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Maximum number of bridges in a path of a given graph

  • Last Updated : 12 Oct, 2021

Given an undirected graph, the task is to count the maximum number of Bridges between any two vertices of the given graph.
Examples: 

Input: 
Graph
1 ------- 2 ------- 3 -------- 4
          |         |
          |         |
          5 ------- 6
Output: 2 
Explanation: 
There are 2 bridges, (1 - 2)
and (3 - 4), in the path from 1 to 4.

Input: 
Graph:
1 ------- 2 ------- 3 ------- 4
Output: 3 
Explanation: 
There are 3 bridges, (1 - 2), (2 - 3)
and (3 - 4) in the path from 1 to 4.

Approach: 
Follow the steps below to solve the problem:  

  • Find all the bridges in the graph and store them in a vector.
  • Removal of all the bridges reduces the graph to small components.
  • These small components do not have any bridges, and they are weakly connected components that do not contain bridges in them.
  • Generate a tree consisting of the nodes connected by bridges, with the bridges as the edges.
  • Now, the maximum bridges in a path between any node are equal to the diameter of this tree.
  • Hence, find the diameter of this tree and print it as the answer.

Below is the implementation of the above approach 
 

C++




// C++ program to find the
// maximum number of bridges
// in any path of the given graph
 
#include <bits/stdc++.h>
using namespace std;
 
const int N = 1e5 + 5;
 
// Stores the nodes
// and their connections
vector<vector<int> > v(N);
 
// Store the tree with
// Bridges as the edges
vector<vector<int> > g(N);
 
// Stores the visited nodes
vector<bool> vis(N, 0);
 
// for finding bridges
vector<int> in(N), low(N);
 
// for Disjoint Set Union
vector<int> parent(N), rnk(N);
// for storing actual bridges
vector<pair<int, int> > bridges;
 
// Stores the number of
// nodes and edges
int n, m;
// For finding bridges
int timer = 0;
 
int find_set(int a)
{
    // Function to find root of
    // the component in which
    // A lies
    if (parent[a] == a)
        return a;
 
    // Doing path compression
    return parent[a]
           = find_set(parent[a]);
}
 
void union_set(int a, int b)
{
    // Function to do union
    // between a and b
    int x = find_set(a), y = find_set(b);
 
    // If both are already in the
    // same component
    if (x == y)
        return;
 
    // If both have same rank,
    // then increase anyone's rank
    if (rnk[x] == rnk[y])
        rnk[x]++;
 
    if (rnk[y] > rnk[x])
        swap(x, y);
 
    parent[y] = x;
}
 
// Function to find bridges
void dfsBridges(int a, int par)
{
    vis[a] = 1;
    // Initialize in time and
    // low value
    in[a] = low[a] = timer++;
 
    for (int i v[a]) {
 
        if (i == par)
            continue;
 
        if (vis[i])
 
            // Update the low value
            // of the parent
            low[a] = min(low[a], in[i]);
        else {
 
            // Perform DFS on its child
            // updating low if the child
            // has connection with any
            // ancestor
            dfsBridges(i, a);
 
            low[a] = min(low[a], low[i]);
 
            if (in[a] < low[i])
 
                // Bridge found
                bridges.push_back(make_pair(i, a));
 
            // Otherwise
            else
 
                // Find union between parent
                // and child as they
                // are in same component
                union_set(i, a);
        }
    }
}
 
// Function to find diameter of the
// tree for storing max two depth child
int dfsDiameter(int a, int par, int& diameter)
{
    int x = 0, y = 0;
    for (int i g[a]) {
        if (i == par)
            continue;
 
        int mx = dfsDiameter(i, a, diameter);
 
        // Finding max two depth
        // from its children
        if (mx > x) {
            y = x;
            x = mx;
        }
        else if (mx > y)
            y = mx;
    }
 
    // Update diameter with the
    // sum of max two depths
    diameter = max(diameter, x + y);
 
    // Return the maximum depth
    return x + 1;
}
 
// Function to find maximum
// bridges bwtween
// any two nodes
int findMaxBridges()
{
 
    for (int i = 0; i <= n; i++) {
        parent[i] = i;
        rnk[i] = 1;
    }
 
    // DFS to find bridges
    dfsBridges(1, 0);
 
