Maximized Network Stream Analyzer 8 — Problem Statement & Solution Guide

GraphsEasyFloyd-Warshall

TimeO(N³)

SpaceO(N²)

Problem Description

You are provided with a square matrix dist of size N x N, where dist[i][j] represents the direct transmission cost from node i to node j in a network. If there is no direct link, the value is -1. Your task is to compute the maximum possible minimum path cost between any pair of nodes (i, j) such that i < j, after optimizing all paths using the Floyd-Warshall algorithm. Specifically, for every pair of distinct nodes, determine the shortest path distance. Then, among all these shortest path distances, find the maximum value. If no path exists between any pair of nodes, return -1. Note: The diagonal elements dist[i][i] are always 0 and should be ignored in the final maximization step.

DSA Pattern Breakdown

"Maximized Network Stream Analyzer 8"

easy

WHY DOES IT MATTER?

All‑pairs shortest‑path (APSP) is a foundational pattern for network reliability, routing, and latency analysis. Knowing the worst‑case shortest distance helps engineers provision resources, detect bottlenecks, and guarantee service‑level agreements.

OPTIMIZATION CHALLENGE

The key insight is recognizing that the problem reduces to APSP, which can be solved in cubic time regardless of the initial sparsity because the input is already an N×N matrix. This eliminates the need for per‑pair BFS/DFS or exponential enumeration.

REAL-WORLD CONNECTION

In distributed systems, each server node can be seen as a graph vertex and the latency between them as edge weight. Floyd‑Warshall models the process of repeatedly updating routing tables so that every node eventually knows the optimal path to every other node, akin to link‑state protocols like OSPF.

When coding, first replace all -1 entries with a large sentinel (e.g., INF) except the diagonal, then run the triple loop. After the DP finishes, a single pass over i<j yields the answer—no extra data structures needed.

COMPLEXITY AT A GLANCE

⏱ Time:O(N³)

💾 Space:O(N²)

Core Theory — Why This Approach?

The Floyd‑Warshall algorithm is a classic dynamic‑programming technique that computes the shortest paths between every pair of vertices in a weighted directed graph. It iteratively improves path estimates by considering each vertex as an intermediate node, updating dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]) for all i, j, k. This approach works in O(N³) time and O(N²) space, making it optimal for dense graphs where an adjacency matrix is already provided.

A naive approach would attempt to enumerate all possible routes between each pair, which explodes combinatorially (O(N!)) and quickly becomes infeasible even for modest N (e.g., N=100). By leveraging the overlapping sub‑problems property—shortest paths through a subset of vertices are reused—the Floyd‑Warshall algorithm collapses the exponential search space into a cubic loop, guaranteeing polynomial runtime while handling negative edge weights (as long as there are no negative cycles). After the all‑pairs shortest paths are known, extracting the maximum of the minimum distances is a simple linear scan over the upper triangle of the matrix.

Interview Questions on This Problem

Q1How would you modify the Floyd‑Warshall algorithm to also detect a negative‑weight cycle in the given network?

After the standard three‑nested loops, inspect the diagonal entries dist[i][i]; if any become negative, a negative‑weight cycle exists reachable from node i.

Q2Given a sparse graph with up to 10⁵ edges but N ≤ 500, would Floyd‑Warshall still be the best choice? Why or why not?

Yes, because N is small (≤500) and the algorithm runs in O(N³) ≈ 125 million operations, which is acceptable; the edge count matters less than the vertex count when using an adjacency matrix.

Q3Explain how you could compute the same "maximum of minimum path costs" using Dijkstra from each node instead of Floyd‑Warshall. What is the time complexity?

Run Dijkstra from every vertex (O(N·(M log N))) to obtain all‑pairs shortest paths, then scan for the maximum. For dense graphs M≈N², this becomes O(N³ log N), slower than Floyd‑Warshall’s O(N³).

Examples

Example 1

Input

dist = [[0, 5, -1], [5, 0, 3], [-1, 3, 0]]

Output

Explanation: Initial matrix: [[0, 5, -1], [5, 0, 3], [-1, 3, 0]]. Step 1: Check intermediate node 0. No changes as direct paths are already optimal or non-existent. Step 2: Check intermediate node 1. Path 0->2 via 1: dist[0][1] + dist[1][2] = 5 + 3 = 8. Since 8 > -1 (no direct path), update dist[0][2] = 8. Similarly, dist[2][0] = 8. Step 3: Check intermediate node 2. Path 0->1 via 2: dist[0][2] + dist[2][1] = 8 + 3 = 11. Since 11 > 5, no update. Path 1->0 via 2: 3 + 8 = 11 > 5, no update. Final shortest paths: dist[0][1]=5, dist[0][2]=8, dist[1][2]=3. Maximum of these values is 8? Wait, let's re-evaluate. The problem asks for the maximum of the shortest paths. The shortest paths are 5, 8, and 3. The maximum is 8. Let me correct the output to 8. Correction: The output should be 8.

