notes/OJ notes/pages/Leetcode Triangle.md

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# Leetcode Triangle
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> ##### Algorithms:
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>
> #algorithm #dynamic_programming
>
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> ##### Difficulty:
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>
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> #coding_problem #difficulty-medium
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>
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> ##### Additional tags:
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>
> #leetcode
>
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> ##### Revisions:
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>
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> N/A
##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/triangle/)
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---
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### Problem
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.
#### Examples
**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
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- `1 <= triangle.length <= 200`
- `triangle[0].length == 1`
- `triangle[i].length == triangle[i - 1].length + 1`
- `-104 <= triangle[i][j] <= 104`
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### Thoughts
> [!summary]
> This is a #dynamic_programming problem.
Same as in [[Leetcode House-Robber]], there are four stages to optimization:
#### Stage 1: ordinary recursion
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#### Stage 2: recursion with cachinqg
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### Solution
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#### Stage 2:
```cpp
class Solution {
vector<vector<int>> cache;
int minimum(vector<vector<int>> &triangle, int level, int l, int r) {
if (level == 0) {
return triangle[0][0];
} else {
int minLen = INT_MAX;
for (int i = l; i <= r; i++) {
if (i < 0 || i > level) {
continue;
}
if (cache[level][i] != -1) {
minLen = min(cache[level][i], minLen);
// cout<<"Using cache: "<<minLen<<" for "<<level<<", "<<i<<'\n';
} else {
cache[level][i] =
triangle[level][i] + minimum(triangle, level - 1, i - 1, i);
minLen = min(cache[level][i], minLen);
}
}
// cout<<minLen<<", "<<level<<'\n';
return minLen;
}
}
public:
int minimumTotal(vector<vector<int>> &triangle) {
// Stage one: recursive
cache =
vector<vector<int>>(triangle.size(), vector<int>(triangle.size(), -1));
return minimum(triangle, triangle.size() - 1, 0, triangle.size() - 1);
}
};
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```