Triangle – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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 <= 200triangle[0].length == 1triangle[i].length == triangle[i - 1].length + 1-104 <= triangle[i][j] <= 104
C++ Triangle LeetCode Solution
class Solution {
public:
int dfs(int i, int j, int n, vector<vector<int>>& triangle) {
if (i == n) return 0;
int lower_left = triangle[i][j] + dfs(i + 1, j, n, triangle);
int lower_right = triangle[i][j] + dfs(i + 1, j + 1, n, triangle);
return min(lower_left, lower_right);
}
int minimumTotal(vector<vector<int>>& triangle) {
int n = triangle.size();
return dfs(0, 0, n, triangle);
}
};
Java Triangle LeetCode Solution
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];
}
Python 3 Triangle LeetCode Solution
class Solution:
def minimumTotal(self, triangle: List[List[int]]) -> int:
def dfs(i, j):
if i == len(triangle):
return 0
lower_left = triangle[i][j] + dfs(i + 1, j)
lower_right = triangle[i][j] + dfs(i + 1, j + 1)
return min(lower_left, lower_right)
return dfs(0, 0)
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