Triangle LeetCode Solution

Problem – Triangle LeetCode Solution

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?

Triangle LeetCode Solution in Java

public int minimumTotal(List<List<Integer>> triangle) {
    int[] A = new int[triangle.size()+1];
    for(int i=triangle.size()-1;i>=0;i--){
        for(int j=0;j<triangle.get(i).size();j++){
            A[j] = Math.min(A[j],A[j+1])+triangle.get(i).get(j);
        }
    }
    return A[0];
}

Triangle LeetCode Solution in C++

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

Triangle LeetCode Solution in Python

def minimumTotal(self, triangle):
    f = [0] * (len(triangle) + 1)
    for row in triangle[::-1]:
        for i in xrange(len(row)):
            f[i] = row[i] + min(f[i], f[i + 1])
    return f[0]]

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