    // If no bridges are found
    if (bridges.empty())
        return 0;
 
    int head = -1;
 
    // Iterate over all bridges
    for (auto& i bridges) {
 
        // Find the endpoints
        int a = find_set(i.first);
        int b = find_set(i.second);
 
        // Generate the tree with
        // bridges as the edges
        g[a].push_back(b);
        g[b].push_back(a);
 
        // Update the head
        head = a;
    }
 
    int diameter = 0;
    dfsDiameter(head, 0, diameter);
 
    // Return the diameter
    return diameter;
}
 
// Driver Code
int main()
{
    /*
     
    Graph =>
 
        1 ---- 2 ---- 3 ---- 4
               |      |
               5 ---- 6
    */
 
    n = 6, m = 6;
 
    v[1].push_back(2);
    v[2].push_back(1);
    v[2].push_back(3);
    v[3].push_back(2);
    v[2].push_back(5);
    v[5].push_back(2);
    v[5].push_back(6);
    v[6].push_back(5);
    v[6].push_back(3);
    v[3].push_back(6);
    v[3].push_back(4);
    v[4].push_back(4);
 
    int ans = findMaxBridges();
 
    cout << ans << endl;
 
    return 0;
}

Java




// Java program to find the
// maximum number of bridges
// in any path of the given graph
import java.util.*;
class GFG{
 
static int N = (int)1e5 + 5;
 
// Stores the nodes
// and their connections
static Vector<Integer> []v =
       new Vector[N];
 
// Store the tree with
// Bridges as the edges
static Vector<Integer> []g =
       new Vector[N];
 
// Stores the visited nodes
static boolean []vis =
       new boolean[N];
 
// For finding bridges
static int []in = new int[N];
static int []low = new int[N];
 
// for Disjoint Set Union
static int []parent = new int[N];
static int []rnk = new int[N];
   
// For storing actual bridges
static Vector<pair> bridges =
       new Vector<>();
 
// Stores the number of
// nodes and edges
static int n, m;
   
// For finding bridges
static int timer = 0;
static int diameter;
   
static class pair
{
  int first, second;
  public pair(int first,
              int second) 
  {
    this.first = first;
    this.second = second;
  }   
}
 
static void swap(int x,
                 int y)
{
  int temp = x;
  x = y;
  y = temp;
}
 
static int find_set(int a)
{
  // Function to find root of
  // the component in which
  // A lies
  if (parent[a] == a)
    return a;
 
  // Doing path compression
  return parent[a] =
         find_set(parent[a]);
}
 
static void union_set(int a, int b)
{
  // Function to do union
  // between a and b
  int x = find_set(a),
      y = find_set(b);
 
  // If both are already
  // in the same component
  if (x == y)
    return;
 
  // If both have same rank,
  // then increase anyone's rank
  if (rnk[x] == rnk[y])
    rnk[x]++;
 
  if (rnk[y] > rnk[x])
    swap(x, y);
 
  parent[y] = x;
}
 
// Function to find bridges
static void dfsBridges(int a,
                       int par)
{
  vis[a] = true;
   
  // Initialize in time and
  // low value
  in[a] = low[a] = timer++;
 
  for (int i : v[a])
  {
    if (i == par)
      continue;
 
    if (vis[i])
 
      // Update the low value
      // of the parent
      low[a] = Math.min(low[a],
                        in[i]);
    else
    {
      // Perform DFS on its child
      // updating low if the child
      // has connection with any
      // ancestor
      dfsBridges(i, a);
 
      low[a] = Math.min(low[a],
                        low[i]);
 
      if (in[a] < low[i])
 
        // Bridge found
        bridges.add(new pair(i, a));
 
      // Otherwise
      else
 
        // Find union between parent
        // and child as they
        // are in same component
        union_set(i, a);
    }
  }
}
 
