Floyd Warshall Algorithm
A medium Graph problem included in Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Graph
- Sheets
- 2
- Core for
- 11 roles
- Platform
- GeeksforGeeks
The problem
Given a weighted directed graph with n vertices, find the shortest path distances between every pair of vertices using the Floyd-Warshall algorithm. The graph may contain negative weights but no negative cycles.
Example 1
- Input
- n = 4, edges = [[0,1,5],[0,3,10],[1,2,3],[2,3,1]], inf = 999999
- Output
- [[0,5,8,9],[999999,0,3,4],[999999,999999,0,1],[999999,999999,999999,0]]
- Why
- The matrix shows shortest distances between all pairs. Missing entries remain infinity.
Example 2
- Input
- n = 3, edges = [[0,1,3],[1,2,4],[0,2,10]], inf = 999999
- Output
- [[0,3,7],[999999,0,4],[999999,999999,0]]
- Why
- The direct path 0->2 is 10, but via 0->1->2 it is 7.
Constraints
- 1 <= n <= 100
- 0 <= edges.length <= n * (n - 1)
- -1000 <= edge weight <= 1000
How to think about it
Updated 2026-09-09Every path between i and j either avoids intermediate vertex k or passes through it. By looping k through all vertices from 0 to n - 1 on the outermost loop, the distance matrix incrementally expands the set of allowed intermediate waypoint nodes, smoothly computing all-pairs shortest paths in-place without allocating extra storage.
Approaches, worst first
Repeated Bellman-Ford from every vertex
time O(V^2 * E) · space O(V^2)
Launch a separate Bellman-Ford single-source shortest path relaxation from each vertex index to populate each row of the distance table. Because edge weights can be negative, repeated Bellman-Ford is necessary without Johnson reweighting, but scanning all edges V times across V origins requires O(V^2 * E) time.
Triple nested dynamic programmingWrite this one
time O(n^3) · space O(n^2)
Initialize an n x n matrix with direct edge costs, 0 along the diagonal, and infinity elsewhere. Iterate k from 0 to n - 1, and for each (i, j) update `dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])`. Outer loop MUST be k.
Where people lose marks · 3
- Placing the intermediate vertex loop k as the innermost loop instead of the outermost loop completely invalidates the dynamic programming substructure.
- Adding to an infinity representation without checking for infinity first triggers numeric overflow wrapping around into negative numbers.
- A negative value along the main diagonal `dist[i][i] < 0` indicates the presence of a negative cycle.
The theory behind it
Graph — the ground this problem stands on. All Graph problems
What Graph is
A graph is a network of individual points, called vertices or nodes, connected by lines called edges. Think of a subway transit map, an electrical circuit, or a web of social friends. Unlike a tree, a graph has no designated top node and no parent-child hierarchy. Connections can run one-way or both ways, and paths can loop back on themselves to form closed cycles.
When to reach for it
Reach for graph algorithms when inputs describe relationships, networks, flights between cities, course prerequisites, or clone networks. Signals include finding the shortest route across unweighted connections, ordering tasks that depend on earlier tasks, counting isolated clusters, or checking whether a path contains an infinite loop. Whenever problems present pairs of related entities and ask for reachability, distances, or dependencies, graph representations apply.
How the pattern works
First convert edge lists into an adjacency list, mapping each node to an array of its neighbors. Choose your exploration strategy based on the goal: use a queue and breadth-first search to find the shortest path in unweighted networks, or use recursion and depth-first search to explore full paths and detect cycles. Because graphs can have loops, always track visited nodes in a set or boolean array. Add nodes to the visited set at the moment they enter the queue so they are never visited twice.
What each operation costs
| Operation | Time |
|---|---|
| visit all nodes and edges via search | O(v + e) |
| topological sort using in-degree counts | O(v + e) |
| shortest path using dijkstra with a min-heap | O((v + e) log v) |
What usually goes wrong with Graph
- Adding a node to the visited set when popping from the queue instead of when pushing, which lets neighboring nodes enqueue duplicate entries and wastes memory.
- Failing to check for cycles in directed graphs when finding prerequisite orders, causing topological sort routines to hang or return incomplete lists.
- Assuming an input graph is fully connected and scanning from only a single starting node, missing disconnected islands and isolated components.
Which roles need this problem
Graph is a core topic for these 11 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 6 more roles, including Performance Engineer, Search Engineer, Information Retrieval Engineer.
Companies that have asked it
Tags taken from the problem's own GeeksforGeeks page — not a copied list.
Track this in your role's order
Pick your target role and all 370 problems — including this one — resequence to what that interview actually asks. Free.
Start freeMore Graph problems
Problem set and role mapping as of .