LeetCode link

Intuition

  • Try DFS. Maybe have a TLE.

solution

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class Solution {
    int min = Integer.MAX_VALUE;
    public int minimumTotal(List<List<Integer>> triangle) {
        this.helper(triangle, 0, 0, 0);
        return min;
    }
    private void helper(List<List<Integer>> triangle, int row, int col, int sum) {
        if (row >= triangle.size()) {
            min = Math.min(min, sum);
            return;
        }
        List<Integer> line = triangle.get(row);
        if (col >= line.size()) {
            return;
        }
        this.helper(triangle, row + 1, col, sum + line.get(col));
        if (col + 1 < line.size()) {
            this.helper(triangle, row + 1, col + 1, sum + line.get(col + 1));
        }
    }
}

problem

With no doubt, TLE

modification

Using DP.

  • Save a 2d-array to save the min total when reach the current position.
  • When reach the last row, compare the min.
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class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int rows = triangle.size();
        int largestCol = triangle.get(rows - 1).size();
        int[][] dp = new int[rows][largestCol];
        List<Integer> firstLine = triangle.get(0);
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < firstLine.size(); i++) {
            dp[0][i] = firstLine.get(i);
            min = Math.min(min, dp[0][i]);
        }
        if (rows <= 1) {
            return min;
        }
        for (int i = 1; i < rows - 1; i++) {
            List<Integer> line = triangle.get(i);
            dp[i][0] = dp[i - 1][0] + line.get(0);
            int j = 1;
            for (j = 1; j < line.size() - 1; j++) {
                dp[i][j] = Math.min(dp[i - 1][j - 1], dp[i - 1][j]) + line.get(j);
            }
            dp[i][j] = dp[i - 1][j - 1] + line.get(j);
        }
        List<Integer> lastLine = triangle.get(rows - 1);
        min = dp[rows - 2][0] + lastLine.get(0);
        int j = 1;
        for (j = 1; j < lastLine.size() - 1; j++) {
            dp[rows - 1][j] = Math.min(dp[rows - 2][j - 1], dp[rows - 2][j]) + lastLine.get(j);
            min = Math.min(min, dp[rows - 1][j]);
        }
        dp[rows - 1][j] = dp[rows - 2][j - 1] + lastLine.get(j);
        min = Math.min(min, dp[rows - 1][j]);
        return min;
    }
}

Follow Up

Using only O(n) extra space, where n is the total number of rows in the triangle

  • This problem assumes that the triangle is from 1(top) to n(bottom), so the follow up actually means the length of the bottom row rather than the number of rows.

solution

  • Can keep a 1d-array which contains current line minimum number as the previous count will never be used again.
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class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int rows = triangle.size();
        int length = triangle.get(rows - 1).size();
        int[] dp = new int[length + 1];
        for (int i = rows - 1; i >= 0; i--) {
            List<Integer> line = triangle.get(i);
            for (int j = 0; j < line.size(); j++) {
                dp[j] = Math.min(dp[j], dp[j + 1]) + line.get(j);
            }
        }
        return dp[0];
    }
}