// Function to find diameter
// of the tree for storing
// max two depth child
static int dfsDiameter(int a,
                       int par)
{
  int x = 0, y = 0;
  for (int i : g[a])
  {
    if (i == par)
      continue;
 
    int mx = dfsDiameter(i, a);
 
    // Finding max two depth
    // from its children
    if (mx > x)
    {
      y = x;
      x = mx;
    }
    else if (mx > y)
      y = mx;
  }
 
  // Update diameter with the
  // sum of max two depths
  diameter = Math.max(diameter,
                      x + y);
 
  // Return the maximum depth
  return x + 1;
}
 
// Function to find maximum
// bridges bwtween
// any two nodes
static int findMaxBridges()
{
  for (int i = 0; i <= n; i++)
  {
    parent[i] = i;
    rnk[i] = 1;
  }
 
  // DFS to find bridges
  dfsBridges(1, 0);
 
  // If no bridges are found
  if (bridges.isEmpty())
    return 0;
 
  int head = -1;
 
  // Iterate over all bridges
  for (pair i : bridges)
  {
    // Find the endpoints
    int a = find_set(i.first);
    int b = find_set(i.second);
 
    // Generate the tree with
    // bridges as the edges
    g[a].add(b);
    g[b].add(a);
 
    // Update the head
    head = a;
  }
 
  diameter = 0;
  dfsDiameter(head, 0);
 
  // Return the diameter
  return diameter;
}
 
// Driver Code
public static void main(String[] args)
{
  /*
 
    Graph =>
 
        1 ---- 2 ---- 3 ---- 4
               |      |
               5 ---- 6
    */
 
  n = 6;
  m = 6;
   
  for (int i = 0; i < v.length; i++)
    v[i] = new Vector<Integer>();
 
  for (int i = 0; i < g.length; i++)
    g[i] = new Vector<Integer>();
   
  v[1].add(2);
  v[2].add(1);
  v[2].add(3);
  v[3].add(2);
  v[2].add(5);
  v[5].add(2);
  v[5].add(6);
  v[6].add(5);
  v[6].add(3);
  v[3].add(6);
  v[3].add(4);
  v[4].add(4);
 
  int ans = findMaxBridges();
  System.out.print(ans + "\n");
}
}
 
// This code is contributed by Princi Singh

Python3




# Python3 program to find the
# maximum number of bridges
# in any path of the given graph
 
N = 100005
   
# Stores the nodes
# and their connections
v = []
 
# Store the tree with
# Bridges as the edges
g = []
 
# Stores the visited nodes
vis = [False]*(N)
 
# For finding bridges
In = [0]*(N)
low = [0]*(N)
 
# for Disjoint Set Union
parent = [0]*(N)
rnk = [0]*(N)
 
# For storing actual bridges
bridges = []
 
# Stores the number of
# nodes and edges
n, m = 6, 6
 
# For finding bridges
timer = 0
diameter = 0
 
def swap(x, y):
  temp = x
  x = y
  y = temp
 
def find_set(a):
  global N, v, g, vis, In, low, parent, rnk, bridges, n, m, timer, diameter
  # Function to find root of
  # the component in which
  # A lies
  if parent[a] == a:
    return a
 
  # Doing path compression
  parent[a] = find_set(parent[a])
  return parent[a]
 
def union_set(a, b):
  global N, v, g, vis, In, low, parent, rnk, bridges, n, m, timer, diameter
  # Function to do union
  # between a and b
  x, y = find_set(a), find_set(b)
 
  # If both are already
  # in the same component
  if x == y:
    return
 
  # If both have same rank,
  # then increase anyone's rank
  if rnk[x] == rnk[y]:
    rnk[x]+=1
 
  if rnk[y] > rnk[x]:
    swap(x, y)
 
  parent[y] = x
 
# Function to find bridges
def dfsBridges(a, par):
  global N, v, g, vis, In, low, parent, rnk, bridges, n, m, timer, diameter
  vis[a] = True
 
