Given an N X M matrix, where ai, j = 1 denotes the cell is not empty, ai, j = 0 denotes the cell is empty and ai, j = 2, denotes that you are standing at that cell. You can move vertically up or down and horizontally left or right to any cell which is empty. The task is to find the minimum number of steps to reach any boundary edge of the matrix. Print -1 if not possible to reach any of the boundary edges.
Note: There will be only one cell with value 2 in the entire matrix.
Examples:
Input: matrix[] = {1, 1, 1, 0, 1} {1, 0, 2, 0, 1} {0, 0, 1, 0, 1} {1, 0, 1, 1, 0} Output: 2 Move to the right and then move upwards to reach the nearest boundary edge. Input: matrix[] = {1, 1, 1, 1, 1} {1, 0, 2, 0, 1} {1, 0, 1, 0, 1} {1, 1, 1, 1, 1} Output: -1
Approach: The problem has been solved in Set-1 using Dynamic Programming. In this article, we will be discussing a method using BFS. Given below are the steps to solve the above problem.
- Find the index of the ‘2’ in the matrix.
- Check if this index is a boundary edge or not, if it is, then no moves are required.
- Insert the index x and index y of ‘2’ in the queue with moves as 0.
- Use a 2-D vis array to mark the visiting positions in the matrix.
- Iterate till the queue is empty or we reach any boundary edge.
- Get the front element(x, y, val = moves) in the queue and mark vis[x][y] as visited. Do all the possible moves(right, left, up and down) possible.
- Re-insert val+1 and their indexes of all the valid moves to the queue.
- If the x and y become the boundary edges any time return val.
- If all the moves are made, and the queue is empty, then it is not possible, hence return -1.
Below is the implementation of the above approach:
// C++ program to find Minimum steps // to reach any of the boundary // edges of a matrix #include <bits/stdc++.h> using namespace std;
#define r 4 #define c 5 // Function to check validity bool check( int i, int j, int n, int m, int mat[r])
{ if (i >= 0 && i < n && j >= 0 && j < m) {
if (mat[i][j] == 0)
return true ;
}
return false ;
} // Function to find out minimum steps int findMinSteps( int mat[r], int n, int m)
{ int indx, indy;
indx = indy = -1;
// Find index of only 2 in matrix
for ( int i = 0; i < n; i++) {
for ( int j = 0; j < m; j++) {
if (mat[i][j] == 2) {
indx = i, indy = j;
break ;
}
}
if (indx != -1)
break ;
}
// Push elements in the queue
queue<pair< int , pair< int , int > > > q;
// Push the position 2 with moves as 0
q.push(make_pair(0, make_pair(indx, indy)));
// If already at boundary edge
if (check(indx, indy, n, m, mat))
return 0;
// Marks the visit
bool vis[r];
memset (vis, 0, sizeof vis);
// Iterate in the queue
while (!q.empty()) {
// Get the front of the queue
auto it = q.front();
// Pop the first element from the queue
q.pop();
// Get the position
int x = it.second.first;
int y = it.second.second;
// Moves
int val = it.first;
// If a boundary edge
if (x == 0 || x == (n - 1) || y == 0 || y == (m - 1)) {
return val;
}
// Marks the visited array
vis[x][y] = 1;
// If a move is possible
if (check(x - 1, y, n, m, mat)) {
// If not visited previously
if (!vis[x - 1][y])
q.push(make_pair(val + 1, make_pair(x - 1, y)));
}
// If a move is possible
if (check(x + 1, y, n, m, mat)) {
// If not visited previously
if (!vis[x + 1][y])
q.push(make_pair(val + 1, make_pair(x + 1, y)));
}
// If a move is possible
if (check(x, y + 1, n, m, mat)) {
// If not visited previously
if (!vis[x][y + 1])
q.push(make_pair(val + 1, make_pair(x, y + 1)));
}
// If a move is possible
if (check(x, y - 1, n, m, mat)) {
// If not visited previously
if (!vis[x][y - 1])
q.push(make_pair(val + 1, make_pair(x, y - 1)));
}
}
return -1;
} // Driver Code int main()
{ int mat[r] = { { 1, 1, 1, 0, 1 },
{ 1, 0, 2, 0, 1 },
{ 0, 0, 1, 0, 1 },
{ 1, 0, 1, 1, 0 } };
cout << findMinSteps(mat, r, c);
} |
// Java program to find Minimum steps // to reach any of the boundary // edges of a matrix import java.util.*;
class GFG {
public static final int R = 4 ;
public static final int C = 5 ;
// Function to check validity
public static boolean check( int i, int j, int n, int m,
