Triangle

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 <= 200
• triangle[0].length == 1
• triangle[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?

Approach - Bottom Up 1D DP

• Need to check the last row separately as base case then do it for the rest
• O(n^2), O(n)

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int n = triangle.size();
        int[] dp = new int[n];
        // base case i.e. last row
        for (int i = 0; i < n; i++) {
            dp[i] = triangle.get(n - 1).get(i);
        }
        //start from second last
        for (int i = n - 2; i >= 0; i--) {
            for (int j = 0; j < triangle.get(i).size(); j++) {
                dp[j] = triangle.get(i).get(j) + Math.min(dp[j], dp[j+1]);
            }
        }
 
        return dp[0];
    }
}