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832. Flipping an Image

Description

Given an n x n binary matrix image, flip the image horizontally, then invert it, and return the resulting image.

To flip an image horizontally means that each row of the image is reversed.

  • For example, flipping [1,1,0] horizontally results in [0,1,1].

To invert an image means that each 0 is replaced by 1, and each 1 is replaced by 0.

  • For example, inverting [0,1,1] results in [1,0,0].

 

Example 1:

Input: image = [[1,1,0],[1,0,1],[0,0,0]]
Output: [[1,0,0],[0,1,0],[1,1,1]]
Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]].
Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]]

Example 2:

Input: image = [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]
Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]].
Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]

 

Constraints:

  • n == image.length
  • n == image[i].length
  • 1 <= n <= 20
  • images[i][j] is either 0 or 1.

Solutions

Solution 1: Two Pointers

We can traverse the matrix, and for each row $\textit{row}$, we use two pointers $i$ and $j$ pointing to the first and last elements of the row, respectively. If $\textit{row}[i] = \textit{row}[j]$, swapping them will keep their values unchanged, so we only need to XOR invert $\textit{row}[i]$ and $\textit{row}[j]$, then move $i$ and $j$ one position towards the center until $i \geq j$. If $\textit{row}[i] \neq \textit{row}[j]$, swapping and then inverting their values will also keep them unchanged, so no operation is needed.

Finally, if $i = j$, we directly invert $\textit{row}[i]$.

The time complexity is $O(n^2)$, where $n$ is the number of rows or columns in the matrix. The space complexity is $O(1)$.

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class Solution:
    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
        n = len(image)
        for row in image:
            i, j = 0, n - 1
            while i < j:
                if row[i] == row[j]:
                    row[i] ^= 1
                    row[j] ^= 1
                i, j = i + 1, j - 1
            if i == j:
                row[i] ^= 1
        return image

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class Solution {
    public int[][] flipAndInvertImage(int[][] image) {
        for (var row : image) {
            int i = 0, j = row.length - 1;
            for (; i < j; ++i, --j) {
                if (row[i] == row[j]) {
                    row[i] ^= 1;
                    row[j] ^= 1;
                }
            }
            if (i == j) {
                row[i] ^= 1;
            }
        }
        return image;
    }
}

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class Solution {
public:
    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& image) {
        for (auto& row : image) {
            int i = 0, j = row.size() - 1;
            for (; i < j; ++i, --j) {
                if (row[i] == row[j]) {
                    row[i] ^= 1;
                    row[j] ^= 1;
                }
            }
            if (i == j) {
                row[i] ^= 1;
            }
        }
        return image;
    }
};

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func flipAndInvertImage(image [][]int) [][]int {
    for _, row := range image {
        i, j := 0, len(row)-1
        for ; i < j; i, j = i+1, j-1 {
            if row[i] == row[j] {
                row[i] ^= 1
                row[j] ^= 1
            }
        }
        if i == j {
            row[i] ^= 1
        }
    }
    return image
}

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/**
 * @param {number[][]} image
 * @return {number[][]}
 */
var flipAndInvertImage = function (image) {
    for (const row of image) {
        let i = 0;
        let j = row.length - 1;
        for (; i < j; ++i, --j) {
            if (row[i] == row[j]) {
                row[i] ^= 1;
                row[j] ^= 1;
            }
        }
        if (i == j) {
            row[i] ^= 1;
        }
    }
    return image;
};

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