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

Time Complexity: Floyd Warshall All Pairs Shortest Path

O(V^3)

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 running time change when the number of points in the graph grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void floydWarshall(int graph[][V], int dist[][V]) {
  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] != INT_MAX && dist[k][j] != INT_MAX && dist[i][k] + dist[k][j] < dist[i][j]) {
          dist[i][j] = dist[i][k] + dist[k][j];
        }
      }
    }
  }
}
    

This code updates shortest distances between all pairs of points by checking if going through an intermediate point is shorter.

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 V times, so the inner operation runs V x V x V = V³ times.

How Execution Grows With Input

As the number of points (V) grows, the number of checks grows very fast because every pair is checked for every possible intermediate point.

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

Pattern observation: The operations grow roughly by the cube of the input size, so tripling the points makes the work about 27 times bigger.

Final Time Complexity

Time Complexity: O(V³)

This means the time needed grows very quickly as the number of points increases, because it checks all pairs through all intermediate points.

Common Mistake

[X] Wrong: "Since it uses loops, it must be O(V²) because there are two loops for pairs."

[OK] Correct: There are actually three nested loops, so the checks happen V x V x V times, making it O(V³), not O(V²).

Interview Connect

Understanding this cubic growth 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? How would the time complexity change?"

Practice

1. What is the main purpose of the Floyd Warshall algorithm in graph theory?

easy

A. To find shortest paths between all pairs of vertices in a weighted graph

B. To find the minimum spanning tree of a graph

C. To detect cycles in a directed graph

D. To perform depth-first search traversal

Solution

1. Step 1: Understand Floyd Warshall algorithm goal

   The Floyd Warshall algorithm is designed to find shortest paths between every pair of vertices in a weighted graph, including negative weights but no negative cycles.

2. Step 2: Compare with other graph algorithms

   Minimum spanning tree algorithms find a tree connecting all vertices with minimum total edge weight, cycle detection algorithms find cycles, and DFS is a traversal method. These are different from Floyd Warshall's goal.

3. Final Answer:

   To find shortest paths between all pairs of vertices in a weighted graph -> Option A

4. Quick Check:

   Floyd Warshall = all pairs shortest paths [OK]

Hint: Floyd Warshall finds shortest paths between all vertex pairs [OK]

Common Mistakes:

• Confusing Floyd Warshall with minimum spanning tree algorithms
• Thinking it only finds shortest path from one source
• Assuming it detects cycles instead of paths

2. Which of the following is the correct way to initialize the distance matrix dist for Floyd Warshall in C++ for a graph with n vertices and adjacency matrix graph?

easy

A. for (int i = 0; i < n; i++) dist[i] = graph[i];

B. for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) dist[i][j] = graph[i][j];

C. dist = graph;

D. dist = new int[n][n];

Solution

1. Step 1: Understand distance matrix initialization

   We need to copy each element from the adjacency matrix graph to dist element-wise because dist is a 2D array and direct assignment is not allowed in C++.

2. Step 2: Analyze each option

   for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) dist[i][j] = graph[i][j]; uses nested loops to copy each element correctly. for (int i = 0; i < n; i++) dist[i] = graph[i]; tries to assign a row pointer which is invalid for 2D arrays. dist = graph; tries direct assignment which is invalid for arrays. dist = new int[n][n]; only allocates memory but does not initialize values.

3. Final Answer:

   for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) dist[i][j] = graph[i][j]; -> Option B

4. Quick Check:

   Element-wise copy needed = for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) dist[i][j] = graph[i][j]; [OK]

Hint: Use nested loops to copy 2D arrays element-wise [OK]

Common Mistakes:

• Trying to assign 2D arrays directly
• Copying only row pointers instead of elements
• Not initializing distance matrix before algorithm

3. What will be the output distance matrix after running Floyd Warshall on the following graph with 3 vertices and adjacency matrix:

int graph[3][3] = {
  {0, 5, 999},
  {50, 0, 15},
  {30, 999, 0}
};

Assuming 999 represents no direct edge.

medium

A. [[0, 5, 20], [50, 0, 15], [30, 35, 0]]

B. [[0, 5, 999], [50, 0, 15], [30, 999, 0]]

C. [[0, 5, 999], [50, 0, 15], [30, 45, 0]]

D. [[0, 5, 20], [45, 0, 15], [30, 35, 0]]

Solution

1. Step 1: Initialize distance matrix

   Start with dist = graph:
   dist = [[0,5,999],[50,0,15],[30,999,0]]