int [][] mat)
{
if (i >= 0 && i < n && j >= 0 && j < m) {
if (mat[i][j] == 0 ) {
return true ;
}
}
return false ;
}
// Function to find out minimum steps
public static int findMinSteps( int [][] mat, int n,
int m)
{
int indx, indy;
indx = indy = - 1 ;
// Find index of only 2 in matrix
for ( int i = 0 ; i < n; i++) {
for ( int j = 0 ; j < m; j++) {
if (mat[i][j] == 2 ) {
indx = i;
indy = j;
break ;
}
}
if (indx != - 1 ) {
break ;
}
}
// Push elements in the queue
Queue< int []> q = new LinkedList< int []>();
// Push the position 2 with moves as 0
q.add( new int [] { indx, indy, 0 });
// If already at boundary edge
if (check(indx, indy, n, m, mat)) {
return 0 ;
}
// Marks the visit
boolean [][] vis = new boolean [n][m];
// Iterate in the queue
while (!q.isEmpty()) {
// Pop the first element from the queue
int [] curr = q.poll();
// Get the position
int x = curr[ 0 ];
int y = curr[ 1 ];
int val = curr[ 2 ];
// If a boundary edge
if (x == 0 || x == (n - 1 ) || y == 0
|| y == (m - 1 )) {
return val;
}
// Marks the visited array
vis[x][y] = true ;
// If a move is possible towards up
if (check(x - 1 , y, n, m, mat)) {
// If not visited previously
if (!vis[x - 1 ][y]) {
q.add( new int [] { x - 1 , y, val + 1 });
}
}
// If a move is possible towards down
if (check(x + 1 , y, n, m, mat)) {
// If not visited previously
if (!vis[x + 1 ][y]) {
q.add( new int [] { x + 1 , y, val + 1 });
}
}
// If a move is possible towards right
if (check(x, y + 1 , n, m, mat)) {
// If not visited previously
if (!vis[x][y + 1 ]) {
q.add( new int [] { x, y + 1 , val + 1 });
}
}
// If a move is possible towards left
if (check(x, y - 1 , n, m, mat)) {
// If not visited previously
if (!vis[x][y - 1 ]) {
q.add( new int [] { x, y - 1 , val + 1 });
}
}
}
return - 1 ;
}
// Driver Code
public static void main(String[] args)
{
int [][] mat = { { 1 , 1 , 1 , 0 , 1 },
{ 1 , 0 , 2 , 0 , 1 },
{ 0 , 0 , 1 , 0 , 1 },
{ 1 , 0 , 1 , 1 , 0 } };
System.out.println(findMinSteps(mat, R, C));
}
} // This code is contributed by Prasad Kandekar(prasad264) |
# Python3 program to find Minimum steps # to reach any of the boundary # edges of a matrix from collections import deque
r = 4
c = 5
# Function to check validity def check(i, j, n, m, mat):
if (i > = 0 and i < n and j > = 0 and j < m):
if (mat[i][j] = = 0 ):
return True
return False
# Function to find out minimum steps def findMinSteps(mat, n, m):
indx = indy = - 1 ;
# Find index of only 2 in matrix
for i in range (n):
for j in range (m):
if (mat[i][j] = = 2 ):
indx = i
indy = j
break
if (indx ! = - 1 ):
break
# Push elements in the queue
q = deque()
# Push the position 2 with moves as 0
q.append([ 0 , indx, indy])
# If already at boundary edge
if (check(indx, indy, n, m, mat)):
return 0
# Marks the visit
vis = [ [ 0 for i in range (r)] for i in range (r)]
# Iterate in the queue
while len (q) > 0 :
# Get the front of the queue
it = q.popleft()
#Pop the first element from the queue
# Get the position
x = it[ 1 ]
y = it[ 2 ]
# Moves
val = it[ 0 ]
# If a boundary edge
if (x = = 0 or x = = (n - 1 ) or y = = 0 or y = = (m - 1 )):
return val
# Marks the visited array
vis[x][y] = 1
# If a move is possible
if (check(x - 1 , y, n, m, mat)):
# If not visited previously
if ( not vis[x - 1 ][y]):
q.append([val + 1 , x - 1 , y])
# If a move is possible
if (check(x + 1 , y, n, m, mat)):
# If not visited previously
if ( not vis[x + 1 ][y]):
q.append([val + 1 , x + 1 , y])
# If a move is possible
if (check(x, y + 1 , n, m, mat)):
# If not visited previously
if ( not vis[x][y + 1 ]):
q.append([val + 1 , x, y + 1 ])
# If a move is possible
if (check(x, y - 1 , n, m, mat)):
# If not visited previously
if ( not vis[x][y - 1 ]):
q.append([val + 1 , x, y - 1 ])
return - 1
# Driver Code mat = [[ 1 , 1 , 1 , 0 , 1 ],
[ 1 , 0 , 2 , 0 , 1 ],
[ 0 , 0 , 1 , 0 , 1 ],
[ 1 , 0 , 1 , 1 , 0 ] ];
print (findMinSteps(mat, r, c))
# This code is contributed by mohit kumar 29 |
// C# program to find Minimum steps // to reach any of the boundary // edges of a matrix using System;
using System.Collections.Generic;
public class GFG {
