Floyd Warshall All Pairs Shortest Path in DSA Java - Time & Space Complexity

Time Complexity: Floyd Warshall All Pairs Shortest Path

O(n³)

Understanding Time Complexity

We want to understand how the time needed to find shortest paths between all pairs of points grows as the number of points increases.

How does the work increase when we add more points to the map?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

for (int k = 0; k < n; k++) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (dist[i][k] + dist[k][j] < dist[i][j]) {
        dist[i][j] = dist[i][k] + dist[k][j];
      }
    }
  }
}
    

This code updates the shortest distances between all pairs of points by checking if going through an intermediate point k gives a shorter path.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The triple nested loops that check and update distances.
• How many times: Each loop runs n times, so the inner operation runs n x n x n = n³ times.

How Execution Grows With Input

As the number of points n grows, the number of checks grows very fast because we check every pair through every possible intermediate point.

Input Size (n)	Approx. Operations
10	1,000
100	1,000,000
1000	1,000,000,000

Pattern observation: When n increases by 10 times, operations increase by 1000 times (cube growth).

Final Time Complexity

Time Complexity: O(n³)

This means the time needed grows very quickly as the number of points increases, roughly proportional to the cube of n.

Common Mistake

[X] Wrong: "Since there are three loops, the time complexity is just O(n) because loops are simple."

[OK] Correct: Each loop runs n times, so the total work multiplies, not adds. The work is n x n x n, which is much bigger than just n.

Interview Connect

Understanding this cubic time complexity helps you explain why Floyd Warshall is good for small graphs but not for very large ones, showing your grasp of algorithm efficiency.

Self-Check

"What if we replaced the triple nested loops with a recursive approach that splits the problem? How might that affect the time complexity?"

Practice

1. What does the Floyd Warshall algorithm primarily compute in a graph?

easy

A. Shortest paths between all pairs of vertices

B. Minimum spanning tree of the graph

C. Depth-first traversal order

D. Maximum flow between two nodes

Solution

1. Step 1: Understand Floyd Warshall purpose

   The Floyd Warshall algorithm is designed to find shortest paths between every pair of vertices in a weighted graph.

2. Step 2: Compare with other graph algorithms

   Minimum spanning tree, depth-first traversal, and maximum flow solve different problems, not all pairs shortest paths.

3. Final Answer:

   Shortest paths between all pairs of vertices -> Option A

4. Quick Check:

   Floyd Warshall = All pairs shortest paths [OK]

Hint: Floyd Warshall = shortest paths for all vertex pairs [OK]

Common Mistakes:

• Confusing with Dijkstra's single-source shortest path
• Thinking it finds minimum spanning tree
• Mixing with traversal algorithms

2. Which of the following is the correct way to initialize the distance matrix dist for Floyd Warshall in Java for a graph with V vertices?

easy

A. for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) dist[i][j] = (i == j) ? 0 : Integer.MAX_VALUE;

B. for (int i = 0; i < V; i++) dist[i] = new int[V];

C. dist = new int[V][V]; for (int i = 0; i < V; i++) dist[i][i] = 1;

D. for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) dist[i][j] = 0;

Solution

1. Step 1: Understand initialization needs

   Distance from a vertex to itself should be 0, and to others should be infinity (represented by Integer.MAX_VALUE).

2. Step 2: Check each option

   for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) dist[i][j] = (i == j) ? 0 : Integer.MAX_VALUE; correctly sets 0 for same vertices and max value for others. for (int i = 0; i < V; i++) dist[i] = new int[V]; only initializes array but not values. dist = new int[V][V]; for (int i = 0; i < V; i++) dist[i][i] = 1; sets diagonal to 1 incorrectly. for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) dist[i][j] = 0; sets all to 0, which is wrong.

