2.3 KiB
2.3 KiB
Leetcode Triangle
2022-07-20 22:59
Algorithms:
#algorithm #dynamic_programming
Difficulty:
#coding_problems #difficulty_medium
Additional tags:
#leetcode
Revisions:
N/A
Related topics:
Links:
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
1 <= triangle.length <= 200triangle[0].length == 1triangle[i].length == triangle[i - 1].length + 1-104 <= triangle[i][j] <= 104
Thoughts
[!summary] This is a #dynamic_programming problem.
Same as in Leetcode House-Robber, there are four stages to optimization:
Stage 1: ordinary recursion
Stage 2: recursion with cachinqg
Solution
Stage 2:
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);
}
};