# 120. Triangle

## Approach 1: Storing row sums using DP

```python
class Solution(object):
    def minimumTotal(self, triangle):
        """
        :type triangle: List[List[int]]
        :rtype: int
        """
        let = len(triangle)
        dp = [[0] * (i+1) for i in range(let)]
        dp[0][0] = triangle[0][0]
        counter = 2
        for i in range(1, let):
            for j in range(counter):
                if j == 0:
                    dp[i][j] = triangle[i][j] + dp[i-1][j]
                elif j == counter - 1:
                    dp[i][j] = triangle[i][j] + dp[i-1][j-1]
                else:
                    dp[i][j] = triangle[i][j] + min(dp[i-1][j-1], dp[i-1][j])
            counter += 1
        return min(dp[-1])
```

As like a standard DP problem, we will store the sums up to the previous row and use it to update the next row. The update condition here being: `dp[i][j] = triangle[i][j] + min(dp[i-1][j-1], dp[i-1][j])` where `(i,j)` are the current co-ordinates of the triangle matrix for which the min sum is being evaluated, and `dp[i][j]` stores the sums up to row i and location j.

`d[element] = element + min(d[left_parent], d[right_parent])`

> **Time Complexity:** *O(N), where N is the total number of numbers in triangle*
>
> **Space Complexity:** *O(N)*

## Approach 1: Same as above but O(1) space

In the above solution it is visible that the for row N we only need information from row N-1 and so on. This way we can optimize memory as below

```python
class Solution(object):
    def minimumTotal(self, triangle):
        """
        https://leetcode.com/problems/triangle/description/
        :type triangle: List[List[int]]
        :rtype: int
        """
        lastRow = triangle[-1]

        for row in reversed(triangle[:-1]):
            tempRow = row[:]
            for i in range(0,len(row)):
                tempRow[i] += min(lastRow[i], lastRow[i+1])
            lastRow = tempRow
        return lastRow[0]
```

> **Time Complexity:** *O(N), where N is the total number of numbers in triangle*
>
> **Space Complexity:** *O(n), where n is the number of rows. N = n(n+1)/2*

 


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://programming.arora-aditya.com/leetcode/dynamic-programming/120.-triangle.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.