation. The objective is to shift all zeros in an integer array to the end while maintaining the relative order of the non-zero elements. Importantly, this must be done in-place, without creating a copy of the array.
Problem
Given an integer array nums, the challenge is to move all 0's to the end of the array while preserving the order of non-zero elements. The operation must be performed in-place.
Examples:
- Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]
- Input: nums = [0]
Output: [0]
2. Solution Steps
1. Initialize a pointer insertPos to track the position where the next non-zero element should be placed.
2. Iterate through the array:
a. When a non-zero element is encountered, place it at the insertPos index and increment insertPos.
3. After all non-zero elements are moved to the front, fill the remaining array with 0's starting from insertPos.
4. The in-place swap ensures that the relative order of non-zero elements is maintained.
3. Code Program
public class MoveZeroes { public static void main(String[] args) { int[] nums = {0, 1, 0, 3, 12}; moveZeroes(nums); for (int num : nums) { System.out.print(num + " "); } } // Function to move zeroes to the end of the array public static void moveZeroes(int[] nums) { int insertPos = 0; for (int num : nums) { if (num != 0) { nums[insertPos++] = num; } } while (insertPos < nums.length) { nums[insertPos++] = 0; } } }
Output:
1 3 12 0 0
Explanation:
1. moveZeroes: This function rearranges the array to move all zeroes to the end.
2. The insertPos variable keeps track of where the next non-zero element should be placed.
3. As the function iterates through the array, each non-zero element is moved to the position indicated by insertPos, and insertPos is incremented.
4. After all non-zero elements are moved to the front, the remaining part of the array is filled with zeroes.
5. This approach maintains the order of non-zero elements and performs the operation in-place, adhering to the problem's constraints.
6. The method efficiently solves the problem with a single pass through the array, resulting in O(n) time complexity and O(1) space complexity.
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