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 position \((i, j)\). Here, position \((i, j)\) refers to the position in row \(i\) and column \(j\) of the triangle (both indexed from \(0\)). We have the following state transition equation:

\[ f[i][j] = \min(f[i + 1][j], f[i + 1][j + 1]) + \text{triangle}[i][j] \]

The answer is \(f[0][0]\).

Python3JavaC++GoTypeScriptRust

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

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int n = triangle.size();
        int[][] f = new int[n + 1][n + 1];
        for (int i = n - 1; i >= 0; --i) {
            for (int j = 0; j <= i; ++j) {
                f[i][j] = Math.min(f[i + 1][j], f[i + 1][j + 1]) + triangle.get(i).get(j);
            }
        }
        return f[0][0];
    }
}

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int n = triangle.size();
        vector<vector<int>> f(n + 1, vector<int>(n + 1, 0));
        for (int i = n - 1; i >= 0; --i) {
            for (int j = 0; j <= i; ++j) {
                f[i][j] = min(f[i + 1][j], f[i + 1][j + 1]) + triangle[i][j];
            }
        }
        return f[0][0];
    }
};

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

function minimumTotal(triangle: number[][]): number {
    const n = triangle.length;
    const f: number[][] = Array.from({ length: n + 1 }, () => Array(n + 1).fill(0));
    for (let i = n - 1; i >= 0; --i) {
        for (let j = 0; j <= i; ++j) {
            f[i][j] = Math.min(f[i + 1][j], f[i + 1][j + 1]) + triangle[i][j];
        }
    }
    return f[0][0];
}

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

Solution 2: Dynamic Programming (Space Optimization)

We notice that the state \(f[i][j]\) only depends on states \(f[i + 1][j]\) and \(f[i + 1][j + 1]\). Therefore, 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 \(O(n)\), where \(n\) is the number of rows in the triangle.