120. Triangle

Description

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

Note:

Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input:

[

[2],

[3, 4],

[6, 5, 7],

[4, 1, 8, 3]

]

Output: 11

Explanation:

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : O(N) where N is the number of rows.
 */

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

Follow up