120. Triangle (Dynamic Programming) 20.3.21 Medium

120. Triangle
Medium

1059

114

Favorite

Share
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.

Solution: ---------------------------------------------------------- "minimun" path sum -> Dynamic Programming obviously
----------------------------------------------------- O(n) T, O(n) S
class Solution(object):
    def minimumTotal(self, triangle):
        """
        :type triangle: List[List[int]]
        :rtype: int
        """
        if not triangle:
            return None
        dp = triangle[:][:]
        for i in range(1, len(triangle)):
            for j in range(len(triangle[i])):
                if j == 0:
                    dp[i][j] = dp[i - 1][j] + triangle[i][j]
                elif j == len(triangle[i]) - 1:
                    dp[i][j] = dp[i - 1][j - 1] + triangle[i][j]
                else:
                    dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - 1]) + triangle[i][j]
        return min(dp[-1])


Java Version no.1: ----------------------------- Simple recursion but TLE because O(n^n) T
class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle == null) {
            return 0;
        }
        return findMin(triangle, 0, 0);
    }
    
    public int findMin(List<List<Integer>> triangle, int currRow, int currIndex) {
        if (currRow == triangle.size() - 1) {
            return triangle.get(currRow).get(currIndex);
        }
        
        return Math.min(findMin(triangle, currRow + 1, currIndex), findMin(triangle, currRow + 1, currIndex + 1)) + \ 
        triangle.get(currRow).get(currIndex);
    }
}


Java Version no.2: -------------------------------- DP => traditional optimization way for backtracking problem
class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle == null) {
            return 0;
        }
        List<List<Integer>> dp = new ArrayList<>(triangle);
        for (int i = 1; i < dp.size(); i ++) {
            for (int j = 0; j < dp.get(i).size(); j ++) {
                if (j == 0) {
                    dp.get(i).set(j, dp.get(i - 1).get(j) + dp.get(i).get(j));
                } else if (j == dp.get(i).size() - 1) {
                    dp.get(i).set(j, dp.get(i - 1).get(j - 1) + dp.get(i).get(j));
                } else {
                    dp.get(i).set(j, Math.min(dp.get(i - 1).get(j - 1), dp.get(i - 1).get(j)) + dp.get(i).get(j));
                }
            }
        }
        return Collections.min(dp.get(dp.size() - 1));
    }
}