Move Zeroes

The problem requires us to move all zeros in the given array to the end while keeping the relative order of non-zero elements the same. Since we have to do this in-place without making a copy of the array, we will use a two-pointer technique.

Approach

  1. Initialize Two Pointers: Start with two pointers, say writePointer and readPointer. writePointer is initialized at 0.
  2. Iterate through the Array: Iterate through the array using readPointer.
    • If nums[readPointer] is not zero, place it at nums[writePointer] and increment writePointer.
  3. Fill the Remaining Part with Zeros: Once all non-zero elements are processed, fill the remaining part of the array with zeros from the writePointer to the end of the array.

Code

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class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        writePointer = 0

        for readPointer in range(len(nums)):
            if nums[readPointer] != 0:
                nums[writePointer] = nums[readPointer]
                writePointer += 1

        for i in range(writePointer, len(nums)):
            nums[i] = 0

Explanation

  • Pointer writePointer keeps track of the position where the next non-zero element should be placed.
  • Pointer readPointer iterates through the array to find non-zero elements.
  • When a non-zero element is found at readPointer, it is placed at writePointer.
  • Zeros naturally get pushed to the end of the array as the non-zero elements are rearranged.

Key Takeaways

  • This approach modifies the array in-place, fulfilling the problem’s requirement.
  • The time complexity of this solution is (O(n)), where (n) is the length of the array, and the space complexity is (O(1)), since no additional space is used.

class Solution: def moveZeroes(self, nums: List[int]) -> None: writePointer = 0

    # Read pointer iterates through the array to find non-zero elements
    for readPointer in range(len(nums)):
        if nums[readPointer] != 0:
            # Place the non-zero element at the writePointer and increment it
            nums[writePointer] = nums[readPointer]
            writePointer += 1

    # Fill the rest of the array with zeros
    for i in range(writePointer, len(nums)):
        nums[i] = 0


## Identifying Problem Isomorphism

"Move Zeroes" can be approximately mapped to "Remove Element".

Reasoning:

Both problems require manipulating a given array in-place. 

1. "Move Zeroes" asks you to move all 0's to the end of an array while maintaining the relative order of the non-zero elements.

2. "Remove Element" requires you to remove all instances of a specific value from the array and return the new length.

The strategy for solving these problems is similar: you can use two-pointer technique where one pointer maintains the position of the next element in the modified array and the other scans through the array.

Both problems emphasize on array manipulation and in-place operations.

"Move Zeroes" is a fairly straightforward problem that introduces the concept of in-place array manipulations. For easier problems, you could start with these:

1. "344. Reverse String": This problem helps you understand the concept of manipulating an array or string in-place, and the idea of using two pointers.
2. "136. Single Number": This problem introduces the concept of bit manipulation, but it can also be solved using a hash map, which is a common data structure in other problems.
3. "88. Merge Sorted Array": This is another problem that requires in-place array manipulation, but with an added complexity of dealing with two sorted arrays.

These cover in-place array manipulation, which will be crucial in solving the "Move Zeroes" problem. 

```python
class Solution:
    def moveZeroes(self, nums: list) -> None:
        slow = 0
        for fast in range(len(nums)):
            if nums[fast] != 0 and nums[slow] == 0:
                nums[slow], nums[fast] = nums[fast], nums[slow]

            if nums[slow] != 0:
                slow += 1

Problem Classification

This problem can be classified under the following categories:

  1. Array Manipulation: The problem is about rearranging the elements of an array.

  2. In-Place Algorithms: You are required to modify the array in-place without creating a new array or copying the existing one.

  3. Two Pointers Approach: The problem’s most efficient solution uses the two-pointers method where one pointer tracks the current element and another one to place the non-zero element.

  4. Order Maintenance: The relative order of the non-zero elements needs to be maintained as per the problem statement. Hence it falls under the order maintenance problems as well.

In terms of difficulty, this problem could be classified as easy-to-intermediate as it requires a fundamental understanding of array manipulation and two pointers strategy but also involves more complex considerations like maintaining relative order and doing the operation in-place.

Language Agnostic Coding Drills

Breaking down this problem and code into its core components can be done as follows:

  1. Understanding Array Traversal: The foundation of this problem is about understanding how to iterate through an array. The traversal in this problem happens linearly, from start to end.

  2. Two Pointers Methodology: In this problem, we use two pointers, slow and fast. Understanding how to use two pointers, one progressing at a slower pace and the other progressing faster is crucial for solving this problem.

  3. In-Place Array Manipulation: The problem requires the solution to be done in-place, that is, the manipulation happens within the given array and no new array is created for the solution. This is a crucial step as it needs a clear understanding of how variables/objects are stored and manipulated in memory.

  4. Condition-Based Manipulation: During the traversal, there is a condition to be checked at each step. If the element at the fast pointer is non-zero and the element at the slow pointer is zero, then swap these elements. This combines understanding of condition checks and the Swap operation.

  5. Incrementation of Pointers: After each iteration, it’s necessary to check whether to increment the slow pointer. In this problem, we only increment the slow pointer when the element at the slow pointer is non-zero. This step involves conditional incrementation based on the state of the array at the slow pointer’s position.

To summarize, the problem-solving approach here involves:

a. Initialize two pointers, slow and fast.

b. Iterate through the array using the fast pointer.

c. If the element at the fast pointer is non-zero and the element at the slow pointer is zero, swap these elements.

d. If the element at the slow pointer is non-zero, increment the slow pointer.

e. Repeat steps b-d until you have traversed the entire array.

At the end of the process, all zeros will have been moved to the end of the array while the relative order of the non-zero elements is maintained.

Targeted Drills in Python

  1. Understanding Array Traversal
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# Array Traversal

arr = [1, 0, 3, 0, 12]

for i in range(len(arr)):
    print(arr[i])
  1. Two Pointers Methodology
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# Two pointers

arr = [1, 0, 3, 0, 12]
slow, fast = 0, 0

while fast < len(arr):
    print("Slow Pointer:", slow, "Fast Pointer:", fast)
    slow += 1
    fast += 2
  1. In-Place Array Manipulation
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# In-place array manipulation

arr = [1, 0, 3, 0, 12]
arr[1], arr[2] = arr[2], arr[1]

print(arr)  # Output: [1, 3, 0, 0, 12]
  1. Condition-Based Manipulation
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# Condition-based Manipulation

arr = [1, 0, 3, 0, 12]
slow = 0

for fast in range(len(arr)):
    if arr[fast] != 0 and arr[slow] == 0:
        arr[slow], arr[fast] = arr[fast], arr[slow]

    print(arr)  # Output after each iteration
  1. Incrementation of Pointers
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# Incrementation of Pointers

arr = [1, 0, 3, 0, 12]
slow = 0

for fast in range(len(arr)):
    if arr[fast] != 0 and arr[slow] == 0:
        arr[slow], arr[fast] = arr[fast], arr[slow]
        
    if arr[slow] != 0:
        slow += 1

    print("Slow Pointer:", slow, "Fast Pointer:", fast, "Array:", arr)

Finally, integrating all the above drills into the final solution:

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class Solution:
    def moveZeroes(self, nums: list) -> None:
        slow = 0
        for fast in range(len(nums)):
            if nums[fast] != 0 and nums[slow] == 0:
                nums[slow], nums[fast] = nums[fast], nums[slow]

            if nums[slow] != 0:
                slow += 1

        return nums

The final integrated solution uses all the components taught in the drills to solve the problem.