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Two dynamic programming techniques to solve Leetcode 120. Triangle. One has space complexity O(n^2). The other is O(n).
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You're provided with a triangle array. Your goal is to find the smallest possible sum of a path from the top of the triangle to the bottom.
You can move to an adjacent number in the row below at each step. Specifically, if you're at index i in the current row, you can move to either index i or index i + 1 in the next row.
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).
Input: triangle = [[-10]]
Output: -10
1 <= triangle.length <= 200.triangle[0].length == 1.triangle[i].length == triangle[i - 1].length + 1.-10^4 <= triangle[i][j] <= 10^4.O(n) extra space, where n is the total number of rows in the triangle?You can store all minimum paths at every positions (i,j) to compute the next ones with this relationship.
minPath[i][j] = triangle[i][j] + min(minPath[i - 1][j - 1], minPath[i - 1][j]);
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int minimumTotal(const vector<vector<int>>& triangle) {
const int n = triangle.size(); // triangle's height
vector<vector<int>> minPath(n);
minPath[0] = triangle[0];
for (int i = 1; i < n; i++) {
const int N = triangle[i].size();
minPath[i].resize(N);
// left most number
minPath[i][0] = triangle[i][0] + minPath[i-1][0];
for (int j = 1; j < N - 1; j++) {
minPath[i][j] = triangle[i][j] + min(minPath[i-1][j-1], minPath[i-1][j]);
}
// right most number
minPath[i][N-1] = triangle[i][N-1] + minPath[i-1][N-2];
}
// pick the min path among the ones (begin -> end)
// go to the bottom (n-1)
return *min_element(minPath[n-1].begin(), minPath[n-1].end());
}
int main() {
vector<vector<int>> triangle{{2},{3,4},{6,5,7},{4,1,8,3}};
cout << minimumTotal(triangle) << endl;
triangle = {{-10}};
cout << minimumTotal(triangle) << endl;
}
Output:
11
-10
This solution finds the minimum path sum from the top to the bottom of a triangle, represented as a vector of vectors. It uses dynamic programming to calculate the minimum path sum.
The algorithm initializes a minPath vector of vectors to store the minimum path sum for each element in the triangle. It starts by setting the first row of minPath to be the same as the first row of the triangle.
Then, it iterates through the rows of the triangle starting from the second row. For each element in the current row, the minimum path sum is calculated by considering the two paths from the previous row leading to that element. It takes the minimum of the two paths and adds the current element's value. This way, it accumulates the minimum path sum efficiently.
The algorithm continues this process until it reaches the last row of the triangle. Finally, it returns the minimum element from the last row of minPath, representing the minimum path sum from top to bottom.
O(n^2), where n is the number of rows in the triangle.O(n^2).You do not need to store all paths for all rows. The computation of the next row only depends on its previous one.
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int minimumTotal(const vector<vector<int>>& triangle) {
const int n = triangle.size();
// store only min path for each row
vector<int> minPath(n);
minPath[0] = triangle[0][0];
for (int i = 1; i < n; i++) {
// right most number
minPath[i] = triangle[i][i] + minPath[i - 1];
for (int j = i - 1; j > 0; j--) {
minPath[j] = triangle[i][j] + min(minPath[j - 1], minPath[j]);
}
// left most number
minPath[0] = triangle[i][0] + minPath[0];
}
return *min_element(minPath.begin(), minPath.end());
}
int main() {
vector<vector<int>> triangle{{2},{3,4},{6,5,7},{4,1,8,3}};
cout << minimumTotal(triangle) << endl;
triangle = {{-10}};
cout << minimumTotal(triangle) << endl;
}
Output:
11
-10
O(n^2), where n is the number of rows in the triangle.O(n).