Discover how computers find the fastest route for you in seconds, even in huge maps!
Why Dijkstra's Algorithm Single Source Shortest Path in DSA Typescript?
Imagine you are in a new city and want to find the shortest way to visit a friend's house from your hotel. You try to remember every street and calculate distances by hand.
Doing this manually is slow and confusing. You might forget some streets or pick a longer path by mistake. It's easy to get lost or waste time.
Dijkstra's Algorithm helps by automatically finding the shortest path from your starting point to all other places. It checks all routes step-by-step and picks the quickest way without missing any option.
let distances = { 'A': 0, 'B': Infinity, 'C': Infinity };
// Manually update distances for each neighbor
// Repeat for all nodesfunction dijkstra(graph, start) {
// Automatically find shortest paths
// Update distances using priority queue
// Return shortest distances
}This algorithm makes it easy to find the fastest route in maps, networks, or any system with connected points.
GPS apps use Dijkstra's Algorithm to quickly show you the fastest way to drive, walk, or bike to your destination.
Manual pathfinding is slow and error-prone.
Dijkstra's Algorithm finds shortest paths efficiently.
It is widely used in navigation and network routing.
Practice
Solution
Step 1: Understand the algorithm's goal
Dijkstra's Algorithm is designed to find the shortest distance from one starting point (source) to every other point in a graph.Step 2: Compare with other graph algorithms
Minimum spanning tree algorithms find a tree connecting all vertices with minimum total weight, cycle detection finds loops, and topological sort orders vertices. These are different tasks.Final Answer:
To find the shortest path from a single source to all other vertices -> Option CQuick Check:
Dijkstra's Algorithm = shortest paths from source [OK]
- Confusing shortest path with minimum spanning tree
- Thinking it detects cycles
- Mixing it with topological sorting
Solution
Step 1: Understand initialization requirements
Distances start as Infinity except the source vertex which is 0.Step 2: Check each code snippet
const dist: number[] = Array(5).fill(Infinity); dist[0] = 0; correctly fills with Infinity and sets source distance to 0. const dist: number[] = new Array(5).map(() => 0); fills with 0 incorrectly. const dist: number[] = Array(5).fill(0); dist[0] = Infinity; fills with 0 and sets source to Infinity, wrong. const dist: number[] = Array(5).fill(null); dist[0] = 0; fills with null, which is invalid for numeric distances.Final Answer:
const dist: number[] = Array(5).fill(Infinity); dist[0] = 0; -> Option AQuick Check:
Distance array init = Infinity except source 0 [OK]
- Using map on new Array without fill (returns undefined)
- Setting all distances to 0 initially
- Using null instead of numeric values
const graph = new Map>();
graph.set(0, [[1, 4], [2, 1]]);
graph.set(1, []);
graph.set(2, [[1, 2]]);
const dist = Array(3).fill(Infinity);
dist[0] = 0;
const visited = new Set<number>();
while (visited.size < 3) {
let u = -1;
let minDist = Infinity;
for (let i = 0; i < 3; i++) {
if (!visited.has(i) && dist[i] < minDist) {
minDist = dist[i];
u = i;
}
}
if (u === -1) break;
visited.add(u);
for (const [v, w] of graph.get(u) || []) {
if (dist[u] + w < dist[v]) {
dist[v] = dist[u] + w;
}
}
}
console.log(dist);Solution
Step 1: Initialize distances and start from source 0
dist = [0, Infinity, Infinity], visited = {}.Step 2: Pick unvisited vertex with smallest dist and update neighbors
Pick 0: update dist[1] = 4, dist[2] = 1. visited = {0}. Pick 2 (dist=1): update dist[1] = min(4, 1+2=3) = 3. visited = {0,2}. Pick 1 (dist=3): no neighbors. visited = {0,1,2}.Final Answer:
[0, 3, 1] -> Option DQuick Check:
Distances after updates = [0, 3, 1] [OK]
- Not updating dist[1] from 4 to 3 via vertex 2
- Picking wrong vertex with smallest distance
- Ignoring visited set causing infinite loops
const graph = new Map>();
graph.set(0, [[1, 7], [2, 9], [5, 14]]);
graph.set(1, [[2, 10], [3, 15]]);
graph.set(2, [[3, 11], [5, 2]]);
graph.set(3, [[4, 6]]);
graph.set(4, [[5, 9]]);
graph.set(5, []);
const dist: number[] = new Array(graph.size);
dist.fill(Infinity);
dist[0] = 0;
const visited = new Set<number>();
while (visited.size < graph.size) {
let u = -1;
let minDist = Infinity;
for (let i = 0; i < graph.size; i++) {
if (!visited.has(i) && dist[i] < minDist) {
minDist = dist[i];
u = i;
}
}
visited.add(u);
for (const [v, w] of graph.get(u) || []) {
if (dist[u] + w < dist[v]) {
dist[v] = dist[u] + w;
}
}
}Solution
Step 1: Analyze the selection of vertex u
If no unvisited vertex has finite dist, u remains -1.Step 2: Check visited addition without validation
Code adds u to visited without checking if u === -1, which can cause adding -1 and errors.Final Answer:
Adding vertex u to visited set before checking if u is -1 -> Option BQuick Check:
Check u !== -1 before visited.add(u) [OK]
- Ignoring u === -1 case causing infinite loops
- Confusing graph.size with array length
- Not initializing dist array properly
Solution
Step 1: Calculate shortest distances step-by-step
Start dist: [0,∞,∞,∞,∞] From 0: dist[1]=2, dist[2]=4 From 1: dist[2]=min(4,2+1=3)=3, dist[3]=2+7=9 From 2: dist[4]=3+3=6 From 3: dist[4]=min(6,9+1=10)=6Step 2: Identify shortest distance to vertex 4
Shortest distance to vertex 4 is 6 via path 0->1->2->4.Final Answer:
6 -> Option AQuick Check:
Shortest path distance to 4 = 6 [OK]
- Ignoring shorter path via vertex 2
- Adding edges incorrectly
- Assuming direct edge is always shortest
Ready to prove it?
Pass the test and this pattern moves unknown
