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

  • Difficulty Level : Expert
  • Last Updated : 27 May, 2021

Given a matrix A[][] of dimension 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 the 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 to fill each cell with distinct values.
  • Traverse matrix A[][] starting from (0, 0). For every cell (i, j), check if its adjacent cells have the same value as A[i][j] or not.
  • If any adjacent cell has the same value, replace that cell in B[][] with the value of B[i][j].
  • The counting 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++




// 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);
}

Java




// 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

Python3




# 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

C#




// 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

Javascript




<script>
 
// Javascript 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
function solve(A, N, M)
{
 
    var B = Array.from(Array(N), ()=> Array(M));
    var c = 1;
    var s = new Set();
 
    // Constructing another matrix
    // consisting of distinct values
    for (var i = 0; i < N; i++) {
        for (var 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 (var i = 0; i < N; i++) {
        for (var 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 (var i = 0; i < N; i++) {
        for (var j = 0; j < M; j++)
            s.add(B[i][j]);
    }
 
    // Return answer
    return s.size;
}
 
// Driver Code
var N = 2, M = 3;
var A = [ [ 2, 1, 3 ],
                [ 1, 1, 2 ]];
// Function Call
document.write( solve(A, N, M));
 
</script>
Output: 
4

 

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




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