Triangle

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

 Notice

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.

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).

Analyse: top to bottom dynamic programming.  

Runtime: 22ms

class Solution {
public:
    /**
     * @param triangle: a list of lists of integers.
     * @return: An integer, minimum path sum.
     */
    int minimumTotal(vector<vector<int> > &triangle) {
        // write your code here
        if (triangle.empty() || triangle[0].empty()) return 0;
        int m = triangle.size();
        vector<int> dp(m, INT_MAX);
        
        dp[0] = triangle[0][0];
        for (int i = 1; i < m; i++) {
            for (int j = i; j > 0; j--) 
                dp[j] = min(dp[j - 1], dp[j]) + triangle[i][j];
            dp[0] += triangle[i][0];
        }
        
        int result = INT_MAX;
        for (int i = 0; i < m; i++)
            result = min(result, dp[i]);
        return result;
    }
};

Analyse: Bottom to top dynamic programming.

Runtime: 38ms

class Solution {
public:
    /**
     * @param triangle: a list of lists of integers.
     * @return: An integer, minimum path sum.
     */
    int minimumTotal(vector<vector<int> > &triangle) {
        // write your code here
        if (triangle.empty() || triangle[0].empty()) return 0;
        int m = triangle.size();
        vector<int> dp = triangle.back();
        
        for (int i = m - 2; i >= 0; i--) {
            for (int j = 0; j <= i; j++) {
                dp[j] = min(dp[j], dp[j + 1]) + triangle[i][j];
            }
        }
        return dp[0];
    }
};