Triangle LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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Triangle 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

Triangle Solutions

Time: O(n)
Space: O(1)

C++

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

Java

 class Solution {
  public int minimumTotal(List<List<Integer>> triangle) {
    for (int i = triangle.size() - 2; i >= 0; --i)
      for (int j = 0; j <= i; ++j)
        triangle.get(i).set(j, triangle.get(i).get(j) + Math.min(triangle.get(i + 1).get(j),
                                                                 triangle.get(i + 1).get(j + 1)));
    return triangle.get(0).get(0);
  }
}

Python

class Solution:
  def minimumTotal(self, triangle: List[List[int]]) -> int:
    for i in reversed(range(len(triangle) - 1)):
      for j in range(i + 1):
        triangle[i][j] += min(triangle[i + 1][j],
                              triangle[i + 1][j + 1])

    return triangle[0][0]

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