What is the time complexity of the Bellman-Ford single-source shortest path algorithm on a completely connected weighted graph of $n$ vertices?

Question

In a binary tree, what is the maximum number of nodes at level k?

Answer 1

Step 1: Bellman-Ford's well-known running time is $O(V \cdot E)$, where $V$ is the number of vertices and $E$ the number of edges.
Step 2: For a complete graph on $n$ vertices, $V = n$ and $E = O(n^2)$ since every pair of vertices is connected.
Step 3: Substituting gives $O(V \cdot E) = O(n \times n^2) = O(n^3)$.
\[ \boxed{O(n^3)} \]

Answer 2

We can also reason about this by placing Bellman-Ford relative to other shortest-path algorithms whose complexities are commonly known:

$O(n^2)$: This is closer to the complexity of computing shortest paths from a single source using a simple array-based Dijkstra on a dense graph, which does far less repeated work than Bellman-Ford's multiple full passes.
$O(n^2 \log n)$: This resembles heap-based Dijkstra's complexity on a sparse-ish graph. Bellman-Ford doesn't use a heap or any ordering structure, so borrowing this complexity class doesn't fit it.
$O(n^3)$: Bellman-Ford is known to be slower than Dijkstra precisely because it blindly repeats $n-1$ rounds of relaxation instead of greedily picking the nearest vertex each time; on a dense/complete graph that repeated brute-force approach lands at $O(n^3)$, a full order of $n$ above Dijkstra's dense-graph cost.
$O(n^3 \log n)$: This would only apply if each of the $O(n^3)$ relaxation steps itself needed logarithmic overhead, e.g. from a data structure lookup, which Bellman-Ford's plain array-based relaxation doesn't require.

Positioning Bellman-Ford as "Dijkstra's cost, but repeated $n$ times over due to no greedy shortcut" lands on one specific class.

Therefore, the correct answer is $O(n^3)$.

Answer 3

We can also reason about this by placing Bellman-Ford relative to other shortest-path algorithms whose complexities are commonly known:

$O(n^2)$: This is closer to the complexity of computing shortest paths from a single source using a simple array-based Dijkstra on a dense graph, which does far less repeated work than Bellman-Ford's multiple full passes.
$O(n^2 \log n)$: This resembles heap-based Dijkstra's complexity on a sparse-ish graph. Bellman-Ford doesn't use a heap or any ordering structure, so borrowing this complexity class doesn't fit it.
$O(n^3)$: Bellman-Ford is known to be slower than Dijkstra precisely because it blindly repeats $n-1$ rounds of relaxation instead of greedily picking the nearest vertex each time; on a dense/complete graph that repeated brute-force approach lands at $O(n^3)$, a full order of $n$ above Dijkstra's dense-graph cost.
$O(n^3 \log n)$: This would only apply if each of the $O(n^3)$ relaxation steps itself needed logarithmic overhead, e.g. from a data structure lookup, which Bellman-Ford's plain array-based relaxation doesn't require.

Positioning Bellman-Ford as "Dijkstra's cost, but repeated $n$ times over due to no greedy shortcut" lands on one specific class.

Therefore, the correct answer is $O(n^3)$.

What is the time complexity of the Bellman-Ford single-source shortest path algorithm on a completely connected weighted graph of $n$ vertices?

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The Correct Option is C

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