Maximal Network Rank

Maximal Network Rank - Problem

Graph Medium

There is an infrastructure of n cities with some number of roads connecting these cities. Each roads[i] = [ai, bi] indicates that there is a bidirectional road between cities ai and bi.

The network rank of two different cities is defined as the total number of directly connected roads to either city. If a road is directly connected to both cities, it is only counted once.

The maximal network rank of the infrastructure is the maximum network rank of all pairs of different cities.

Given the integer n and the array roads, return the maximal network rank of the entire infrastructure.

Input & Output

Example 1 — Basic Network

$ Input: n = 4, roads = [[0,1],[0,3],[1,2],[1,3]]

› Output: 4

💡 Note: The network rank of cities 0 and 1 is 4 as there are 4 roads that are connected to either 0 or 1. The road between 0 and 1 is only counted once.

Example 2 — Simple Network

$ Input: n = 5, roads = [[0,1],[0,3],[1,2],[1,3],[2,3],[2,4]]

› Output: 5

💡 Note: There are 5 roads that are connected to either 1 or 2. Since 1 and 2 are not directly connected, we don't subtract anything.

Example 3 — No Roads

$ Input: n = 8, roads = []

› Output: 0

💡 Note: The maximal network rank is 0 since there are no roads.

Constraints

2 ≤ n ≤ 100
0 ≤ roads.length ≤ n × (n-1) / 2
roads[i].length == 2
0 ≤ a_i, b_i ≤ n-1
a_i ≠ b_i
Each pair of cities has at most one road connecting them

Visualization

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Asked in

G Google 12 f Facebook 8 M Microsoft 6 a Amazon 4

The key insight is that network rank equals degree[city1] + degree[city2] - (1 if directly connected). Best approach pre-calculates degrees and uses a set for O(1) direct connection lookup. Time: O(n² + m), Space: O(n + m)

Common Approaches

✓ Sliding Window

⏱️ Time: N/A Space: N/A

Binary Search Answer

⏱️ Time: N/A Space: N/A

Brute Force - Count Roads for Each Pair

⏱️ Time: O(n² × m) Space: O(1)

Check every possible pair of cities and for each pair, iterate through all roads to count how many are connected to either city. If a road connects both cities directly, count it only once.

Optimized - Degree Calculation + Direct Connection Check

⏱️ Time: O(n² + m) Space: O(n + m)

First pass calculates the degree (number of roads) for each city and builds an adjacency set. Second pass checks all city pairs using pre-calculated degrees and fast direct connection lookup.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

int solution(int* nums, int numsSize, int k) {
    qsort(nums, numsSize, sizeof(int), compare);
    int maxBeauty = 1;
    int left = 0;
    
    for (int right = 1; right < numsSize; right++) {
        // Check if nums[left] and nums[right] can have overlapping ranges
        while (nums[right] - nums[left] > 2 * k) {
            left++;
        }
        int current = right - left + 1;
        if (current > maxBeauty) {
            maxBeauty = current;
        }
    }
    
    return maxBeauty;
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array - handle the format [1,2,3,4]
    int nums[100000];
    int numsSize = 0;
    
    // Find the opening bracket and skip it
    char* start = strchr(line, '[');
    if (start) {
        start++; // Skip '['
        char* end = strchr(start, ']');
        if (end) {
            *end = '\0'; // Replace ']' with null terminator
        }
        
        // Parse comma-separated numbers
        char* token = strtok(start, ",");
        while (token != NULL && numsSize < 100000) {
            // Skip whitespace
            while (*token == ' ') token++;
            nums[numsSize++] = atoi(token);
            token = strtok(NULL, ",");
        }
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(nums, numsSize, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Smart Actions

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