Min number of moves to traverse entire Matrix through connected cells with equal values

Given a matrix A[][] of dimensions N*M, the task is to find the minimum number of moves required to traverse the entire matrix starting from (0, 0) by traversing connected cells with equal values at every step. 

From a cell (i, j), cells (i + 1, j), (i – 1, j), (i, j – 1) and (i, j + 1) can be connected. 
 

Examples: 

Input: arr[][] = {{1, 1, 2, 2, 1}, {1, 1, 2, 2, 1}, {1, 1, 1, 3, 2}} 
Output:
Explanation: 
Minimum 5 moves are required to traverse the matrix. 
First move: Starting from [0, 0], traverse cells [0, 1], [1, 1], [1, 0], [2, 0], [2, 1], [2, 2] as all these cells have same value 1. 
Second move: Traverse cells [0, 2], [0, 3], [1, 3], [1, 2] as all these cells have value 2. 
Third move: Traverse cells [0, 4], [1, 4] containing 1. 
Fourth move: Traverse [2, 3] containing 3. 
Fifth move: Traverse [2, 4] containing 4.

Input: arr[][] = {{2, 1, 3}, {1, 1, 2}} 
Output:
Explanation: 
Minimum 4 moves are required to cover this 2-D array 
First move: Traverse only [0, 0] as no other connected cell has value 2. 
Second move: Traverse cells [0, 1], [1, 1], [1, 0] as these cells contain value 1. 
Third move: Traverse cell [0, 2] containing 3. 
Fourth move: Traverse cell [1, 2] containing 2. 



Approach: 
Follow the steps below to solve the problem: 

  • Create another matrix fill each cell with distinct values.
  • Traverse matrix A[][] starting from (0, 0). For every cell (i, j), check if its adjacent cells have same value as A[i][j] or not.
  • If any adjacent cell has same value, replace that cell in B[][] with the value of B[i][j].
  • Count of remaining distinct elements in the B[][] matrix after completing the traversal of A[][], gives the required answer.

Below is the implementation of the above approach: 

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the
// minimum number of moves
// to traverse a given matrix
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum
// number of moves to traverse
// the given matrix
int solve(int A[][10], int N, int M)
{
 
    int B[N][M];
    int c = 1;
    set<int> s;
 
    // Constructing another matrix
    // consisting of distinct values
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++)
            B[i][j] = c++;
    }
 
    // Updating the array B by checking
    // the values of A that if there are
    // same values connected
    // through an edge or not
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
 
            // Check for boundary
            // condition of the matrix
            if (i != 0) {
 
                // If adjacent cells have
                // same value
                if (A[i - 1][j] == A[i][j])
                    B[i - 1][j] = B[i][j];
            }
 
            // Check for boundary
            // condition of the matrix
            if (i != N - 1) {
 
                // If adjacent cells have
                // same value
                if (A[i + 1][j] == A[i][j])
                    B[i + 1][j] = B[i][j];
            }
 
            // Check for boundary
            // condition of the matrix
            if (j != 0) {
 
                // If adjacent cells have
                // same value
                if (A[i][j - 1] == A[i][j])
                    B[i][j - 1] = B[i][j];
            }
 
            // Check for boundary
            // condition of the matrix
            if (j != M - 1) {
 
                // If adjacent cells have
                // same value
                if (A[i][j + 1] == A[i][j])
                    B[i][j + 1] = B[i][j];
            }
        }
    }
 
    // Store all distinct elements
    // in a set
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++)
            s.insert(B[i][j]);
    }
 
    // Return answer
    return s.size();
}
 
// Driver Code
int main()
{
    int N = 2, M = 3;
    int A[][10] = { { 2, 1, 3 },
                    { 1, 1, 2 } };
    // Function Call
    cout << solve(A, N, M);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the
// minimum number of moves
// to traverse a given matrix
import java.util.*;
 
class GFG{
 
// Function to find the minimum
// number of moves to traverse
// the given matrix
static int solve(int A[][], int N,
                            int M)
{
    int [][]B = new int[N][M];
    int c = 1;
     
    HashSet<Integer> s = new HashSet<Integer>();
 
    // Constructing another matrix
    // consisting of distinct values
    for(int i = 0; i < N; i++)
    {
       for(int j = 0; j < M; j++)
          B[i][j] = c++;
    }
 
    // Updating the array B by checking
    // the values of A that if there are
    // same values connected
    // through an edge or not
    for(int i = 0; i < N; i++)
    {
       for(int j = 0; j < M; j++)
       {
            
          // Check for boundary
          // condition of the matrix
          if (i != 0)
          {
               
              // If adjacent cells have
              // same value
              if (A[i - 1][j] == A[i][j])
                  B[i - 1][j] = B[i][j];
          }
           
          // Check for boundary
          // condition of the matrix
          if (i != N - 1)
          {
               
              // If adjacent cells have
              // same value
              if (A[i + 1][j] == A[i][j])
                  B[i + 1][j] = B[i][j];
          }
           
