714334. Min Time
Min Time
Summary
The Min Time problem involves finding the minimum time it takes to visit a set of cities, given the distance between each pair of cities. This is a classic optimization problem that requires a dynamic programming approach.
Detailed Explanation
To solve this problem, we can start by representing the distance matrix as an adjacency list or adjacency matrix. The goal is to find the minimum total time required to visit all cities and return to the starting city.
The algorithm works by initializing a 2D array dp of size n x n, where n is the number of cities, with all values set to infinity. Then, for each city i, we calculate the minimum time it takes to visit all cities starting from i. This is done by iterating through all possible paths and updating the dp array accordingly.
Here's a step-by-step breakdown of the solution:
- Initialize an adjacency list or adjacency matrix representing the distance between each pair of cities.
- Create a 2D array
dpof sizen x n, wherenis the number of cities, with all values set to infinity. - Iterate through each city
i. - For each city
i, calculate the minimum time it takes to visit all cities starting fromi. This can be done by iterating through all possible paths and updating thedparray accordingly. - Return the minimum value in the last row of the
dparray, which represents the minimum total time required to visit all cities and return to the starting city.
Time complexity: O(n^2) Space complexity: O(n^2)
Optimized Solutions
Java
public class MinTime {
public static int minTime(int[][] graph) {
int n = graph.length;
int[] dp = new int[n];
for (int i = 0; i < n; i++) {
dp[i] = Integer.MAX_VALUE;
}
dp[0] = 0;
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (graph[j][i] != -1) {
dp[i] = Math.min(dp[i], dp[j] + graph[j][i]);
}
}
}
return dp[n-1];
}
}