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Problem solution · Java

Maximal Network Rank

Maximal Network Rank: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
65 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Maximal Network Rank, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 65 lines of Java from the credited upstream file 1615.java.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximal Network Rank · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maximalNetworkRank(int n, int[][] roads) {    int[] degrees = new int[n];     for (int[] road : roads) {      final int u = road[0];      final int v = road[1];      ++degrees[u];      ++degrees[v];    }     // Find the first maximum and the second maximum degrees.    int maxDegree1 = 0;    int maxDegree2 = 0;    for (final int degree : degrees)      if (degree > maxDegree1) {        maxDegree2 = maxDegree1;        maxDegree1 = degree;      } else if (degree > maxDegree2) {        maxDegree2 = degree;      }     // There can be multiple nodes with `maxDegree1` or `maxDegree2`.    // Find the counts of such nodes.    int countMaxDegree1 = 0;    int countMaxDegree2 = 0;    for (final int degree : degrees)      if (degree == maxDegree1)        ++countMaxDegree1;      else if (degree == maxDegree2)        ++countMaxDegree2;     if (countMaxDegree1 == 1) {      // 1. If there is only one node with degree = `maxDegree1`, then we'll      // need to use the node with degree = `maxDegree2`. The answer in general      // will be (maxDegree1 + maxDegree2), but if the two nodes that we're      // considering are connected, then we'll have to subtract 1.      final int edgeCount = getEdgeCount(roads, degrees, maxDegree1, maxDegree2) +                            getEdgeCount(roads, degrees, maxDegree2, maxDegree1);      return maxDegree1 + maxDegree2 - (countMaxDegree2 == edgeCount ? 1 : 0);    } else {      // 2. If there are more than one node with degree = `maxDegree1`, then we      // can consider `maxDegree1` twice, and we don't need to use `maxDegree2`.      // The answer in general will be 2 * maxDegree1, but if the two nodes that      // we're considering are connected, then we'll have to subtract 1.      final int edgeCount = getEdgeCount(roads, degrees, maxDegree1, maxDegree1);      final int maxPossibleEdgeCount = countMaxDegree1 * (countMaxDegree1 - 1) / 2;      return 2 * maxDegree1 - (maxPossibleEdgeCount == edgeCount ? 1 : 0);    }  }   // Returns the number of edges (u, v) where degress[u] == degreeU and  // degrees[v] == degreeV.  private int getEdgeCount(int[][] roads, int[] degrees, int degreeU, int degreeV) {    int edgeCount = 0;    for (int[] road : roads) {      final int u = road[0];      final int v = road[1];      if (degrees[u] == degreeU && degrees[v] == degreeV)        ++edgeCount;    }    return edgeCount;  }}

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