LeetCode in Kotlin
120. Triangle
Medium
Given a triangle array, return the minimum path sum from top to bottom.
For each step, you may move to an adjacent number of the row below. More formally, if you are on index i on the current row, you may move to either index i or index i + 1 on the next row.
Example 1:
Input: triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
Output: 11
Explanation: The triangle looks like:
2
3 4
6 5 7
4 1 8 3
The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).
Example 2:
Input: triangle = [[-10]]
Output: -10
Constraints:
1 <= triangle.length <= 200triangle[0].length == 1triangle[i].length == triangle[i - 1].length + 1-104 <= triangle[i][j] <= 104
Follow up: Could you do this using only O(n) extra space, where n is the total number of rows in the triangle?
Solution
class Solution {
fun minimumTotal(triangle: List<List<Int>>): Int {
if (triangle.isEmpty()) {
return 0
}
val dp = Array(triangle.size) { IntArray(triangle[triangle.size - 1].size) }
for (temp in dp) {
temp.fill(-10001)
}
return dfs(triangle, dp, 0, 0)
}
private fun dfs(triangle: List<List<Int>>, dp: Array<IntArray>, row: Int, col: Int): Int {
if (row >= triangle.size) {
return 0
}
if (dp[row][col] != -10001) {
return dp[row][col]
}
val sum = (
triangle[row][col] +
Math.min(
dfs(triangle, dp, row + 1, col),
dfs(triangle, dp, row + 1, col + 1)
)
)
dp[row][col] = sum
return sum
}
}