README_EN.md

120. Triangle

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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

Dynamic programming.

Python3

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

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

Java

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

C++

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

Go

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

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

TypeScript

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

Rust

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

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