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
Source: Asked in Directi
Algorithm: A dynamic programming approach is given below:
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 */
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)
This article is contributed by Ayush Jauhari. 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.
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Improved By : vt_m
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