Triangle
Problem 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 74 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?
Solve it on LeetCode
Approaches
1. Recursive Brute Force
Intuition:
Start from the top of the triangle, at each step, move to one of the two adjacent numbers on the row below. Sum the paths and return the minimum path sum recursively.
Code:
- Time Complexity: O(2^n) - each element can lead to two possibilities.
- Space Complexity: O(n) - recursion depth.
2. Memoization
Intuition:
To avoid recalculating the same values, we store the results of subproblems in a memoization table.
Code:
- Time Complexity: O(n^2) - all cells are computed once.
- Space Complexity: O(n^2) - memoization table storage.
3. Bottom-Up Dynamic Programming
Intuition:
Start filling the DP table from the bottom of the triangle towards the top, by considering the potential next moves in the row below.
Code:
- Time Complexity: O(n^2) - each cell is computed once.
- Space Complexity: O(n^2) - DP table storage.
4. Space Optimized Dynamic Programming
Intuition:
Instead of keeping a full DP table, keep a single array representing the current row computations, reducing space usage.
Code:
- Time Complexity: O(n^2) - each cell is computed once.
- Space Complexity: O(n) - reused single array storage.