LeetCode 120. Triangle

Description

https://leetcode.com/problems/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

Follow up: Could you do this using only O(n) extra space, where n is the total number of rows in the triangle?

Explanation

Track the minimum path sum at each element. At each row, the j th element’s minimum path sum could either get from the previous row j – 1 element or j – 1 element.

Python Solution

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        dp = [[] for i in range(len(triangle))]
 
       
        for i in range(len(triangle)):
            if i == 0:
                dp[i].append(triangle[i][0])            
            else:
                for j in range(len(triangle[i])):  
                    if j == 0:
                        dp[i].append(dp[i - 1][j] + triangle[i][j])
                    elif j == len(triangle[i]) - 1:                        
                        dp[i].append(dp[i - 1][j - 1] + triangle[i][j])                    
                    else:                        
                        dp[i].append(min(dp[i - 1][j - 1], dp[i - 1][j]) + triangle[i][j])
            
        
        min_path_sum = sys.maxsize
        
        for path_sum in dp[-1]:
            min_path_sum = min(min_path_sum, path_sum)
            
        return min_path_sum 
        
        
  • Time Complexity: O(N^2).
  • Space Complexity: O(N^2).

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