Example 2

Input

dist = [[0, 2, 1], [2, 0, 4], [1, 4, 0]]

Output

Explanation: Initial matrix: [[0, 2, 1], [2, 0, 4], [1, 4, 0]]. Step 1: Intermediate 0. Path 1->2 via 0: dist[1][0] + dist[0][2] = 2 + 1 = 3. Since 3 < 4, update dist[1][2] = 3. Similarly dist[2][1] = 3. Step 2: Intermediate 1. Path 0->2 via 1: dist[0][1] + dist[1][2] = 2 + 3 = 5. Since 5 > 1, no update. Step 3: Intermediate 2. Path 0->1 via 2: dist[0][2] + dist[2][1] = 1 + 3 = 4. Since 4 > 2, no update. Final shortest paths: dist[0][1]=2, dist[0][2]=1, dist[1][2]=3. Maximum of these values is 3. Let me correct the output to 3.

Example 3

Input

dist = [[0, -1, -1], [-1, 0, -1], [-1, -1, 0]]

Output

-1

Explanation: No direct connections exist between any distinct nodes. Floyd-Warshall does not create new paths. All off-diagonal elements remain -1. Since no valid path exists between any pair, the result is -1.

Constraints

2 <= N <= 100
dist[i][i] == 0 for all i
dist[i][j] == dist[j][i] for all i, j
-1 <= dist[i][j] <= 10^4 for i != j

Optimal Approach & Strategy

Apply Floyd‑Warshall to compute all‑pairs shortest paths in O(N³) time, then perform a single O(N²) scan to find the maximum shortest distance.

Brute Force Approach

Enumerate every simple path between each pair (i, j) and pick the cheapest, then take the maximum over all pairs; this is exponential in N and impossible for N>10.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N³)

function solution(matrix) {
   const n = matrix.length;
   const dp = Array.from({length: n}, () => Array.from({length: n}, () => Infinity));
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           if (i === j) dp[i][j] = 0;
           else dp[i][j] = matrix[i][j];
       }
   }
   for (let k = 0; k < n; k++) {
       for (let i = 0; i < n; i++) {
           for (let j = 0; j < n; j++) {
               dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k][j]);
           }
       }
   }
   let sum = 0;
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           sum += dp[i][j];
       }
   }
   return sum;
}

class Solution {
   public:
       int solution(vector<vector<int>>& matrix) {
           int n = matrix.size();
           vector<vector<int>> dp(n, vector<int>(n, INT_MAX));
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < n; j++) {
                   if (i == j) dp[i][j] = 0;
                   else dp[i][j] = matrix[i][j];
               }
           }
           for (int k = 0; k < n; k++) {
               for (int i = 0; i < n; i++) {
                   for (int j = 0; j < n; j++) {
                       dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
                   }
               }
           }
           int sum = 0;
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < n; j++) {
                   sum += dp[i][j];
               }
           }
           return sum;
       }
   };

class Solution {
   public int solution(int[][] matrix) {
       int n = matrix.length;
       int[][] dp = new int[n][n];
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < n; j++) {
               if (i == j) dp[i][j] = 0;
               else dp[i][j] = matrix[i][j];
           }
       }
       for (int k = 0; k < n; k++) {
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < n; j++) {
                   dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k][j]);
               }
           }
       }
       int sum = 0;
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < n; j++) {
               sum += dp[i][j];
           }
       }
       return sum;
   }
}

def solution(matrix):
   n = len(matrix)
   dp = [[float('inf')] * n for _ in range(n)]
   for i in range(n):
       for j in range(n):
           if i == j: dp[i][j] = 0
           else: dp[i][j] = matrix[i][j]
   for k in range(n):
       for i in range(n):
           for j in range(n):
               dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j])
   sum = 0
   for i in range(n):
       for j in range(n):
           sum += dp[i][j]
   return sum

function solution(matrix) {
   const n = matrix.length;
   const dp = Array.from({length: n}, () => Array.from({length: n}, () => Infinity));
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           if (i === j) dp[i][j] = 0;
           else dp[i][j] = matrix[i][j];
       }
   }
   for (let k = 0; k < n; k++) {
       for (let i = 0; i < n; i++) {
           for (let j = 0; j < n; j++) {
               dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k][j]);
           }
       }
   }
   let sum = 0;
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           sum += dp[i][j];
       }
   }
   return sum;
}

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Maximized Network Stream Analyzer 8 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Graphs

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Maximized Network Stream Analyzer 8 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor