Triangle

Description

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?

Code

DP with memoization

Time Complexity: $O(n^2)$, Space Complexity: $O(n^2)$

class Solution {
    int memo[201][201];
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        memset(memo, -1, sizeof(memo));
        return dfs(0, 0, triangle);
    }
 
    int dfs(int i, int j, vector<vector<int>>& triangle) {
 
        if(i >= triangle.size()) {
            return 0; 
        }
        if(j >= triangle[i].size()) {
            return 1001;
        }
 
        if(memo[i][j] != -1) return memo[i][j];
 
        int new_i = i + 1;
        int new_j_l = j, new_j_r = j + 1;
        int pathSum = triangle[i][j];
        pathSum = min(pathSum + dfs(new_i, new_j_r, triangle), pathSum + dfs(new_i, new_j_l, triangle));
        return memo[i][j] = pathSum;
 
    }
};

Bottom Up DP

Time Complexity: $O(n)$, Space Complexity: $O(1)$

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int n = triangle.size();
        for(int i = n - 2; i >= 0; i--) {
            for(int j = 0; j < triangle[i].size(); j++) {
                triangle[i][j] = triangle[i][j] + min(triangle[i + 1][j], triangle[i + 1][j + 1]);
            }
        }
        return triangle[0][0];
    }
 
};

# Bottom Up DP - Start from the top

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int n = triangle.size();
        for(int i = 0; i < n - 1; i++) {
            vector<int> tmp(triangle[i + 1].size(), INT_MAX);
            for(int j = 0; j < triangle[i].size(); j++) {
                tmp[j] = min(tmp[j], triangle[i][j] + triangle[i + 1][j]);
                tmp[j + 1] = min(tmp[j + 1], triangle[i][j] + triangle[i + 1][j + 1]);
            }
            triangle[i + 1] = tmp;
        }
        return *min_element(triangle[n - 1].begin(), triangle[n - 1].end());
    }
};

Source

• Triangle - LeetCode