Minimum cells required to reach destination with jumps equal to cell values

Given a m x n matrix mat[][] containing positive integers. The problem is to reach to the cell (m-1, n-1) from the cell (0, 0) by following the given constraints. From a cell (i, j) one can move ‘exactly’ a distance of ‘mat[i][j]’ to the right (in the same row) or to below (in the same column) only if the movement takes to a cell within matrix boundaries.
For example: Given mat[1][1] = 4, then one can move to cells mat[1][5] and mat[5][1] only if these cells exists in the matrix. Following the constraints check whether one can reach cell (m-1, n-1) from (0, 0). 1If one can reach then print the minimum number of cells required to be covered during the movement else print “-1”.

Examples:

Input : mat[][] = { {2, 3, 2, 1, 4},
                    {3, 2, 5, 8, 2},
                    {1, 1, 2, 2, 1}  }
Output : 4
The movement and cells covered are as follows:
(0, 0)->(0, 2)
          |
        (2, 2)->(2, 4)

Input : mat[][] = { {2, 4, 2},
                {5, 3, 8},
            {1, 1, 1} }
Output : 3

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Algorithm: A dynamic programming approach is given below:

Below is the implementation of above approach:

C++

// C++ implementation to count minimum cells required 
// to be covered to reach destination 
#include <bits/stdc++.h> 
  
using namespace std; 
  
#define SIZE 100 
  
// function to count minimum cells required 
// to be covered to reach destination 
int minCells(int mat[SIZE][SIZE], int m, int n) 
{ 
    // to store min cells required to be 
    // covered to reach a particular cell 
    int dp[m][n]; 
  
    // initially no cells can be reached 
    for (int i = 0; i < m; i++) 
        for (int j = 0; j < n; j++) 
            dp[i][j] = INT_MAX; 
  
    // base case 
    dp[0][0] = 1; 
  
    // building up the dp[][] matrix 
    for (int i = 0; i < m; i++) { 
        for (int j = 0; j < n; j++) { 
  
            // dp[i][j] != INT_MAX denotes that cell (i, j) 
            // can be reached from cell (0, 0) and the other 
            // half of the condition finds the cell on the 
            // right that can be reached from (i, j) 
            if (dp[i][j] != INT_MAX && (j + mat[i][j]) < n 
                && (dp[i][j] + 1) < dp[i][j + mat[i][j]]) 
                dp[i][j + mat[i][j]] = dp[i][j] + 1; 
  
            // the other half of the condition finds the cell 
            // right below that can be reached from (i, j) 
            if (dp[i][j] != INT_MAX && (i + mat[i][j]) < m 
                && (dp[i][j] + 1) < dp[i + mat[i][j]][j]) 
                dp[i + mat[i][j]][j] = dp[i][j] + 1; 
        } 
    } 
  
    // it true then cell (m-1, n-1) can be reached 
    // from cell (0, 0) and returns the minimum 
    // number of cells covered 
    if (dp[m - 1][n - 1] != INT_MAX) 
        return dp[m - 1][n - 1]; 
  
    // cell (m-1, n-1) cannot be reached from 
    // cell (0, 0) 
    return -1; 
} 
  
// Driver program to test above 
int main() 
{ 
    int mat[SIZE][SIZE] = { { 2, 3, 2, 1, 4 }, 
                            { 3, 2, 5, 8, 2 }, 
                            { 1, 1, 2, 2, 1 } }; 
  
    int m = 3, n = 5; 
    cout << "Minimum number of cells = "
         << minCells(mat, m, n); 
  
    return 0; 
} 

Java

// Java implementation to count minimum 
// cells required to be covered to reach 
// destination 
class MinCellsDestination 
{ 
    static final int SIZE=100; 
     
    // function to count minimum cells required 
    // to be covered to reach destination 
    static int minCells(int mat[][], int m, int n) 
    { 
        // to store min cells required to be 
        // covered to reach a particular cell 
        int dp[][] = new int[m][n]; 
        
        // initially no cells can be reached 
        for (int i = 0; i < m; i++) 
            for (int j = 0; j < n; j++) 
                dp[i][j] = Integer.MAX_VALUE; 
        
        // base case 
        dp[0][0] = 1; 
        
        // building up the dp[][] matrix 
        for (int i = 0; i < m; i++) { 
            for (int j = 0; j < n; j++) { 
        
