Today I have done leetcode's June LeetCoding Challenge with cpp.

June LeetCoding Challenge 13

Description

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
• -10^4 <= triangle[i][j] <= 10^4

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 {
public:
  int minimumTotal(vector<vector<int>>& triangle) {
    int n = triangle.size();
    vector<int> dp(n, INT_MAX);
    dp[0] = triangle.front().front();
    for(int i = 1; i < n; ++i) {
      for(int j = i; j >= 0; --j) {
        if(j == i) dp[j] = dp[j - 1];
        else if(j) dp[j] = min(dp[j], dp[j - 1]);
        dp[j] += triangle[i][j];
      }
    }
    return *min_element(dp.begin(), dp.end());
  }
};

// Accepted
// 44/44 cases passed (2 ms)
// Your runtime beats 98.01 % of cpp submissions
// Your memory usage beats 70.75 % of cpp submissions (8.6 MB)