Data Modelling & AI Data Structure & Algorithm

Maximum number of bridges in a path of a given graph

23 June 2025

0

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 between
// 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 between
// 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 between
# 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 between
// 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 between
    // 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:

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

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