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Number of pair of positions in matrix which are not accessible

  • Difficulty Level : Hard
  • Last Updated : 22 Jun, 2021

Given a positive integer N. Consider a matrix of N X N. No cell can be accessible from any other cell, except the given pair cell in the form of (x1, y1), (x2, y2) i.e there is a path (accessible) between (x2, y2) to (x1, y1). The task is to find the count of pairs (a1, b1), (a2, b2) such that cell (a2, b2) is not accessible from (a1, b1).
Examples: 
 

Input : N = 2
Allowed path 1: (1, 1) (1, 2)
Allowed path 2: (1, 2) (2, 2)
Output : 6
Cell (2, 1) is not accessible from any cell
and no cell is accessible from it.

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(1, 1) - (2, 1)
(1, 2) - (2, 1)
(2, 2) - (2, 1)
(2, 1) - (1, 1)
(2, 1) - (1, 2)
(2, 1) - (2, 2)

Consider each cell as a node, numbered from 1 to N*N. Each cell (x, y) can be map to number using (x – 1)*N + y. Now, consider each given allowed path as an edge between nodes. This will form a disjoint set of the connected component. Now, using Depth First Traversal or Breadth First Traversal, we can easily find the number of nodes or size of a connected component, say x. Now, count of non-accessible paths are x*(N*N – x). This way we can find non-accessible paths for each connected path.
Below is implementation of this approach: 
 

C++




// C++ program to count number of pair of positions
// in matrix which are not accessible
#include<bits/stdc++.h>
using namespace std;
 
// Counts number of vertices connected in a component
// containing x. Stores the count in k.
void dfs(vector<int> graph[], bool visited[],
                               int x, int *k)
{
    for (int i = 0; i < graph[x].size(); i++)
    {
        if (!visited[graph[x][i]])
        {
            // Incrementing the number of node in
            // a connected component.
            (*k)++;
 
            visited[graph[x][i]] = true;
            dfs(graph, visited, graph[x][i], k);
        }
    }
}
 
// Return the number of count of non-accessible cells.
int countNonAccessible(vector<int> graph[], int N)
{
    bool visited[N*N + N];
    memset(visited, false, sizeof(visited));
 
    int ans = 0;
    for (int i = 1; i <= N*N; i++)
    {
        if (!visited[i])
        {
            visited[i] = true;
 
            // Initialize count of connected
            // vertices found by DFS starting
            // from i.
            int k = 1;
            dfs(graph, visited, i, &k);
 
            // Update result
            ans += k * (N*N - k);
        }
    }
    return ans;
}
 
// Inserting the edge between edge.
void insertpath(vector<int> graph[], int N, int x1,
                             int y1, int x2, int y2)
{
    // Mapping the cell coordinate into node number.
    int a = (x1 - 1) * N + y1;
    int b = (x2 - 1) * N + y2;
 
    // Inserting the edge.
    graph[a].push_back(b);
    graph[b].push_back(a);
}
 
// Driven Program
int main()
{
    int N = 2;
 
    vector<int> graph[N*N + 1];
 
    insertpath(graph, N, 1, 1, 1, 2);
    insertpath(graph, N, 1, 2, 2, 2);
 
    cout << countNonAccessible(graph, N) << endl;
    return 0;
}

Java




// Java program to count number of
// pair of positions in matrix
// which are not accessible
import java.util.*;
class GFG
{
 
static int k;
 
// Counts number of vertices connected
// in a component containing x.
// Stores the count in k.
static void dfs(Vector<Integer> graph[],
                boolean visited[], int x)
{
    for (int i = 0; i < graph[x].size(); i++)
    {
        if (!visited[graph[x].get(i)])
        {
            // Incrementing the number of node in
            // a connected component.
            (k)++;
 
            visited[graph[x].get(i)] = true;
            dfs(graph, visited, graph[x].get(i));
        }
    }
}
 
// Return the number of count of non-accessible cells.
static int countNonAccessible(Vector<Integer> graph[], int N)
{
    boolean []visited = new boolean[N * N + N];
 
    int ans = 0;
    for (int i = 1; i <= N * N; i++)
    {
        if (!visited[i])
        {
            visited[i] = true;
 
            // Initialize count of connected
            // vertices found by DFS starting
            // from i.
            int k = 1;
            dfs(graph, visited, i);
 
            // Update result
            ans += k * (N * N - k);
        }
    }
    return ans;
}
 
// Inserting the edge between edge.
static void insertpath(Vector<Integer> graph[],
                       int N, int x1, int y1,
                       int x2, int y2)
{
    // Mapping the cell coordinate into node number.
    int a = (x1 - 1) * N + y1;
    int b = (x2 - 1) * N + y2;
 
    // Inserting the edge.
    graph[a].add(b);
    graph[b].add(a);
}
 
// Driver Code
public static void main(String args[])
{
    int N = 2;
 
    Vector[] graph = new Vector[N * N + 1];
    for (int i = 1; i <= N * N; i++)
        graph[i] = new Vector<Integer>();
    insertpath(graph, N, 1, 1, 1, 2);
    insertpath(graph, N, 1, 2, 2, 2);
 
    System.out.println(countNonAccessible(graph, N));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program to count number of pair of
# positions in matrix which are not accessible
 
# Counts number of vertices connected in a
# component containing x. Stores the count in k.
def dfs(graph,visited, x, k):
    for i in range(len(graph[x])):
        if (not visited[graph[x][i]]):
             
            # Incrementing the number of node 
            # in a connected component.
            k[0] += 1
 
            visited[graph[x][i]] = True
            dfs(graph, visited, graph[x][i], k)
 
# Return the number of count of
# non-accessible cells.
def countNonAccessible(graph, N):
    visited = [False] * (N * N + N)
 
    ans = 0
    for i in range(1, N * N + 1):
        if (not visited[i]):
            visited[i] = True
 
            # Initialize count of connected
            # vertices found by DFS starting
            # from i.
            k = [1]
            dfs(graph, visited, i, k)
 
            # Update result
            ans += k[0] * (N * N - k[0])
    return ans
 
# Inserting the edge between edge.
def insertpath(graph, N, x1, y1, x2, y2):
     
    # Mapping the cell coordinate
    # into node number.
    a = (x1 - 1) * N + y1
    b = (x2 - 1) * N + y2
 
    # Inserting the edge.
    graph[a].append(b)
    graph[b].append(a)
 
# Driver Code
if __name__ == '__main__':
 
    N = 2
 
    graph = [[] for i in range(N*N + 1)]
 
    insertpath(graph, N, 1, 1, 1, 2)
    insertpath(graph, N, 1, 2, 2, 2)
 
    print(countNonAccessible(graph, N))
 
# This code is contributed by PranchalK

C#




// C# program to count number of
// pair of positions in matrix
// which are not accessible
using System;
using System.Collections.Generic;
 
class GFG
{
static int k;
 
// Counts number of vertices connected
// in a component containing x.
// Stores the count in k.
static void dfs(List<int> []graph,
                bool []visited, int x)
{
    for (int i = 0; i < graph[x].Count; i++)
    {
        if (!visited[graph[x][i]])
        {
            // Incrementing the number of node in
            // a connected component.
            (k)++;
 
            visited[graph[x][i]] = true;
            dfs(graph, visited, graph[x][i]);
        }
    }
}
 
// Return the number of count
// of non-accessible cells.
static int countNonAccessible(List<int> []graph,
                                   int N)
{
    bool []visited = new bool[N * N + N];
 
    int ans = 0;
    for (int i = 1; i <= N * N; i++)
    {
        if (!visited[i])
        {
            visited[i] = true;
 
            // Initialize count of connected
            // vertices found by DFS starting
            // from i.
            int k = 1;
            dfs(graph, visited, i);
 
            // Update result
            ans += k * (N * N - k);
        }
    }
    return ans;
}
 
// Inserting the edge between edge.
static void insertpath(List<int> []graph,
                            int N, int x1, int y1,
                            int x2, int y2)
{
    // Mapping the cell coordinate into node number.
    int a = (x1 - 1) * N + y1;
    int b = (x2 - 1) * N + y2;
 
    // Inserting the edge.
    graph[a].Add(b);
    graph[b].Add(a);
}
 
// Driver Code
public static void Main(String []args)
{
    int N = 2;
 
    List<int>[] graph = new List<int>[N * N + 1];
    for (int i = 1; i <= N * N; i++)
        graph[i] = new List<int>();
    insertpath(graph, N, 1, 1, 1, 2);
    insertpath(graph, N, 1, 2, 2, 2);
 
    Console.WriteLine(countNonAccessible(graph, N));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript program to count number of
// pair of positions in matrix
// which are not accessible
 
let k;
 
// Counts number of vertices connected
// in a component containing x.
// Stores the count in k.
function dfs(graph,visited,x)
{
    for (let i = 0; i < graph[x].length; i++)
    {
        if (!visited[graph[x][i]])
        {
            // Incrementing the number of node in
            // a connected component.
            (k)++;
   
            visited[graph[x][i]] = true;
            dfs(graph, visited, graph[x][i]);
        }
    }
}
 
// Return the number of count of non-accessible cells.
function countNonAccessible(graph,N)
{
    let visited = new Array(N * N + N);
   
    let ans = 0;
    for (let i = 1; i <= N * N; i++)
    {
        if (!visited[i])
        {
            visited[i] = true;
   
            // Initialize count of connected
            // vertices found by DFS starting
            // from i.
            let k = 1;
            dfs(graph, visited, i);
   
            // Update result
            ans += k * (N * N - k);
        }
    }
    return ans;
}
 
// Inserting the edge between edge.
function insertpath(graph,N,x1,y1,x2,y2)
{
    // Mapping the cell coordinate into node number.
    let a = (x1 - 1) * N + y1;
    let b = (x2 - 1) * N + y2;
   
    // Inserting the edge.
    graph[a].push(b);
    graph[b].push(a);
}
 
// Driver Code
let N = 2;
   
let graph = new Array(N * N + 1);
for (let i = 1; i <= N * N; i++)
    graph[i] = [];
insertpath(graph, N, 1, 1, 1, 2);
insertpath(graph, N, 1, 2, 2, 2);
 
document.write(countNonAccessible(graph, N));
 
 
// This code is contributed by rag2127
 
</script>

Output:  

6

Time Complexity : O(N * N).
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