public static readonly int R = 4;
public static readonly int C = 5;
// Function to check validity
public static bool Check( int i, int j, int n, int m,
int [, ] mat)
{
if (i >= 0 && i < n && j >= 0 && j < m) {
if (mat[i, j] == 0) {
return true ;
}
}
return false ;
}
// Function to find out minimum steps
public static int findMinSteps( int [, ] mat, int n,
int m)
{
int indx, indy;
indx = indy = -1;
// Find index of only 2 in matrix
for ( int i = 0; i < n; i++) {
for ( int j = 0; j < m; j++) {
if (mat[i, j] == 2) {
indx = i;
indy = j;
break ;
}
}
if (indx != -1) {
break ;
}
}
// Push elements in the queue
Queue< int []> q = new Queue< int []>();
// Push the position 2 with moves as 0
q.Enqueue( new int [] { indx, indy, 0 });
// If already at boundary edge
if (Check(indx, indy, n, m, mat)) {
return 0;
}
// Marks the visit
bool [, ] vis = new bool [n, m];
// Iterate in the queue
while (q.Count != 0) {
// Get the front of the queue
int [] curr = q.Dequeue();
// Get the position
int x = curr[0];
int y = curr[1];
// Moves
int val = curr[2];
// If a boundary edge
if (x == 0 || x == (n - 1) || y == 0
|| y == (m - 1)) {
return val;
}
// Marks the visited array
vis[x, y] = true ;
// If a move is possible
if (Check(x - 1, y, n, m, mat)) {
// If not visited previously
if (!vis[x - 1, y]) {
q.Enqueue(
new int [] { x - 1, y, val + 1 });
}
}
// If a move is possible
if (Check(x + 1, y, n, m, mat)) {
// If not visited previously
if (!vis[x + 1, y]) {
q.Enqueue(
new int [] { x + 1, y, val + 1 });
}
}
// If a move is possible
if (Check(x, y + 1, n, m, mat)) {
// If not visited previously
if (!vis[x, y + 1]) {
q.Enqueue(
new int [] { x, y + 1, val + 1 });
}
}
// If a move is possible
if (Check(x, y - 1, n, m, mat)) {
// If not visited previously
if (!vis[x, y - 1]) {
q.Enqueue(
new int [] { x, y - 1, val + 1 });
}
}
}
return -1;
}
// Driver Code
static public void Main( string [] args)
{
int [, ] mat = { { 1, 1, 1, 0, 1 },
{ 1, 0, 2, 0, 1 },
{ 0, 0, 1, 0, 1 },
{ 1, 0, 1, 1, 0 } };
Console.WriteLine(findMinSteps(mat, 4, 5));
}
} // This code is contributed by Prasad Kandekar(prasad264) |
<script> // JavaScript program to find Minimum steps // to reach any of the boundary // edges of a matrix const r = 4 const c = 5 // Function to check validity function check(i, j, n, m, mat){
if (i >= 0 && i < n && j >= 0 && j < m){
if (mat[i][j] == 0)
return true
}
return false
} // Function to find out minimum steps function findMinSteps(mat, n, m){
let indx = -1, indy = -1;
// Find index of only 2 in matrix
for (let i=0;i<n;i++){
for (let j=0;j<m;j++){
if (mat[i][j] == 2){
indx = i
indy = j
break
}
}
if (indx != -1){
break
}
}
// Push elements in the queue
let q = []
// Push the position 2 with moves as 0
q.push([0, indx, indy])
// If already at boundary edge
if (check(indx, indy, n, m, mat))
return 0
// Marks the visit
let vis = new Array(r);
for (let i=0;i<r;i++){
vis[i] = new Array(r).fill(0);
}
// Iterate in the queue
while (q.length > 0){
// Get the front of the queue
let it = q.shift()
//Pop the first element from the queue
// Get the position
let x = it[1]
let y = it[2]
// Moves
let val = it[0]
// If a boundary edge
if (x == 0 || x == (n - 1) || y == 0 || y == (m - 1))
return val
// Marks the visited array
vis[x][y] = 1
// If a move is possible
if (check(x - 1, y, n, m, mat)){
// If not visited previously
if (vis[x - 1][y] == 0)
q.push([val + 1, x - 1, y])
}
// If a move is possible
if (check(x + 1, y, n, m, mat)){
// If not visited previously
if (vis[x + 1][y] == 0)
q.push([val + 1, x + 1, y])
}
// If a move is possible
if (check(x, y + 1, n, m, mat)){
// If not visited previously
if (vis[x][y + 1] == 0)
q.push([val + 1, x, y + 1])
}
// If a move is possible
if (check(x, y - 1, n, m, mat)){
// If not visited previously
if (vis[x][y - 1] == 0)
q.push([val + 1, x, y - 1])
}
}
return -1
} // Driver Code let mat = [[1, 1, 1, 0, 1 ], [1, 0, 2, 0, 1 ],
[0, 0, 1, 0, 1 ],
[1, 0, 1, 1, 0 ]];
document.write(findMinSteps(mat, r, c)) // This code is contributed by shinjanpatra </script> |
2
Complexity Analysis:
- Time Complexity: O(N^2)
- Auxiliary Space: O(N^2)