2. Step 2: Update distances using Floyd Warshall

   Check if going through vertex 0 improves paths:
   dist[1][2] = min(15, dist[1][0]+dist[0][2]) = min(15, 50+999)=15 (no change)
   dist[2][1] = min(999, dist[2][0]+dist[0][1]) = min(999, 30+5)=35 (updated)
   Check through vertex 1:
   dist[0][2] = min(999, dist[0][1]+dist[1][2]) = min(999, 5+15)=20 (updated)
   dist[2][2] = min(0, dist[2][1]+dist[1][2]) = min(0, 35+15)=0 (no change)
   Check through vertex 2:
   dist[1][0] = min(50, dist[1][2]+dist[2][0]) = min(50, 15+30)=45 (updated)

3. Final Answer:

   [[0, 5, 20], [45, 0, 15], [30, 35, 0]] -> Option D

4. Quick Check:

   Updated shortest paths matrix = [[0, 5, 20], [45, 0, 15], [30, 35, 0]] [OK]

Hint: Check all intermediate vertices to update shortest paths [OK]

Common Mistakes:

• Not updating paths through intermediate vertices
• Confusing no edge (999) with zero
• Missing updates for reverse paths

4. Identify the error in this Floyd Warshall code snippet and select the fix:

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];
    }
  }
}

medium

A. Check if dist[i][k] and dist[k][j] are not infinity before adding

B. No error; code is correct

C. Change loop order to i, j, k

D. Use max instead of min in the if condition

Solution

1. Step 1: Understand potential overflow in addition

   If dist[i][k] or dist[k][j] is infinity (representing no path), adding them causes incorrect results or overflow.

2. Step 2: Fix by checking for infinity before addition

   Before adding dist[i][k] + dist[k][j], ensure both are not infinity to avoid invalid updates.

3. Final Answer:

   Check if dist[i][k] and dist[k][j] are not infinity before adding -> Option A

4. Quick Check:

   Prevent invalid addition by checking infinity [OK]

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

Common Mistakes:

• Ignoring infinity checks causing wrong updates
• Changing loop order incorrectly
• Using max instead of min for shortest path

5. Given a weighted directed graph with 4 vertices and adjacency matrix:

int graph[4][4] = {
  {0, 3, 999, 7},
  {8, 0, 2, 999},
  {5, 999, 0, 1},
  {2, 999, 999, 0}
};

Use Floyd Warshall to find the shortest path distance from vertex 0 to vertex 3 after all updates. What is the shortest distance?

hard

A. 8

B. 7

C. 6

D. 9

Solution

1. Step 1: Initialize distance matrix

   Start with dist = graph:
   dist = [[0,3,999,7],[8,0,2,999],[5,999,0,1],[2,999,999,0]]

2. Step 2: Update distances through intermediate vertices

   Check paths through vertex 0:
   dist[1][3] = min(999, dist[1][0]+dist[0][3]) = min(999, 8+7)=15
   dist[2][1] = min(999, dist[2][0]+dist[0][1]) = min(999, 5+3)=8
   dist[3][1] = min(999, dist[3][0]+dist[0][1]) = min(999, 2+3)=5
   Check paths through vertex 1:
   dist[0][2] = min(999, dist[0][1]+dist[1][2]) = min(999, 3+2)=5
   dist[0][3] = min(7, dist[0][1]+dist[1][3]) = min(7, 3+15)=7
   dist[3][2] = min(999, dist[3][1]+dist[1][2]) = min(999, 5+2)=7
   Check paths through vertex 2:
   dist[1][3] = min(15, dist[1][2]+dist[2][3]) = min(15, 2+1)=3 (updated)
   dist[0][3] = min(7, dist[0][2]+dist[2][3]) = min(7, 5+1)=6 (updated)
   Check paths through vertex 3:
   dist[2][0] = min(5, dist[2][3]+dist[3][0]) = min(5, 1+2)=3 (updated)

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Shortest path 0 to 3 = 6 [OK]

Hint: Update distances through all vertices stepwise [OK]

Common Mistakes:

• Not updating paths through all intermediate vertices
• Missing shorter path updates
• Confusing direct edge with shortest path