Flipping an Image

The problem essentially has two parts:

  1. Flip the Image Horizontally: To flip an image horizontally means to reverse the order of elements in each row. For example, if a row in the matrix is [1, 0, 1], flipping this row would result in [1, 0, 1].

  2. Invert the Image: After flipping each row, you need to invert each element in each row, which simply means changing every 1 to 0 and every 0 to 1. For example, if a flipped row is [1, 0, 1], inverting this row would result in [0, 1, 0].

Let’s take an example from the problem to understand this better:

Input: image = [[1,1,0],[1,0,1],[0,0,0]] Output: [[1,0,0],[0,1,0],[1,1,1]]

Here’s what happens step-by-step:

  • First, each row is reversed:

    • [1,1,0] becomes [0,1,1]
    • [1,0,1] becomes [1,0,1]
    • [0,0,0] becomes [0,0,0]

  • Then, each element in each row is inverted:

    • [0,1,1] becomes [1,0,0]
    • [1,0,1] becomes [0,1,0]
    • [0,0,0] becomes [1,1,1]

So, the final output is [[1,0,0],[0,1,0],[1,1,1]], which matches the expected output.

The challenge lies in implementing these steps efficiently in code. The task requires understanding of array manipulation and binary number manipulation (since we’re just dealing with 0s and 1s).

Sure, let’s define the solution for this problem:

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class Solution:
    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
        # Reverse each row
        flipped = [row[::-1] for row in image]
        
        # Invert the image (Replace 0s by 1s and 1s by 0s)
        inverted = [[1 - pixel for pixel in row] for row in flipped]
        
        return inverted

Here’s what the code does:

  1. It flips each row by using list slicing with a step of -1, which reverses the row.
  2. It then inverts the image by iterating through each pixel in each row. If a pixel’s value is 1, it becomes 0 (because 1 - 1 = 0), and if it’s 0, it becomes 1 (because 1 - 0 = 1).
  3. Finally, it returns the resulting image.

Here is the code broken down into more steps for better understanding:

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class Solution:
    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
        # Initialize an empty list for the flipped image
        flipped = []
        # Loop over each row in the image
        for row in image:
            # Reverse the row and append it to the flipped image
            reversed_row = row[::-1]
            flipped.append(reversed_row)

        # Initialize an empty list for the inverted image
        inverted = []
        # Loop over each row in the flipped image
        for row in flipped:
            # Initialize an empty list for the inverted row
            inverted_row = []
            # Loop over each pixel in the row
            for pixel in row:
                # If the pixel is 1, make it 0, and if it's 0, make it 1
                inverted_pixel = 1 - pixel
                # Append the inverted pixel to the inverted row
                inverted_row.append(inverted_pixel)
            # Append the inverted row to the inverted image
            inverted.append(inverted_row)

        # Return the inverted image
        return inverted

No need to use extra space. Just perform the operations on the input image and return it

Alright, so we need to modify the image in-place, without using any additional space. Here’s how we can do that:

  1. Flip Image Horizontally: Python list’s reverse method can help to flip the image horizontally. We can loop through the rows and call row.reverse() to reverse each row.

  2. Invert Image: To invert an image, we just need to switch 0s to 1s and 1s to 0s. Since we’re dealing with binary numbers, we can use 1 - x to invert a number x. So, we can go through each row again, and replace each element x with 1 - x.

Here’s how it would look in Python:

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class Solution:
    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
        for row in image:
            # Step 1: Flip the row
            row.reverse()
            
            # Step 2: Invert the row
            for i in range(len(row)):
                row[i] = 1 - row[i]

        return image

This code directly modifies the input list (image), and doesn’t use any additional space. The reverse method reverses a list in-place, and 1 - x is used to invert each element. The changes are done directly on the original list.

Q&A

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class Solution {
  public: vector < vector > flipAndInvertImage(vector < vector > & image) {
    int n = image.size();
    // int m=image[0].size();
    vector < vector > v;
    for (int i = 0; i < n; i++) {
      int k = 0;
      for (int j = n - 1; j >= 0; j--) {
        if (image[i][j] == 1) {
          v[i][k] = 0;
          k++;
        } else {
          v[i][k] = 1;
          k++;
        }
      }
    }
    return v;

  }
};

what’s the mistake in my code?(error int 9th line)

The error in your code is because you are trying to access v[i][k] when it has not yet been initialized.

The vector<vector<int>> v is currently empty and has no elements. You can’t access v[i][k] directly without either initializing the size of v with the same size as image or using the push_back() function to add elements.

Here’s how you can modify your code to fix the error. Notice how we use push_back to add elements to v:

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class Solution {
public:
    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& image) {
        int n = image.size();
        vector<vector<int>> v;

        for(int i = 0; i < n; i++){
            vector<int> row;
            for(int j = n - 1; j >= 0; j--){
                if(image[i][j] == 1){
                    row.push_back(0);
                } else {
                    row.push_back(1);
                }
            }
            v.push_back(row);
        }
        return v;
    }
};

This modified version of your code works as follows: for each row, we create a new empty vector row. We populate row by iterating over the corresponding row in image in reverse order, pushing 0 if the original number was 1, and 1 otherwise. After we’re done with a row, we add it to the v vector with v.push_back(row).

Reverse and Toggle

reverse every row. toggle every value.

Compare the i th and n - i - 1 th in a row. The “trick” is that if the values are not the same, but you swap and flip, nothing will change. So if they are same, we toggle both, otherwise we do nothing.

The trick you mentioned is based on the following fact: if a number and its inverse are swapped, the overall state remains the same.

Let’s break down what you said:

  1. “reverse every row”: This is the first operation. For each row, we iterate over the row and swap the elements at positions i and n - i - 1, where i is the current index and n is the length of the row.

