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
Approach:
- Initialize a 2D array dp of size m x n with all values as INT_MAX.
- Set dp[0][0] to 1, as the first cell is already reached.
- Iterate through each cell of the matrix, and check if the current cell can be reached from any cell already reached, i.e., if dp[i][j] != INT_MAX. If it can be reached, update the minimum cells required to reach the current cell from the previous cell by checking the cells to the right and bottom of the current cell.
- Return dp[m-1][n-1] if it is not equal to INT_MAX, else return -1.
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 destinationint 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 aboveint 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// destinationimport java.util.*;import java.io.*;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=100MAX=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 = 3n = 5print("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// destinationusing 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 destinationfunction 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.?> |
Javascript
<script>// Javascript implementation to count minimum// cells required to be covered to reach// destinationlet SIZE=100; // 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 let dp = new Array(m); // Loop to create 2D array using 1D array for (var i = 0; i < dp.length; i++) { dp[i] = new Array(2); } // initially no cells can be reached for (let i = 0; i < m; i++) for (let j = 0; j < n; j++) dp[i][j] = Number.MAX_VALUE; // base case dp[0][0] = 1; // building up the dp[][] matrix for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { // dp[i][j] != LET_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] != Number.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] != Number.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] != Number.MAX_VALUE) return dp[m - 1][n - 1]; // cell (m-1, n-1) cannot be reached from // cell (0, 0) return -1; } // driver function let mat = [[ 2, 3, 2, 1, 4 ], [ 3, 2, 5, 8, 2 ], [ 1, 1, 2, 2, 1 ]]; let m = 3, n = 5; document.write("Minimum number of cells" + " = " + minCells(mat, m, n)); </script> |
Output
Minimum number of cells = 4
Time Complexity: O(m*n)
Auxiliary Space: O(m*n)
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