Shortest Cycle Containing Target Node - Google Top Interview Questions
Problem Statement :
You are given a two-dimensional list of integers graph representing a directed graph as an adjacency list. You are also given an integer target.
Return the length of a shortest cycle that contains target. If a solution does not exist, return -1.
Constraints
n, m ≤ 250 where n and m are the number of rows and columns in graph
Example 1
Input
Visualize
graph = [
[1],
[2],
[0]
]
target = 0
Output
3
Explanation
The nodes 0 -> 1 -> 2 -> 0 form a cycle
Example 2
Input
Visualize
graph = [
[1],
[2],
[4],
[],
[0]
]
target = 3
Output
-1
Solution :
Solution in C++ :
vector<int> color;
vector<long long int> dist;
vector<vector<int>> adj;
int dest;
long long int dfs(int u) {
if (u == dest) {
return 0;
}
color[u] = 1; // gray
long long int mindist = INT_MAX;
for (int v : adj[u]) {
if (color[v] == 0) {
mindist = min(mindist, 1 + dfs(v));
} else if (color[v] == 2) {
mindist = min(mindist, 1 + dist[v]);
}
}
color[u] = 2; // black
return dist[u] = mindist;
}
int solve(vector<vector<int>>& graph, int target) {
adj = graph;
dest = target;
color.assign(graph.size(), 0);
dist.assign(graph.size(), 0);
long long int res = INT_MAX;
for (int i : graph[target]) {
res = min(res, 1 + dfs(i));
}
if (res == INT_MAX) return -1;
return res;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] graph, int target) {
boolean[] visited = new boolean[graph.length];
Queue<Integer> q = new LinkedList<>();
q.add(target);
int level = 0;
while (!q.isEmpty()) {
int size = q.size();
for (int i = 0; i < size; i++) {
int tmp = q.remove();
visited[tmp] = true;
for (int neighbor : graph[tmp]) {
if (neighbor == target)
return level + 1;
else if (!visited[neighbor])
q.add(neighbor);
}
}
level++;
}
return -1;
}
}
Solution in Python :
class Solution:
def solve(self, graph, target):
visited = set()
layer = [target]
length = 0
while layer:
length += 1
next_layer = []
for u in layer:
for v in graph[u]:
if v == target:
return length
if v in visited:
continue
visited.add(v)
next_layer.append(v)
layer = next_layer
return -1
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