120. Triangle¶

LeetCode Problem View on GitHub

120. Triangle

Medium

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?

Solution¶

class Solution {
    private int dp[][];
    public int minimumTotal(List<List<Integer>> triangle) {
        int n = triangle.size();
        dp = new int[n + 1][triangle.get(n - 1).size() + 1];
        for (int current[] : dp) 
            Arrays.fill(current, -(int)(1e9));
        return solve(0, 0, triangle);
    }
    private int solve(int row, int col, List<List<Integer>> triangle) {
        if (row >= triangle.size())
            return 0;

        if (col >= triangle.get(row).size()) {
            return Integer.MAX_VALUE / 10;
        }

        if (dp[row][col] != -(int)(1e9))
            return dp[row][col];

        int op1 = triangle.get(row).get(col) + solve(row + 1, col, triangle);
        int op2 = triangle.get(row).get(col) + solve(row + 1, col + 1, triangle);       

        return dp[row][col] = Math.min(op1, op2);
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here