  # Initialize in time and
  # low value
  timer += 1
  In[a], low[a] = timer, timer
 
  for i in range(len(v[a])):
    if v[a][i] == par:
      continue
 
    if vis[v[a][i]]:
      # Update the low value
      # of the parent
      low[a] = min(low[a], In[v[a][i]])
    else:
      # Perform DFS on its child
      # updating low if the child
      # has connection with any
      # ancestor
      dfsBridges(v[a][i], a)
 
      low[a] = min(low[a], low[v[a][i]])
 
      if In[a] < low[v[a][i]]:
        # Bridge found
        bridges.append([v[a][i], a])
 
      # Otherwise
      else:
        # Find union between parent
        # and child as they
        # are in same component
        union_set(v[a][i], a)
 
# Function to find diameter
# of the tree for storing
# max two depth child
def dfsDiameter(a, par):
  global N, v, g, vis, In, low, parent, rnk, bridges, n, m, timer, diameter
  x, y = 0, 0
  for i in range(len(g[a])):
    if g[a][i] == par:
      continue
 
    mx = dfsDiameter(g[a][i], a)
 
    # Finding max two depth
    # from its children
    if mx > x:
      y = x
      x = mx
 
    elif mx > y:
      y = mx
 
  # Update diameter with the
  # sum of max two depths
  diameter = max(diameter, x + y)
 
  # Return the maximum depth
  return x + 1
 
# Function to find maximum
# bridges bwtween
# any two nodes
def findMaxBridges():
  global N, v, g, vis, In, low, parent, rnk, bridges, n, m, timer, diameter
  for i in range(n + 1):
    parent[i] = i
    rnk[i] = 1
 
  # DFS to find bridges
  dfsBridges(1, 0);
 
  # If no bridges are found
  if len(bridges) == 0:
    return 0
 
  head = -1
 
  # Iterate over all bridges
  for i in range(len(bridges)):
    # Find the endpoints
    a = find_set(bridges[i][0])
    b = find_set(bridges[i][1])
 
    # Generate the tree with
    # bridges as the edges
    g[a].append(b)
    g[b].append(a)
 
    # Update the head
    head = a
 
  diameter = 0
  dfsDiameter(head, 0)
 
  # Return the diameter
  return diameter
 
"""
   
  Graph =>
 
      1 ---- 2 ---- 3 ---- 4
             |      |
             5 ---- 6
"""
 
for i in range(N):
  v.append([])
 
for i in range(N):
  g.append([])
 
v[1].append(2)
v[2].append(1)
v[2].append(3)
v[3].append(2)
v[2].append(5)
v[5].append(2)
v[5].append(6)
v[6].append(5)
v[6].append(3)
v[3].append(6)
v[3].append(4)
v[4].append(4)
 
ans = findMaxBridges()
print(ans)
 
# This code is contributed by suresh07.

C#




// C# program to find the
// maximum number of bridges
// in any path of the given graph
using System;
using System.Collections.Generic;
class GFG{
 
static int N = (int)1e5 + 5;
 
// Stores the nodes
// and their connections
static List<int> []v =
       new List<int>[N];
 
// Store the tree with
// Bridges as the edges
static List<int> []g =
       new List<int>[N];
 
// Stores the visited nodes
static bool []vis =
       new bool[N];
 
// For finding bridges
static int []init = new int[N];
static int []low = new int[N];
 
// for Disjoint Set Union
static int []parent = new int[N];
static int []rnk = new int[N];
   
// For storing actual bridges
static List<pair> bridges =
       new List<pair>();
 
// Stores the number of
// nodes and edges
static int n, m;
   
// For finding bridges
static int timer = 0;
static int diameter;
   
class pair
{
  public int first, second;
  public pair(int first,
              int second) 
  {
    this.first = first;
    this.second = second;
  }   
}
 
static void swap(int x,
                 int y)
{
  int temp = x;
  x = y;
  y = temp;
}
 
static int find_set(int a)
{
  // Function to find root of
  // the component in which
  // A lies
  if (parent[a] == a)
    return a;
 
  // Doing path compression
  return parent[a] =
         find_set(parent[a]);
}
 
static void union_set(int a,
                      int b)
{
  // Function to do union
  // between a and b
  int x = find_set(a),
      y = find_set(b);
 
  // If both are already
  // in the same component
  if (x == y)
    return;
 
