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 a value of 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 can be solved using a Dynamic Programming approach. Given below is the algorithm to solve the above problem.
- Find the position which has ‘2’ in the matrix.
- Initialize two 2-D arrays of size the same as the matrix. The dp[][] which stores the minimum number of steps to reach any index i, j and vis[][] marks if any particular i, j position has been visited or not previously.
- Call the recursive function which has the base case as follows:
- if the traversal at any point reaches any of the boundary edges return 0.
- if the position of the points n, m has stored the minimum number of steps previously, then return dp[n][m].
- Call the recursion again with all possible four moves that can be done from the position n, m. The moves are only possible if mat[n][m] is 0 and the position has not been visited previously.
- Store the minimal of the four moves.
- If the recursion returns any value less than 1e9, which we had stored as the maximum value, then there is an answer, else it does not have an answer.
Below is the implementation of the above approach:
C++
// 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 col 5// Function to find out minimum stepsint findMinSteps(int mat[r][col], int n, int m, int dp[r][col], bool vis[r][col]){ // boundary edges reached if (n == 0 || m == 0 || n == (r - 1) || m == (col - 1)) { return 0; } // already had a route through this // point, hence no need to re-visit if (dp[n][m] != -1) return dp[n][m]; // visiting a position vis[n][m] = true; int ans1, ans2, ans3, ans4; ans1 = ans2 = ans3 = ans4 = 1e9; // vertically up if (mat[n - 1][m] == 0) { if (!vis[n - 1][m]) ans1 = 1 + findMinSteps(mat, n - 1, m, dp, vis); } // horizontally right if (mat[n][m + 1] == 0) { if (!vis[n][m + 1]) ans2 = 1 + findMinSteps(mat, n, m + 1, dp, vis); } // horizontally left if (mat[n][m - 1] == 0) { if (!vis[n][m - 1]) ans3 = 1 + findMinSteps(mat, n, m - 1, dp, vis); } // vertically down if (mat[n + 1][m] == 0) { if (!vis[n + 1][m]) ans4 = 1 + findMinSteps(mat, n + 1, m, dp, vis); } // minimum of every path dp[n][m] = min(ans1, min(ans2, min(ans3, ans4))); return dp[n][m];}// Function that returns the minimum stepsint minimumSteps(int mat[r][col], int n, int m){ // index to store the location at // which you are standing int twox = -1; int twoy = -1; // find '2' in the matrix for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (mat[i][j] == 2) { twox = i; twoy = j; break; } } if (twox != -1) break; } // Initialize dp matrix with -1 int dp[r][col]; memset(dp, -1, sizeof dp); // Initialize vis matrix with false bool vis[r][col]; memset(vis, false, sizeof vis); // Call function to find out minimum steps // using memoization and recursion int res = findMinSteps(mat, twox, twoy, dp, vis); // if not possible if (res >= 1e9) return -1; else return res;}// Driver Codeint main(){ int mat[r][col] = { { 1, 1, 1, 0, 1 }, { 1, 0, 2, 0, 1 }, { 0, 0, 1, 0, 1 }, { 1, 0, 1, 1, 0 } }; cout << minimumSteps(mat, r, col);} |
Java
// Java program to find Minimum steps// to reach any of the boundary// edges of a matrixclass Solution{static final int r=4,c=5;// Function to find out minimum stepsstatic int findMinSteps(int mat[][], int n, int m, int dp[][], boolean vis[][]){ // boundary edges reached if (n == 0 || m == 0 || n == (r - 1) || m == (c - 1)) { return 0; } // already had a route through this // point, hence no need to re-visit if (dp[n][m] != -1) return dp[n][m]; // visiting a position vis[n][m] = true; int ans1, ans2, ans3, ans4; ans1 = ans2 = ans3 = ans4 = (int)1e9; // vertically up if (mat[n - 1][m] == 0) { if (!vis[n - 1][m]) ans1 = 1 + findMinSteps(mat, n - 1, m, dp, vis); } // horizontally right if (mat[n][m + 1] == 0) { if (!vis[n][m + 1]) ans2 = 1 + findMinSteps(mat, n, m + 1, dp, vis); } // horizontally left if (mat[n][m - 1] == 0) { if (!vis[n][m - 1]) ans3 = 1 + findMinSteps(mat, n, m - 1, dp, vis); } // vertically down if (mat[n + 1][m] == 0) { if (!vis[n + 1][m]) ans4 = 1 + findMinSteps(mat, n + 1, m, dp, vis); } // minimum of every path dp[n][m] = Math.min(ans1, Math.min(ans2, Math.min(ans3, ans4))); return dp[n][m];} // Function that returns the minimum stepsstatic int minimumSteps(int mat[][], int n, int m){ // index to store the location at // which you are standing int twox = -1; int twoy = -1; // find '2' in the matrix for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (mat[i][j] == 2) { twox = i; twoy = j; break; } } if (twox != -1) break; } // Initialize dp matrix with -1 int dp[][]=new int[r][r]; for(int j=0;j<r;j++) for(int i=0;i<r;i++)dp[j][i]=-1; // Initialize vis matrix with false boolean vis[][]= new boolean[r][r]; for(int j=0;j<r;j++) for(int i=0;i<r;i++)vis[j][i]=false; // Call function to find out minimum steps // using memoization and recursion int res = findMinSteps(mat, twox, twoy, dp, vis); // if not possible if (res >= 1e9) return -1; else return res;} // Driver Codepublic 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( minimumSteps(mat, r, c));}}//contributed by Arnab Kundu |
Python
# Python program to find Minimum steps# to reach any of the boundary# edges of a matrixr=4col=5 # Function to find out minimum stepsdef findMinSteps(mat, n, m, dp,vis): # boundary edges reached if (n == 0 or m == 0 or n == (r - 1) or m == (col - 1)): return 0 # already had a route through this # point, hence no need to re-visit if (dp[n][m] != -1): return dp[n][m] # visiting a position vis[n][m] = True ans1, ans2, ans3, ans4=10**9,10**9,10**9,10**9 # vertically up if (mat[n - 1][m] == 0): if (vis[n - 1][m]==False): ans1 = 1 + findMinSteps(mat, n - 1, m, dp, vis) # horizontally right if (mat[n][m + 1] == 0): if (vis[n][m + 1]==False): ans2 = 1 + findMinSteps(mat, n, m + 1, dp, vis) # horizontally left if (mat[n][m - 1] == 0): if (vis[n][m - 1]==False): ans3 = 1 + findMinSteps(mat, n, m - 1, dp, vis) # vertically down if (mat[n + 1][m] == 0): if (vis[n + 1][m]==False): ans4 = 1 + findMinSteps(mat, n + 1, m, dp, vis) # minimum of every path dp[n][m] = min(ans1, min(ans2, min(ans3, ans4))) return dp[n][m] # Function that returns the minimum stepsdef minimumSteps(mat, n, m): # index to store the location at # which you are standing twox = -1 twoy = -1 # find '2' in the matrix for i in range(n): for j in range(m): if (mat[i][j] == 2): twox = i twoy = j break if (twox != -1): break # Initialize dp matrix with -1 dp=[[-1 for i in range(col)] for i in range(r)] # Initialize vis matrix with false vis=[[False for i in range(col)] for i in range(r)] # Call function to find out minimum steps # using memoization and recursion res = findMinSteps(mat, twox, twoy, dp, vis) # if not possible if (res >= 10**9): return -1 else: return res # Driver Codemat= [ [ 1, 1, 1, 0, 1 ], [ 1, 0, 2, 0, 1 ], [ 0, 0, 1, 0, 1 ], [ 1, 0, 1, 1, 0 ] ]print(minimumSteps(mat, r, col))#this is contributed by Mohit kumar 29 |
C#
// C# program to find Minimum steps // to reach any of the boundary // edges of a matrixusing System;class Solution { static int r=4,c=5; // Function to find out minimum steps static int findMinSteps(int [,]mat, int n, int m, int [,]dp, bool [,]vis) { // boundary edges reached if (n == 0 || m == 0 || n == (r - 1) || m == (c - 1)) { return 0; } // already had a route through this // point, hence no need to re-visit if (dp[n,m] != -1) return dp[n,m]; // visiting a position vis[n,m] = true; int ans1, ans2, ans3, ans4; ans1 = ans2 = ans3 = ans4 = (int)1e9; // vertically up if (mat[n - 1,m] == 0) { if (!vis[n - 1,m]) ans1 = 1 + findMinSteps(mat, n - 1, m, dp, vis); } // horizontally right if (mat[n,m + 1] == 0) { if (!vis[n,m + 1]) ans2 = 1 + findMinSteps(mat, n, m + 1, dp, vis); } // horizontally left if (mat[n,m - 1] == 0) { if (!vis[n,m - 1]) ans3 = 1 + findMinSteps(mat, n, m - 1, dp, vis); } // vertically down if (mat[n + 1,m] == 0) { if (!vis[n + 1,m]) ans4 = 1 + findMinSteps(mat, n + 1, m, dp, vis); } // minimum of every path dp[n,m] = Math.Min(ans1, Math.Min(ans2, Math.Min(ans3, ans4))); return dp[n,m]; } // Function that returns the minimum steps static int minimumSteps(int [,]mat, int n, int m) { // index to store the location at // which you are standing int twox = -1; int twoy = -1; // find '2' in the matrix for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (mat[i,j] == 2) { twox = i; twoy = j; break; } } if (twox != -1) break; } // Initialize dp matrix with -1 int [,]dp = new int[r,r]; for(int j=0;j<r;j++) for(int i=0;i<r;i++) dp[j,i]=-1; // Initialize vis matrix with false bool [,]vis= new bool [r,r]; for(int j=0;j<r;j++) for(int i=0;i<r;i++) vis[j,i]=false; // Call function to find out minimum steps // using memoization and recursion int res = findMinSteps(mat, twox, twoy, dp, vis); // if not possible if (res >= 1e9) return -1; else return res; } // Driver Code public static void Main() { int [,]mat = { { 1, 1, 1, 0, 1 }, { 1, 0, 2, 0, 1 }, { 0, 0, 1, 0, 1 }, { 1, 0, 1, 1, 0 }, }; Console.WriteLine(minimumSteps(mat, r, c)); } // This code is contributed by Ryuga} |
Javascript
<script> // Javascript program to find Minimum steps // to reach any of the boundary // edges of a matrix let r=4,c=5; // Function to find out minimum steps function findMinSteps(mat, n, m, dp, vis) { // boundary edges reached if (n == 0 || m == 0 || n == (r - 1) || m == (c - 1)) { return 0; } // already had a route through this // point, hence no need to re-visit if (dp[n][m] != -1) return dp[n][m]; // visiting a position vis[n][m] = true; let ans1, ans2, ans3, ans4; ans1 = ans2 = ans3 = ans4 = 1e9; // vertically up if (mat[n - 1][m] == 0) { if (!vis[n - 1][m]) ans1 = 1 + findMinSteps(mat, n - 1, m, dp, vis); } // horizontally right if (mat[n][m + 1] == 0) { if (!vis[n][m + 1]) ans2 = 1 + findMinSteps(mat, n, m + 1, dp, vis); } // horizontally left if (mat[n][m - 1] == 0) { if (!vis[n][m - 1]) ans3 = 1 + findMinSteps(mat, n, m - 1, dp, vis); } // vertically down if (mat[n + 1][m] == 0) { if (!vis[n + 1][m]) ans4 = 1 + findMinSteps(mat, n + 1, m, dp, vis); } // minimum of every path dp[n][m] = Math.min(ans1, Math.min(ans2, Math.min(ans3, ans4))); return dp[n][m]; } // Function that returns the minimum steps function minimumSteps(mat, n, m) { // index to store the location at // which you are standing let twox = -1; let twoy = -1; // find '2' in the matrix for (let i = 0; i < n; i++) { for (let j = 0; j < m; j++) { if (mat[i][j] == 2) { twox = i; twoy = j; break; } } if (twox != -1) break; } // Initialize dp matrix with -1 let dp = new Array(r); for(let j=0;j<r;j++) { dp[j] = new Array(r); for(let i=0;i<r;i++) { dp[j][i]=-1; } } // Initialize vis matrix with false let vis = new Array(r); for(let j=0;j<r;j++) { vis[j] = new Array(r); for(let i=0;i<r;i++) { vis[j][i]=false; } } // Call function to find out minimum steps // using memoization and recursion let res = findMinSteps(mat, twox, twoy, dp, vis); // if not possible if (res >= 1e9) return -1; else return res; } let mat = [ [ 1, 1, 1, 0, 1 ], [ 1, 0, 2, 0, 1 ], [ 0, 0, 1, 0, 1 ], [ 1, 0, 1, 1, 0 ] ]; document.write( minimumSteps(mat, r, c));// This code is contributed by suresh07.</script> |
Output
2
Complexity Analysis:
- Time Complexity: O(N^2), as we are using a for loop to traverse N times and in each traversal, we are calling the function findMinSteps which costs O(N) time, so the effective time will be O(N*N).
- Auxiliary Space: O(N^2), as we are using extra space for the dp matrix.
Minimum steps to reach any of the boundary edges of a matrix | Set-2
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