# 120. Triangle

## Description

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?

## Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ as the minimum path sum from the bottom of the triangle to the position $(i, j)$. Here, the position $(i, j)$ refers to the position in the $i$th row and $j$th column of the triangle (both starting from $0$). Then we have the following state transition equation:

$f[i][j] = \min(f[i + 1][j], f[i + 1][j + 1]) + triangle[i][j]$

The answer is $f[0][0]$.

We notice that the state $f[i][j]$ is only related to the states $f[i + 1][j]$ and $f[i + 1][j + 1]$, so we can use a one-dimensional array instead of a two-dimensional array, reducing the space complexity from $O(n^2)$ to $O(n)$.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the number of rows in the triangle.

Furthermore, we can directly reuse the triangle as the f array, so there is no need to create an additional f array, reducing the space complexity to $O(1)$.

• 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) {
int x = triangle.get(i).get(j);
int y = Math.min(triangle.get(i + 1).get(j), triangle.get(i + 1).get(j + 1));
triangle.get(i).set(j, x + y);
}
}
return triangle.get(0).get(0);
}
}

• class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
for (int i = triangle.size() - 2; ~i; --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];
}
};

• class Solution:
def minimumTotal(self, triangle: List[List[int]]) -> int:
n = len(triangle)
for i in range(n - 2, -1, -1):
for j in range(i + 1):
triangle[i][j] = (
min(triangle[i + 1][j], triangle[i + 1][j + 1]) + triangle[i][j]
)
return triangle[0][0]


• func minimumTotal(triangle [][]int) int {
for i := len(triangle) - 2; i >= 0; i-- {
for j := 0; j <= i; j++ {
triangle[i][j] += min(triangle[i+1][j], triangle[i+1][j+1])
}
}
return triangle[0][0]
}

• function minimumTotal(triangle: number[][]): number {
for (let i = triangle.length - 2; ~i; --i) {
for (let j = 0; j <= i; ++j) {
triangle[i][j] += Math.min(triangle[i + 1][j], triangle[i + 1][j + 1]);
}
}
return triangle[0][0];
}


• impl Solution {
pub fn minimum_total(triangle: Vec<Vec<i32>>) -> i32 {
let mut triangle = triangle;
for i in (0..triangle.len() - 1).rev() {
for j in 0..=i {
triangle[i][j] += triangle[i + 1][j].min(triangle[i + 1][j + 1]);
}
}
triangle[0][0]
}
}