  // If both have same rank,
  // then increase anyone's rank
  if (rnk[x] == rnk[y])
    rnk[x]++;
 
  if (rnk[y] > rnk[x])
    swap(x, y);
 
  parent[y] = x;
}
 
// Function to find bridges
static void dfsBridges(int a,
                       int par)
{
  vis[a] = true;
   
  // Initialize in time and
  // low value
  init[a] = low[a] = timer++;
 
  foreach (int i in v[a])
  {
    if (i == par)
      continue;
 
    if (vis[i])
 
      // Update the low value
      // of the parent
      low[a] = Math.Min(low[a],
                        init[i]);
    else
    {
      // Perform DFS on its child
      // updating low if the child
      // has connection with any
      // ancestor
      dfsBridges(i, a);
 
      low[a] = Math.Min(low[a],
                        low[i]);
 
      if (init[a] < low[i])
 
        // Bridge found
        bridges.Add(new pair(i, a));
 
      // Otherwise
      else
 
        // Find union between parent
        // and child as they
        // are in same component
        union_set(i, a);
    }
  }
}
 
// Function to find diameter
// of the tree for storing
// max two depth child
static int dfsDiameter(int a,
                       int par)
{
  int x = 0, y = 0;
  foreach (int i in g[a])
  {
    if (i == par)
      continue;
 
    int mx = dfsDiameter(i, a);
 
    // Finding max two depth
    // from its children
    if (mx > x)
    {
      y = x;
      x = mx;
    }
    else if (mx > y)
      y = mx;
  }
 
  // Update diameter with the
  // sum of max two depths
  diameter = Math.Max(diameter,
                      x + y);
 
  // Return the
  // maximum depth
  return x + 1;
}
 
// Function to find maximum
// bridges bwtween
// any two nodes
static int findMaxBridges()
{
  for (int i = 0; i <= n; i++)
  {
    parent[i] = i;
    rnk[i] = 1;
  }
 
  // DFS to find bridges
  dfsBridges(1, 0);
 
  // If no bridges are found
  if (bridges.Count == 0)
    return 0;
 
  int head = -1;
 
  // Iterate over all bridges
  foreach (pair i in bridges)
  {
    // Find the endpoints
    int a = find_set(i.first);
    int b = find_set(i.second);
 
    // Generate the tree with
    // bridges as the edges
    g[a].Add(b);
    g[b].Add(a);
 
    // Update the head
    head = a;
  }
 
  diameter = 0;
  dfsDiameter(head, 0);
 
  // Return the diameter
  return diameter;
}
 
// Driver Code
public static void Main(String[] args)
{
  /*
 
    Graph =>
 
        1 ---- 2 ---- 3 ---- 4
               |      |
               5 ---- 6
    */
 
  n = 6;
  m = 6;
   
  for (int i = 0; i < v.Length; i++)
    v[i] = new List<int>();
 
  for (int i = 0; i < g.Length; i++)
    g[i] = new List<int>();
   
  v[1].Add(2);
  v[2].Add(1);
  v[2].Add(3);
  v[3].Add(2);
  v[2].Add(5);
  v[5].Add(2);
  v[5].Add(6);
  v[6].Add(5);
  v[6].Add(3);
  v[3].Add(6);
  v[3].Add(4);
  v[4].Add(4);
 
  int ans = findMaxBridges();
  Console.Write(ans + "\n");
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
    // Javascript program to find the
    // maximum number of bridges
    // in any path of the given graph
     
    let N = 1e5 + 5;
  
    // Stores the nodes
    // and their connections
    let v = new Array(N);
 
    // Store the tree with
    // Bridges as the edges
    let g = new Array(N);
 
    // Stores the visited nodes
    let vis = new Array(N);
 
    // For finding bridges
    let In = new Array(N);
    let low = new Array(N);
 
    // for Disjoint Set Union
    let parent = new Array(N);
    let rnk = new Array(N);
 
    // For storing actual bridges
    let bridges = [];
 
    // Stores the number of
    // nodes and edges
    let n, m;
 
