This is the introductory dynamic programming question.
First, you start off with the bottom row:
vector<int> dp = triangle.back();
Then, you make your way up the triangle,
iterating through each of the rows above and updating our dp array using the equation:
dp[j] = triangle[i][j] + min(dp[j], dp[j+1]);
// where i = row of the triangle
// j = specific index of a row of the triangle
Adding in the for loops, we get:
for (int i = triangle.size()-2; i >= 0; --i) {
for (int j = 0; j < triangle[i].size(); ++j) {
dp[j] = triangle[i][j] + min(dp[j], dp[j+1]);
}
}
Eventually, our answer will be present at dp[0] which we return:
return dp[0];
code
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
vector<int> dp = triangle.back();
for (int i = triangle.size()-2; i >= 0; --i) {
for (int j = 0; j < triangle[i].size(); ++j) {
dp[j] = triangle[i][j] + min(dp[j], dp[j+1]);
}
}
return dp[0];
}
};
complexity
- Time:
O(n^2)wherendenotes the number of elements insidetriangle- Space:
O(n)
time taken
- 4 minutes and 46 seconds