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832. Flipping an Image.md

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832. Flipping an Image

leetcode Daily Challenge on November 10th, 2020.

leetcode-cn Daily Challenge on Febrary 24th, 2021.

Difficulty : Easy

Related Topics : Array

Given a binary matrix A, we want to 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: [[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: [[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]]

Notes:

  • 1 <= A.length = A[0].length <= 20
  • 0 <= A[i][j] <= 1

Solution

  • mine
    • Java
      • Runtime: 0 ms, faster than 100.00%, Memory Usage: 39.7 MB, less than 8.49% of Java online submissions 
        public int[][] flipAndInvertImage(int[][] A) {
    for(int i = 0; i < A.length; i++){
        A[i] = flipAndInvert(A[i]);
    }
    return A;
}

public int[] flipAndInvert(int[] a){
    int start = 0;
    int end = a.length - 1;
    int temp;
    while(start < end){
        temp = a[start];
        a[start] = a[end];
        a[end] = temp;
        start++;
        end--;
    }

    for(int i = 0; i < a.length; i++){
        a[i] = a[i] == 0 ? 1 : 0;
    }
    return a;
}
  • the most votes
  • Runtime: 0 ms, faster than 100.00%, Memory Usage: 38.9 MB, less than 95.01% of Java online submissions 
    public int[][] flipAndInvertImage(int[][] A) {
    int n = A.length;
    for (int[] row : A)
        for (int i = 0; i * 2 < n; i++)
            if (row[i] == row[n - i - 1])
                row[i] = row[n - i - 1] ^= 1;
    return A;
}
  • Runtime: 0 ms, faster than 100.00%, Memory Usage: 38.9 MB, less than 95.01% of Java online submissions 
    // O(N)time
// O(1)space
public int[][] flipAndInvertImage(int[][] A) {
    int C = A[0].length;
    for (int[] row: A)
        for (int i = 0; i < (C + 1) / 2; ++i) {
            int tmp = row[i] ^ 1;
            row[i] = row[C - 1 - i] ^ 1;
            row[C - 1 - i] = tmp;
        }

    return A;
}