https://leetcode.com/problems/triangle/
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?
🤞java
class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
int n = triangle.size();
if(n == 1) return triangle.get(0).get(0);
//starting from the 2nd row
for(int row = 1; row < n; row++){
for(int col = 0; col <= row; col++){
int minAbove = Integer.MAX_VALUE;
//the cell above (if it exists)
if(col < row){
minAbove = triangle.get(row-1).get(col);
}
//then compare with the cell above to the left (if it exists)
if(col > 0){
minAbove = Math.min(minAbove, triangle.get(row-1).get(col-1));
}
//update the current cell
triangle.get(row).set(col, minAbove+triangle.get(row).get(col));
}
}
//find the min in the last row
int minTotal = Integer.MAX_VALUE;
for(int val:triangle.get(n-1)){
minTotal = Math.min(minTotal, val);
}
return minTotal;
}
}
Time: O(n^2)
Space: O(1)