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Triangle

Medium Dynamic Programming 2D DP - max/min of last row Original on LeetCode

Triangle is a medium dynamic programming problem solved with the 2d dp - max/min of last row pattern. The best approach, optimal (bottom-up in one array), runs in O(n²) time and O(n) space. Below are 2 approaches in Java, from top-down recursion up.

Problem

Given a triangle of numbers, find the minimum path sum from top to bottom. From index j in one row you may move to index j or j + 1 in the next row.

Examples

Example 1

Input
triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
Output
11
Why
2 + 3 + 5 + 1.

Example 2

Input
triangle = [[-10]]
Output
-10

Constraints

  • 1 <= rows <= 200; row i has i + 1 values.
  • From index j you can move to j or j + 1 in the next row.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
Top-down recursionO(2ⁿ)O(n)
Optimal (bottom-up in one array)O(n²)O(n)

1Top-down recursion

TimeO(2ⁿ)
SpaceO(n)

best(r, j) = value + min(best(r + 1, j), best(r + 1, j + 1)).

  1. Base: the last row returns its own value.
Java
class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        return best(triangle, 0, 0);
    }

    private int best(List<List<Integer>> t, int r, int j) {
        int v = t.get(r).get(j);
        if (r == t.size() - 1) return v;
        return v + Math.min(best(t, r + 1, j), best(t, r + 1, j + 1));
    }
}

2Optimal (bottom-up in one array)

TimeO(n²)
SpaceO(n)

Copy the last row into dp. For each row above, dp[j] = value + min(dp[j], dp[j + 1]). Reading from the bottom means there is one answer at the top and no need to take a minimum over a row.

  1. dp = copy of the last row.
  2. For r from n - 2 down to 0, for j in 0..r: dp[j] = t[r][j] + min(dp[j], dp[j + 1]).
  3. 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 r = n - 1; r >= 0; r--)
            for (int j = 0; j <= r; j++)
                dp[j] = triangle.get(r).get(j) + Math.min(dp[j], dp[j + 1]);
        return dp[0];
    }
}

Edge cases to test

  • Single row
  • Negative numbers

Hints

Hint 1

Work from the bottom row up: every cell becomes its value plus the smaller of its two children. The top cell is the answer.

FAQ

What is the best time complexity for Triangle?

Optimal (bottom-up in one array) runs in O(n²) time and O(n) extra space.

Which pattern does Triangle use?

It is a dynamic programming problem that uses the 2d dp - max/min of last row pattern. Other problems with the same pattern: Minimum Falling Path Sum, Geek's Training.

Is there a brute force solution for Triangle?

Yes. Top-down recursion takes O(2ⁿ) time and O(n) space. best(r, j) = value + min(best(r + 1, j), best(r + 1, j + 1)).

Which edge cases should I test for Triangle?

Single row; Negative numbers.