LeetCode 1548: The Most Similar Path in a Graph Solution
Master LeetCode problem 1548 (The Most Similar Path in a Graph), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1548. The Most Similar Path in a Graph
Problem Explanation
Explanation
This problem involves finding the most similar path in a graph. Given a list of strings representing the cities in the graph and a 2D array edges representing the connections between cities, we need to find the path that minimizes the total difference in names between the path and the target path.
To solve this problem, we can use dynamic programming with backtracking. We will maintain a 2D DP array where dp[i][j] represents the minimum difference between the first i cities in the path and the first j cities in the target path.
At each step, we will iterate over the possible next cities in the path and update the DP array based on the minimum difference. We will keep track of the previous city in the path to reconstruct the final path.
Finally, we will backtrack from the last city in the target path to reconstruct the most similar path.
Time complexity: O(n * m * m), where n is the number of cities in the path and m is the number of cities in the target path. Space complexity: O(n * m).
Solution Code
class Solution {
public List<Integer> mostSimilar(int n, int[][] roads, String[] names, String[] targetPath) {
int m = targetPath.length;
int[][] dp = new int[m + 1][n];
int[][] prev = new int[m + 1][n];
for (int i = 0; i < m; i++) {
Arrays.fill(dp[i + 1], Integer.MAX_VALUE);
for (int j = 0; j < n; j++) {
for (int k = 0; k < n; k++) {
int diff = names[j].equals(targetPath[i]) ? 0 : 1;
if (dp[i][k] + diff < dp[i + 1][j]) {
dp[i + 1][j] = dp[i][k] + diff;
prev[i + 1][j] = k;
}
}
}
}
int minDiff = Integer.MAX_VALUE;
int endCity = -1;
for (int j = 0; j < n; j++) {
if (dp[m][j] < minDiff) {
minDiff = dp[m][j];
endCity = j;
}
}
List<Integer> path = new ArrayList<>();
for (int i = m; i > 0; i--) {
path.add(endCity);
endCity = prev[i][endCity];
}
Collections.reverse(path);
return path;
}
}Try It Yourself
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