                // dp[i][j] != INT_MAX denotes that cell 
                // (i, j) can be reached from cell (0, 0) 
                // and the other half of the condition 
                // finds the cell on the right that can 
                // be reached from (i, j) 
                if (dp[i][j] != Integer.MAX_VALUE &&  
                   (j + mat[i][j]) < n && (dp[i][j] + 1) 
                   < dp[i][j + mat[i][j]]) 
                    dp[i][j + mat[i][j]] = dp[i][j] + 1; 
        
                // the other half of the condition finds 
                // the cell right below that can be  
                // reached from (i, j) 
                if (dp[i][j] != Integer.MAX_VALUE &&  
                   (i + mat[i][j]) < m && (dp[i][j] + 1) 
                   < dp[i + mat[i][j]][j]) 
                    dp[i + mat[i][j]][j] = dp[i][j] + 1; 
            } 
        } 
        
        // it true then cell (m-1, n-1) can be reached 
        // from cell (0, 0) and returns the minimum 
        // number of cells covered 
        if (dp[m - 1][n - 1] != Integer.MAX_VALUE) 
            return dp[m - 1][n - 1]; 
        
        // cell (m-1, n-1) cannot be reached from 
        // cell (0, 0) 
        return -1; 
    } 
      
    // Driver code 
    public static void main(String args[]) 
    { 
         int mat[][] = { { 2, 3, 2, 1, 4 }, 
                         { 3, 2, 5, 8, 2 }, 
                         { 1, 1, 2, 2, 1 }}; 
    
        int m = 3, n = 5; 
        System.out.println("Minimum number of cells" + 
                          " = " + minCells(mat, m, n)); 
    } 
} 
/* This code is contributed by Danish Kaleem */

Python3

# Python3 implementation to count minimum cells required  
# to be covered to reach destination  
SIZE=100 
MAX=10000000
# function to count minimum cells required  
# to be covered to reach destination  
def minCells( mat,  m,  n):  
    # to store min cells required to be  
    # covered to reach a particular cell  
    dp=[[MAX for i in range(n)]for i in range(m)]  
    
    # initially no cells can be reached  
      
    # base case  
    dp[0][0] = 1
    
    # building up the dp[][] matrix  
    for i in range(m): 
        for j in range(n):   
            # dp[i][j] != MAX denotes that cell (i, j)  
            # can be reached from cell (0, 0) and the other  
            # half of the condition finds the cell on the  
            # right that can be reached from (i, j)  
            if (dp[i][j] != MAX and
             (j + mat[i][j]) < n and 
             (dp[i][j] + 1) < dp[i][j + mat[i][j]]):  
                dp[i][j + mat[i][j]] = dp[i][j] + 1
    
            # the other half of the condition finds the cell  
            # right below that can be reached from (i, j)  
            if (dp[i][j] != MAX and (i + mat[i][j]) < m  
                and (dp[i][j] + 1) < dp[i + mat[i][j]][j]): 
                dp[i + mat[i][j]][j] = dp[i][j] + 1
  
    # it true then cell (m-1, n-1) can be reached  
    # from cell (0, 0) and returns the minimum  
    # number of cells covered  
    if (dp[m - 1][n - 1] != MAX): 
        return dp[m - 1][n - 1]  
    
    # cell (m-1, n-1) cannot be reached from  
    # cell (0, 0)  
    return -1
    
# Driver program to test above  
mat= [ [ 2, 3, 2, 1, 4 ],  
       [ 3, 2, 5, 8, 2 ], 
       [ 1, 1, 2, 2, 1 ]]  
    
m = 3
n = 5 
print("Minimum number of cells = ", 
     minCells(mat, m, n)) 
#this code is contributed by sahilshelangia 

C#

// C# implementation to count minimum 
// cells required to be covered to reach 
// destination 
using System; 
  
class GFG { 
      
    //static int SIZE=100; 
      
    // function to count minimum cells required 
    // to be covered to reach destination 
    static int minCells(int [,]mat, int m, int n) 
    { 
          
        // to store min cells required to be 
        // covered to reach a particular cell 
        int [,]dp = new int[m,n]; 
      
        // initially no cells can be reached 
        for (int i = 0; i < m; i++) 
            for (int j = 0; j < n; j++) 
                dp[i,j] = int.MaxValue; 
      
        // base case 
        dp[0,0] = 1; 
      
