LeetCode in Kotlin

1289. Minimum Falling Path Sum II

Hard

Given an n x n integer matrix grid, return the minimum sum of a falling path with non-zero shifts.

A falling path with non-zero shifts is a choice of exactly one element from each row of grid such that no two elements chosen in adjacent rows are in the same column.

Example 1:

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

Output: 13

Explanation: The possible falling paths are: [1,5,9], [1,5,7], [1,6,7], [1,6,8], [2,4,8], [2,4,9], [2,6,7], [2,6,8], [3,4,8], [3,4,9], [3,5,7], [3,5,9] The falling path with the smallest sum is [1,5,7], so the answer is 13.

Example 2:

Input: grid = [[7]]

Output: 7

Constraints:

• n == grid.length == grid[i].length
• 1 <= n <= 200
• -99 <= grid[i][j] <= 99

Solution

class Solution {
    fun minFallingPathSum(grid: Array<IntArray>): Int {
        val n = grid[0].size
        var prev = IntArray(n)
        var curr = IntArray(n)
        var prevMinOne = 0
        var prevMinTwo = 0
        for (ints in grid) {
            var currMinOne = Int.MAX_VALUE
            var currMinTwo = Int.MAX_VALUE
            for (j in 0 until n) {
                val prevMin = if (prev[j] == prevMinOne) prevMinTwo else prevMinOne
                curr[j] = ints[j] + prevMin
                if (curr[j] < currMinOne) {
                    currMinTwo = currMinOne
                    currMinOne = curr[j]
                } else if (curr[j] < currMinTwo) {
                    currMinTwo = curr[j]
                }
            }
            prevMinOne = currMinOne
            prevMinTwo = currMinTwo
            // reuse curr array, avoid new int[] in every row
            val temp = prev
            prev = curr
            curr = temp
        }
        return prevMinOne
    }
}

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