LeetCode in Kotlin

832. Flipping an Image

Easy

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.

Solution

class Solution {
    fun flipAndInvertImage(image: Array<IntArray>): Array<IntArray> {
        val m = image.size
        val n = image[0].size
        val result = Array(m) { IntArray(n) }
        for (i in 0 until m) {
            val flipped = reverse(image[i])
            result[i] = invert(flipped)
        }
        return result
    }

    private fun invert(flipped: IntArray): IntArray {
        val result = IntArray(flipped.size)
        for (i in flipped.indices) {
            if (flipped[i] == 0) {
                result[i] = 1
            } else {
                result[i] = 0
            }
        }
        return result
    }

    private fun reverse(nums: IntArray): IntArray {
        var i = 0
        var j = nums.size - 1
        while (i < j) {
            val tmp = nums[i]
            nums[i] = nums[j]
            nums[j] = tmp
            i++
            j--
        }
        return nums
    }
}

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