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Move Zeroes to End

Certification: Basic Level Accuracy: 66.67% Submissions: 3 Points: 5

Write a Java program to move all zeroes in an array to the end while maintaining the relative order of the non-zero elements. The operation should be done in-place with minimal operations.

Example 1
     
  • Input: nums = [0, 1, 0, 3, 12]
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  • Output: [1, 3, 12, 0, 0]
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  • Explanation: 
       
    • Step 1: Maintain a pointer for the position where next non-zero element should be placed.
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    • Step 2: Iterate through the array. When a non-zero element is found, place it at the pointer position.
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    • Step 3: After all non-zero elements are positioned correctly, fill the remaining positions with zeroes.
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    • Step 4: The array now has all zeroes at the end while preserving the order of non-zero elements.
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    •
Example 2
     
  • Input: nums = [0, 0, 1]
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  • Output: [1, 0, 0]
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  • Explanation: 
       
    • Step 1: Maintain a pointer for the position where next non-zero element should be placed.
      •  
    • Step 2: Iterate through the array. The only non-zero element 1 is placed at index 0.
      •  
    • Step 3: Fill the remaining positions (indices 1 and 2) with zeroes.
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    • Step 4: The array now has all zeroes at the end while preserving the order of non-zero elements.
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Constraints
     
  • 1 ≤ nums.length ≤ 10^4
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  • -2^31 ≤ nums[i] ≤ 2^31 - 1
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  • You must do this in-place without making a copy of the array
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  • Minimize the total number of operations
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  • Time Complexity: O(n) where n is the length of the array
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  • Space Complexity: O(1)
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Hint
Solution

Solution Hints

    
  
  • Use a two-pointer technique to solve this problem
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  • Keep track of the position where the next non-zero element should be placed
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  • Iterate through the array once
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  • When you find a non-zero element, place it at the tracked position and increment the position
    • 
  
  • After processing all elements, fill the remaining positions with zeroes
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Steps to solve by this approach:

 Step 1: Initialize a pointer to keep track of where the next non-zero element should go.
 Step 2: Iterate through the array with another pointer.
 Step 3: If the current element is non-zero, place it at the position of the first pointer and increment it.
 Step 4: After processing all elements, all non-zero elements are now at the beginning of the array in their original order.
 Step 5: Fill the remaining positions with zeroes.
 Step 6: The array now has all zeroes moved to the end while maintaining the relative order of non-zero elements.
 Step 7: Time complexity is O(n) as we only need to iterate through the array twice.

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