        // building up the dp[][] matrix 
        for (int i = 0; i < m; i++) { 
            for (int j = 0; j < n; j++) { 
      
                // dp[i][j] != INT_MAX denotes that 
                // cell (i, j) can be reached from 
                // cell (0, 0) and the other half 
                // of the condition finds the cell 
                // on the right that can be reached 
                // from (i, j) 
                if (dp[i,j] != int.MaxValue &&  
                (j + mat[i,j]) < n && (dp[i,j] + 1) 
                < dp[i,j + mat[i,j]]) 
                    dp[i,j + mat[i,j]] = dp[i,j] + 1; 
      
                // the other half of the condition 
                // finds the cell right below that 
                // can be reached from (i, j) 
                if (dp[i,j] != int.MaxValue &&  
                (i + mat[i,j]) < m && (dp[i,j] + 1) 
                < dp[i + mat[i,j],j]) 
                    dp[i + mat[i,j],j] = dp[i,j] + 1; 
            } 
        } 
      
        // it true then cell (m-1, n-1) can be 
        // reached from cell (0, 0) and returns 
        // the minimum number of cells covered 
        if (dp[m - 1,n - 1] != int.MaxValue) 
            return dp[m - 1,n - 1]; 
      
        // cell (m-1, n-1) cannot be reached from 
        // cell (0, 0) 
        return -1; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        int [,]mat = { { 2, 3, 2, 1, 4 }, 
                       { 3, 2, 5, 8, 2 }, 
                       { 1, 1, 2, 2, 1 } }; 
  
        int m = 3, n = 5; 
        Console.WriteLine("Minimum number of "
            + "cells = " + minCells(mat, m, n)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP implementation to count 
// minimum cells required to be  
// covered to reach destination 
  
// function to count minimum 
// cells required to be  
// covered to reach destination 
function minCells( $mat, $m, $n) 
{ 
      
    // to store min cells 
    // required to be 
    // covered to reach  
    // a particular cell 
    $dp =array(array()); 
  
    // initially no cells  
    // can be reached 
    for($i = 0; $i < $m; $i++) 
        for($j = 0; $j < $n; $j++) 
            $dp[$i][$j] = PHP_INT_MAX; 
  
    // base case 
    $dp[0][0] = 1; 
  
    // building up the dp[][] matrix 
    for($i = 0; $i < $m; $i++) 
    { 
        for($j = 0; $j < $n; $j++)  
        { 
  
            // dp[i][j] != INT_MAX 
            // denotes that cell (i, j) 
            // can be reached from cell 
            // (0, 0) and the other half  
            // of the condition finds the 
            // cell on the right that can  
            // be reached from (i, j) 
            if ($dp[$i][$j] != PHP_INT_MAX and
                        ($j + $mat[$i][$j]) <$n
                         and ($dp[$i][$j] + 1) <  
                     $dp[$i][$j + $mat[$i][$j]]) 
                       
                $dp[$i][$j + $mat[$i][$j]] =  
                           $dp[$i][$j] + 1; 
  
            // the other half of the  
            // condition finds the cell 
            // right below that can be 
            // reached from (i, j) 
            if ($dp[$i][$j] != PHP_INT_MAX and 
                      ($i + $mat[$i][$j]) < $m
                      and ($dp[$i][$j] + 1) <  
                      $dp[$i +$mat[$i][$j]][$j]) 
                        
                $dp[$i + $mat[$i][$j]][$j] = $dp[$i][$j] + 1; 
        } 
    } 
  
    // it true then cell  
    // (m-1, n-1) can be reached 
    // from cell (0, 0) and  
    // returns the minimum 
    // number of cells covered 
    if ($dp[$m - 1][$n - 1] != PHP_INT_MAX) 
        return $dp[$m - 1][$n - 1]; 
  
    // cell (m-1, n-1) cannot  
    // be reached from 
    // cell (0, 0) 
    return -1; 
} 
  
    // Driver Code 
    $mat = array(array(2, 3, 2, 1, 4), 
                 array(3, 2, 5, 8, 2), 
                 array(1, 1, 2, 2, 1)); 
  
    $m = 3; $n = 5; 
    echo "Minimum number of cells = "
        , minCells($mat, $m, $n); 
  
// This code is contributed by anuj_67. 
?> 

Output:

Minimum number of cells = 4

Time Complexity: O(m*n)
Auxiliary Space: O(m*n)

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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