LeetCode – Triangle (Java)

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[     [2],    [3,4],   [6,5,7],  [4,1,8,3]]

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Note: Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

Top-Down Approach (Wrong Answer!)

This solution gets wrong answer! I will try to make it work later.

public class Solution {    public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {        int[] temp = new int[triangle.size()];int minTotal = Integer.MAX_VALUE; for(int i=0; i< temp.length; i++){temp[i] = Integer.MAX_VALUE;} if (triangle.size() == 1) {return Math.min(minTotal, triangle.get(0).get(0));} int first = triangle.get(0).get(0); for (int i = 0; i < triangle.size() - 1; i++) {for (int j = 0; j <= i; j++) { int a, b; if(i==0 && j==0){a = first + triangle.get(i + 1).get(j);b = first + triangle.get(i + 1).get(j + 1); }else{a = temp[j] + triangle.get(i + 1).get(j);b = temp[j] + triangle.get(i + 1).get(j + 1); } temp[j] = Math.min(a, temp[j]);temp[j + 1] = Math.min(b, temp[j + 1]);}} for (int e : temp) {if (e < minTotal)minTotal = e;} return minTotal;    }}

Bottom-Up (Good Solution)

We can actually start from the bottom of the triangle.

public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {int[] total = new int[triangle.size()];int l = triangle.size() - 1; for (int i = 0; i < triangle.get(l).size(); i++) {total[i] = triangle.get(l).get(i);} // iterate from last second rowfor (int i = triangle.size() - 2; i >= 0; i--) {for (int j = 0; j < triangle.get(i + 1).size() - 1; j++) {total[j] = triangle.get(i).get(j) + Math.min(total[j], total[j + 1]);}} return total[0];}