3. Final Answer:

   for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) dist[i][j] = (i == j) ? 0 : Integer.MAX_VALUE; -> Option A

4. Quick Check:

   Initialize dist[i][i] = 0, others = infinity [OK]

Hint: Set zero for same nodes, max value for others [OK]

Common Mistakes:

• Not setting distance to self as zero
• Setting all distances to zero
• Forgetting to initialize the matrix fully

3. What will be the output distance matrix after running Floyd Warshall on this graph with 3 vertices and edges: 0->1=4, 1->2=3, 0->2=10?

int V = 3;
int[][] dist = {
  {0, 4, 10},
  {Integer.MAX_VALUE, 0, 3},
  {Integer.MAX_VALUE, Integer.MAX_VALUE, 0}
};
// Floyd Warshall algorithm applied here

medium

A. [[0, 4, 3], [∞, 0, 3], [∞, ∞, 0]]

B. [[0, 4, 10], [∞, 0, 3], [∞, ∞, 0]]

C. [[0, 7, 10], [∞, 0, 3], [∞, ∞, 0]]

D. [[0, 4, 7], [∞, 0, 3], [∞, ∞, 0]]

Solution

1. Step 1: Initial distances

   Start with dist matrix: 0->1=4, 1->2=3, 0->2=10, others infinity or 0 on diagonal.

2. Step 2: Update using intermediate vertices

   Check if path 0->1->2 is shorter than 0->2 directly: 4 + 3 = 7 < 10, so update dist[0][2] = 7.

3. Final Answer:

   [[0, 4, 7], [∞, 0, 3], [∞, ∞, 0]] -> Option D

4. Quick Check:

   0->2 updated to 7 via 1 [OK]

Hint: Check if indirect path is shorter than direct [OK]

Common Mistakes:

• Not updating indirect shorter paths
• Confusing infinity with zero
• Ignoring intermediate vertex updates

4. Identify the bug in this Floyd Warshall code snippet:

for (int k = 0; k < V; k++) {
  for (int i = 0; i < V; i++) {
    for (int j = 0; j < V; j++) {
      if (dist[i][k] + dist[k][j] < dist[i][j]) {
        dist[i][j] = dist[i][k] + dist[k][j];
      }
    }
  }
}

Assume dist uses Integer.MAX_VALUE for infinity.

medium

A. Missing initialization of dist matrix

B. Loop variables are in wrong order

C. No check for Integer.MAX_VALUE before addition causing overflow

D. Should use '>' instead of '<' in if condition

Solution

1. Step 1: Understand addition with infinity

   Adding Integer.MAX_VALUE + any positive number causes overflow and incorrect results.

2. Step 2: Check code for safety

   Code lacks checks to ensure dist[i][k] and dist[k][j] are not Integer.MAX_VALUE before adding.

3. Final Answer:

   No check for Integer.MAX_VALUE before addition causing overflow -> Option C

4. Quick Check:

   Check for infinity before addition to avoid overflow [OK]

Hint: Always check for infinity before adding distances [OK]

Common Mistakes:

• Ignoring overflow from Integer.MAX_VALUE + number
• Assuming loops order causes error
• Confusing comparison operator direction

5. Given a graph with 4 vertices and edges:
0->1=5, 0->3=10, 1->2=3, 2->3=1
Use Floyd Warshall to find the shortest distance from vertex 0 to 3 after all updates.
What is the shortest distance from 0 to 3?

hard

A. 10

B. 9

C. 8

D. 7

Solution

1. Step 1: Initial distances

   Direct edges: 0->1=5, 0->3=10, 1->2=3, 2->3=1. Other distances are infinity or 0 on diagonal.

2. Step 2: Check indirect paths

   Path 0->1->2->3 distance = 5 + 3 + 1 = 9, which is less than direct 0->3=10, so update shortest distance to 9.

3. Final Answer:

   9 -> Option B

4. Quick Check:

   0->1->2->3 = 9 < 10 [OK]

Hint: Sum edges along path to find shortest distance [OK]

Common Mistakes:

• Ignoring multi-step paths
• Choosing direct edge without checking alternatives
• Adding edges incorrectly