    // For finding bridges
    let timer = 0;
    let diameter;
 
    function swap(x, y)
    {
      let temp = x;
      x = y;
      y = temp;
    }
 
    function find_set(a)
    {
      // Function to find root of
      // the component in which
      // A lies
      if (parent[a] == a)
        return a;
 
      // Doing path compression
      return parent[a] = find_set(parent[a]);
    }
 
    function union_set(a, b)
    {
      // Function to do union
      // between a and b
      let x = find_set(a), y = find_set(b);
 
      // If both are already
      // in the same component
      if (x == y)
        return;
 
      // If both have same rank,
      // then increase anyone's rank
      if (rnk[x] == rnk[y])
        rnk[x]++;
 
      if (rnk[y] > rnk[x])
        swap(x, y);
 
      parent[y] = x;
    }
 
    // Function to find bridges
    function dfsBridges(a, par)
    {
      vis[a] = true;
 
      // Initialize in time and
      // low value
      In[a] = low[a] = timer++;
 
      for (let i = 0; i < v[a].length; i++)
      {
        if (v[a][i] == par)
          continue;
 
        if (vis[v[a][i]])
 
          // Update the low value
          // of the parent
          low[a] = Math.min(low[a], In[v[a][i]]);
        else
        {
          // Perform DFS on its child
          // updating low if the child
          // has connection with any
          // ancestor
          dfsBridges(v[a][i], a);
 
          low[a] = Math.min(low[a], low[v[a][i]]);
 
          if (In[a] < low[v[a][i]])
 
            // Bridge found
            bridges.push([v[a][i], a]);
 
          // Otherwise
          else
 
            // Find union between parent
            // and child as they
            // are in same component
            union_set(v[a][i], a);
        }
      }
    }
 
    // Function to find diameter
    // of the tree for storing
    // max two depth child
    function dfsDiameter(a, par)
    {
      let x = 0, y = 0;
      for (let i = 0; i < g[a].length; i++)
      {
        if (g[a][i] == par)
          continue;
 
        let mx = dfsDiameter(g[a][i], a);
 
        // Finding max two depth
        // from its children
        if (mx > x)
        {
          y = x;
          x = mx;
        }
        else if (mx > y)
          y = mx;
      }
 
      // Update diameter with the
      // sum of max two depths
      diameter = Math.max(diameter,
                          x + y);
 
      // Return the maximum depth
      return x + 1;
    }
 
    // Function to find maximum
    // bridges bwtween
    // any two nodes
    function findMaxBridges()
    {
      for (let i = 0; i <= n; i++)
      {
        parent[i] = i;
        rnk[i] = 1;
      }
 
      // DFS to find bridges
      dfsBridges(1, 0);
 
      // If no bridges are found
      if (bridges.length == 0)
        return 0;
 
      let head = -1;
 
      // Iterate over all bridges
      for (let i = 0; i < bridges.length; i++)
      {
        // Find the endpoints
        let a = find_set(bridges[i][0]);
        let b = find_set(bridges[i][1]);
 
        // Generate the tree with
        // bridges as the edges
        g[a].push(b);
        g[b].push(a);
 
        // Update the head
        head = a;
      }
 
      diameter = 0;
      dfsDiameter(head, 0);
 
      // Return the diameter
      return diameter;
    }
     
    /*
  
      Graph =>
 
          1 ---- 2 ---- 3 ---- 4
                 |      |
                 5 ---- 6
      */
 
    n = 6;
    m = 6;
 
    for (let i = 0; i < v.length; i++)
      v[i] = [];
 
    for (let i = 0; i < g.length; i++)
      g[i] = [];
 
    v[1].push(2);
    v[2].push(1);
    v[2].push(3);
    v[3].push(2);
    v[2].push(5);
    v[5].push(2);
    v[5].push(6);
    v[6].push(5);
    v[6].push(3);
    v[3].push(6);
    v[3].push(4);
    v[4].push(4);
 
    let ans = findMaxBridges();
    document.write(ans + "</br>");
 
// This code is contributed by mukesh07.
</script>
Output: 
2

 

Time Complexity: O(N + M) 
Auxiliary Space: O(N + M)




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