  2. “toggle every value”: This is the second operation. Here, you flip the binary digit - if it’s a 1, you change it to 0 and vice versa.

Now, the trick:

  1. “So if they are same, we toggle both, otherwise we do nothing”: This is where the trick comes in. Since swapping an element and its inverse doesn’t change the state, you don’t need to swap them if they are different. If they are the same, you flip them both.

Let’s illustrate this with a short Python code snippet:

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class Solution:
    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
        n = len(image)
        for i in range(n):
            for j in range((n + 1) // 2):
                if image[i][j] == image[i][n - j - 1]:
                    image[i][j] = image[i][n - j - 1] = 1 - image[i][j]
        return image

Here, we iterate over each row i from 0 to n - 1. For each row, we iterate over the first half of the elements (up to (n + 1) // 2). If an element and its corresponding element from the end of the row are the same, we flip them. If they are different, we leave them as is. This gives us the desired result.

This approach has a time complexity of O(N^2), where N is the length of image. It does not require any additional space, so its space complexity is O(1).

The “trick” is that if the values are not the same, but you swap and flip, nothing will change.

Start: 0 and 1 Swap: 1 and 0 Invert: 0 and 1

Exactly! That’s the crucial insight.

Let’s illustrate this with a concrete example:

Suppose we have a row as follows: [0, 1, 0, 1].

If we first flip and then invert the row, we get:

  1. Swap: Swap elements at positions i and n - i - 1. The row becomes: [1, 0, 1, 0].
  2. Invert: Change every 0 to 1 and vice versa. The row becomes: [0, 1, 0, 1].

As we can see, after swapping and inverting, the row is the same as before. Therefore, if the elements at positions i and n - i - 1 are different, we don’t need to do anything.

However, if the elements at positions i and n - i - 1 are the same, we need to flip them, because after swapping and inverting, they would still be the same, and we want them to be inverted.

This is why in the solution code, we only flip the elements when they are the same:

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if image[i][j] == image[i][n - j - 1]:
    image[i][j] = image[i][n - j - 1] = 1 - image[i][j]

May I ask why i*2 < n is the condition?

The condition i*2 < n is used in the context of iterating over the rows of the image and performing the swap and invert operation within each row.

Here’s why this specific condition is used:

We’re iterating over each row from both ends towards the center. The variable i starts at 0 (the first element) and n - i - 1 starts at n - 1 (the last element) and they move towards each other.

The condition i*2 < n ensures that we stop when we’ve processed all pairs of elements in the row. In the case where the row has an odd length, this condition also makes sure that the middle element of the row (which doesn’t have a pair) is correctly processed.

Here’s an illustration:

Suppose we have the row [0, 1, 0, 1, 0]. It has 5 elements (so n is 5) and n - i - 1 is 5 - i - 1.

  • When i is 0, i*2 is 0 and n - i - 1 is 4.
  • When i is 1, i*2 is 2 and n - i - 1 is 3.
  • When i is 2, i*2 is 4 and n - i - 1 is 2.

At this point, we’ve processed all pairs and the single middle element, and i*2 is equal to n, so we stop.

So, in short, the condition i*2 < n is a way to make sure we process all elements in each row exactly once. It works for both even and odd-length rows.

Explain:

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    def flipAndInvertImage(self, A):
        return [[1 ^ i for i in reversed(row)] for row in A]

The function flipAndInvertImage takes in a two-dimensional list A and returns a new two-dimensional list where each row is reversed and all the bits are flipped.

Here’s a step-by-step breakdown of how it works:

  1. for row in A: This outer loop goes through each row in the two-dimensional list A.

  2. reversed(row): This function call reverses the order of elements in the current row. For example, if row is [1, 0, 1], reversed(row) would be [1, 0, 1].

  3. 1 ^ i for i in reversed(row): This inner loop goes through each element i in the reversed row. The ^ operator is the bitwise XOR operator. When i is 1, 1 ^ i is 0, and when i is 0, 1 ^ i is 1. This effectively flips all the bits in the row.

  4. [[1 ^ i for i in reversed(row)] for row in A]: This is a list comprehension that creates a new two-dimensional list. Each element of this new list is a list that corresponds to a reversed and bit-flipped row of A.

So, for example, if A is [[1, 1, 0], [1, 0, 1], [0, 0, 0]], flipAndInvertImage(A) would return [[1, 0, 0], [0, 1, 0], [1, 1, 1]].

Could you help me explain what is the meaning of “row[i] = row[n - i - 1] ^= 1”? I am a little confused about the operator “^= " and how it edits row[i] and row[n-1-i] at the same time. Many thanks!

The expression row[i] = row[n - i - 1] ^= 1 involves bitwise XOR assignment (^=) and normal assignment (=) operators. Let’s break it down:

  1. row[n - i - 1] ^= 1 : The bitwise XOR assignment operator (^=) performs a bitwise XOR operation on row[n - i - 1] and 1, and then assigns the result back to row[n - i - 1]. In other words, it flips the value of row[n - i - 1].

    Here’s how the bitwise XOR operation works for flipping:

    • If row[n - i - 1] is 1, then 1 XOR 1 results in 0.
    • If row[n - i - 1] is 0, then 0 XOR 1 results in 1.

  2. row[i] = row[n - i - 1] : After the XOR operation, the updated value of row[n - i - 1] is then assigned to row[i].

    So, if the values at row[i] and row[n - i - 1] were different initially, this operation would first flip the value at row[n - i - 1] and then copy it to row[i]. As a result, the two values become the same.

Please note that this operation is based on the assumption that the initial values at row[i] and row[n - i - 1] are either 0 or 1. The flipping operation ^= 1 wouldn’t make sense for other initial values.