What if one small road closure could split your whole city in two? Let's find those roads fast!
Why Bridges in Graph Tarjan's Algorithm in DSA Typescript?
Imagine you have a city map with roads connecting different neighborhoods. You want to find roads that, if closed, would split the city into disconnected parts. Doing this by checking every road manually would be like walking every street and guessing which ones are critical.
Manually checking each road to see if it disconnects the city is slow and confusing. You might miss some roads or waste time checking roads that don't matter. It's easy to make mistakes and hard to keep track of all connections.
Tarjan's algorithm helps find these critical roads, called bridges, quickly and correctly. It uses a smart way to explore the city map once, tracking how neighborhoods connect, so it spots exactly which roads are bridges without extra work.
for each road in roads: remove road check if city is disconnected restore road
function findBridges(graph) {
// use Tarjan's algorithm to find bridges in one DFS
}This lets you quickly find all critical connections in a network, helping prevent failures or plan repairs efficiently.
Network engineers use this to find cables or connections in the internet that, if broken, would isolate parts of the network, so they can strengthen or monitor those points.
Manual checking of bridges is slow and error-prone.
Tarjan's algorithm finds all bridges efficiently in one pass.
Knowing bridges helps keep networks and systems reliable.
Practice
Solution
Step 1: Understand what a bridge is in a graph
A bridge is an edge that, when removed, increases the number of connected components in the graph.Step 2: Identify Tarjan's algorithm goal
Tarjan's algorithm is used to find such edges efficiently by using DFS and tracking discovery and low times.Final Answer:
To find edges that, if removed, increase the number of connected components -> Option DQuick Check:
Bridges = edges increasing components when removed [OK]
- Confusing bridges with cycles
- Thinking it finds shortest paths
- Mixing with topological sorting
Solution
Step 1: Understand array declaration with initial values in TypeScript
To create an array of numbers with size n and initialize all values to -1, use new Array(n).fill(-1).Step 2: Check type correctness
TypeScript syntax requires disc to be number[] for an array of numbers.Final Answer:
let disc: number[] = new Array(n).fill(-1); -> Option CQuick Check:
Array<number> with fill(-1) = correct initialization [OK]
- Using wrong type annotation
- Not initializing array values
- Filling with string instead of number
const graph = [[1, 2], [0, 2], [0, 1, 3], [2, 4], [3]]; // Edges: 0-1, 0-2, 1-2, 2-3, 3-4 // Find bridges const bridges = findBridges(graph); console.log(bridges);
Solution
Step 1: Analyze the graph edges and connectivity
The graph edges are 0-1, 0-2, 1-2 forming a triangle, and 2-3, 3-4 forming a chain. The triangle edges are not bridges because removing one does not disconnect the graph.Step 2: Identify bridges
Edges 2-3 and 3-4 are bridges because removing either disconnects node 4 or node 3 from the rest.Final Answer:
[[2, 3], [3, 4]] -> Option AQuick Check:
Bridges disconnect graph = [[2,3],[3,4]] [OK]
- Including edges inside cycles as bridges
- Missing chain edges as bridges
- Confusing node indices
function dfs(u: number, parent: number) {
visited[u] = true;
disc[u] = low[u] = time++;
for (const v of graph[u]) {
if (!visited[v]) {
dfs(v, u);
low[u] = Math.min(low[u], low[v]);
if (low[v] > disc[u]) {
bridges.push([u, v]);
}
} else if (v !== parent) {
low[u] = Math.min(low[u], disc[v]);
}
}
}Solution
Step 1: Check initialization of visited array
The code uses 'visited[u]' but does not show initialization. Without initializing, visited may be undefined causing runtime errors.Step 2: Verify other parts
The condition 'low[v] > disc[u]' is correct for bridges. The parent check is present. The time increment is correct.Final Answer:
The 'visited' array is not initialized before DFS -> Option BQuick Check:
Uninitialized visited array causes errors [OK]
- Changing correct bridge condition
- Ignoring visited initialization
- Removing parent check
Solution
Step 1: Analyze graph structure
Edges 0-1, 1-2, 2-0 form a triangle (cycle). Edges 3-4, 4-5, 5-3 form another cycle. Edge 1-3 connects these two cycles.Step 2: Identify bridges
Only edge 1-3 is a bridge because removing it disconnects the two cycles. Other edges are part of cycles and not bridges.Final Answer:
[[1, 3]] -> Option AQuick Check:
Bridge connects two cycles = [[1, 3]] [OK]
- Marking cycle edges as bridges
- Missing the connecting bridge edge
- Confusing node pairs
Ready to prove it?
Pass the test and this pattern moves unknown