          // Check for boundary
          // condition of the matrix
          if (j != 0)
          {
               
              // If adjacent cells have
              // same value
              if (A[i][j - 1] == A[i][j])
                  B[i][j - 1] = B[i][j];
          }
           
          // Check for boundary
          // condition of the matrix
          if (j != M - 1)
          {
               
              // If adjacent cells have
              // same value
              if (A[i][j + 1] == A[i][j])
                  B[i][j + 1] = B[i][j];
          }
       }
    }
 
    // Store all distinct elements
    // in a set
    for(int i = 0; i < N; i++)
    {
       for(int j = 0; j < M; j++)
          s.add(B[i][j]);
    }
 
    // Return answer
    return s.size();
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 2, M = 3;
    int A[][] = { { 2, 1, 3 },
                  { 1, 1, 2 } };
                   
    // Function Call
    System.out.print(solve(A, N, M));
}
}
 
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to find the
# minimum number of moves
# to traverse a given matrix
 
# Function to find the minimum
# number of moves to traverse
# the given matrix
def solve(A, N, M):
   
    B = []
    c = 1
    s = set()
     
    # Constructing another matrix
    # consisting of distinct values
    for i in range(N):
        new = []
        for j in range(M):
            new.append(c)
            c = c + 1
             
        B.append(new)
 
    # Updating the array B by checking
    # the values of A that if there are
    # same values connected
    # through an edge or not
    for i in range(N):
        for j in range(M):
   
            # Check for boundary
            # condition of the matrix
            if i != 0:
   
                # If adjacent cells have
                # same value
                if A[i - 1][j] == A[i][j]:
                    B[i - 1][j] = B[i][j]
   
            # Check for boundary
            # condition of the matrix
            if (i != N - 1):
   
                # If adjacent cells have
                # same value
                if A[i + 1][j] == A[i][j]:
                    B[i + 1][j] = B[i][j]
   
            # Check for boundary
            # condition of the matrix
            if (j != 0):
   
                # If adjacent cells have
                # same value
                if A[i][j - 1] == A[i][j]:
                    B[i][j - 1] = B[i][j]
   
            # Check for boundary
            # condition of the matrix
            if (j != M - 1):
   
                # If adjacent cells have
                # same value
                if (A[i][j + 1] == A[i][j]): 
                    B[i][j + 1] = B[i][j]
   
    # Store all distinct elements
    # in a set
    for i in range(N):
        for j in range(M):
            s.add(B[i][j])
             
    # Return answer
    return len(s)
 
# Driver code
N = 2
M = 3
A = [ [ 2, 1, 3 ], [ 1, 1, 2 ] ]
 
# Function call
print(solve(A, N, M))
 
# This code is contributed by divyeshrabadiya07

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find the
// minimum number of moves
// to traverse a given matrix
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the minimum
// number of moves to traverse
// the given matrix
static int solve(int [,]A, int N,
                           int M)
{
    int [,]B = new int[N, M];
    int c = 1;
     
    HashSet<int> s = new HashSet<int>();
 
    // Constructing another matrix
    // consisting of distinct values
    for(int i = 0; i < N; i++)
    {
       for(int j = 0; j < M; j++)
          B[i, j] = c++;
    }
 
    // Updating the array B by checking
    // the values of A that if there are
    // same values connected
    // through an edge or not
    for(int i = 0; i < N; i++)
    {
       for(int j = 0; j < M; j++)
       {
            
          // Check for boundary
          // condition of the matrix
          if (i != 0)
          {
               
              // If adjacent cells have
              // same value
              if (A[i - 1, j] == A[i, j])
                  B[i - 1, j] = B[i, j];
          }
           
          // Check for boundary
          // condition of the matrix
          if (i != N - 1)
          {
 
              // If adjacent cells have
              // same value
              if (A[i + 1, j] == A[i, j])
                  B[i + 1, j] = B[i, j];
          }
           
          // Check for boundary
          // condition of the matrix
          if (j != 0)
          {
               
              // If adjacent cells have
              // same value
              if (A[i, j - 1] == A[i, j])
                  B[i, j - 1] = B[i, j];
          }
           
          // Check for boundary
          // condition of the matrix
          if (j != M - 1)
          {
               
              // If adjacent cells have
              // same value
              if (A[i, j + 1] == A[i, j])
                  B[i, j + 1] = B[i, j];
          }
       }
    }
 
    // Store all distinct elements
    // in a set
    for(int i = 0; i < N; i++)
    {
       for(int j = 0; j < M; j++)
          s.Add(B[i, j]);
    }
 
    // Return answer
    return s.Count;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 2, M = 3;
    int [,]A = { { 2, 1, 3 },
                 { 1, 1, 2 } };
                     
    // Function Call
    Console.Write(solve(A, N, M));
}
}
 
// This code is contributed by 29AjayKumar

chevron_right


Output: 

4

 

Time Complexity: O(N2
Auxiliary Space: O(N2)
 

competitive-programming-img




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : 29AjayKumar