LeetCode in Kotlin

1260. Shift 2D Grid

Easy

Given a 2D grid of size m x n and an integer k. You need to shift the grid k times.

In one shift operation:

• Element at grid[i][j] moves to grid[i][j + 1].
• Element at grid[i][n - 1] moves to grid[i + 1][0].
• Element at grid[m - 1][n - 1] moves to grid[0][0].

Return the 2D grid after applying shift operation k times.

Example 1:

Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1

Output: [[9,1,2],[3,4,5],[6,7,8]]

Example 2:

Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4

Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]

Example 3:

Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9

Output: [[1,2,3],[4,5,6],[7,8,9]]

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m <= 50
• 1 <= n <= 50
• -1000 <= grid[i][j] <= 1000
• 0 <= k <= 100

Solution

class Solution {
    fun shiftGrid(grid: Array<IntArray>, k: Int): List<List<Int>> {
        val flat = IntArray(grid.size * grid[0].size)
        var index = 0
        for (ints in grid) {
            for (j in grid[0].indices) {
                flat[index++] = ints[j]
            }
        }
        val mode = k % flat.size
        var readingIndex = flat.size - mode
        if (readingIndex == flat.size) {
            readingIndex = 0
        }
        val result: MutableList<List<Int>> = ArrayList()
        for (i in grid.indices) {
            val eachRow: MutableList<Int> = ArrayList()
            for (j in grid[0].indices) {
                eachRow.add(flat[readingIndex++])
                if (readingIndex == flat.size) {
                    readingIndex = 0
                }
            }
            result.add(eachRow)
        }
        return result
    }
}

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