Suppose we have a triangle. We have to find the minimum path sum from top to the bottom. In each step we can move to adjacent numbers on the row below.
For example, if the following triangle is like
[ [2], [3,4], [6,5,7], [4,1,8,3] ]
The minimum path sum from top to bottom is 11 (2 + 3 + 5 + 1 = 11).
Let us see the steps
- Create one table to use in Dynamic programming approach.
- n := size of triangle
- for i := n – 2 down to 0
- for j := 0 to i
- dp[j] := triangle[i, j] + minimum of dp[j] and dp[j + 1]
- for j := 0 to i
- return dp[0]
Let us see the following implementation to get better understanding
Example
class Solution {
public:
void printVector(vector <int>& v){
for(int i = 0; i < v.size(); i++)cout << v[i] << " ";
cout << endl;
}
int minimumTotal(vector<vector<int>>& triangle) {
vector <int> dp(triangle.back());
int n = triangle.size();
for(int i = n - 2; i >= 0; i--){
for(int j = 0; j <= i; j++){
dp[j] = triangle[i][j] + min(dp[j], dp[j + 1]);
}
// printVector(dp);
}
return dp[0];
}
};Input
[[2],[3,4],[6,5,7],[4,1,8,3]]
Output
11