Find minimum steps required to reach the end of a matrix | Set – 1

12 Sep, 2021
Given a 2d-matrix consisting of positive integers, the task is to find the minimum number of steps required to reach the end(leftmost-bottom cell) of the matrix. If we are at cell (i, j) we can go to cells (i, j+arr[i][j]) or (i+arr[i][j], j). We can not go out of bounds. If no path exists, print -1.
Examples: 
Input : 
mat[][] = {{2, 1, 2},
           {1, 1, 1},
           {1, 1, 1}}
Output : 2
Explanation : The path will be {0, 0} -> {0, 2} -> {2, 2}
Thus, we are reaching end in two steps.

Input :
mat[][] = {{1, 1, 2},
           {1, 1, 1},
           {2, 1, 1}}
Output : 3
A simple solution is to explore all possible solutions. This will take exponential time.Better approach: We can use dynamic programming to solve this problem in polynomial time.Let’s decide the states of ‘dp’. We will build up our solution on 2d DP. Let’s say we are at cell {i, j}. We will try to find the minimum number of steps required to reach the cell (n-1, n-1) from this cell. We only have two possible paths i.e. to cells {i, j+arr[i][j]} or {i+arr[i][j], j}.A simple recurrence relation will be: 
dp[i][j] = 1 + min(dp[i+arr[i]][j], dp[i][j+arr[i][j]])
Below is the implementation of the above idea: 
C++
Java
Python3
C#
Javascript
// C++ program to implement above approach #include <bits/stdc++.h>#define n 3using namespace std; // 2d array to store// states of dpint dp[n][n]; // array to determine whether// a state has been solved beforeint v[n][n]; // Function to find the minimum number of// steps to reach the end of matrixint minSteps(int i, int j, int arr[][n]){    // base cases    if (i == n - 1 and j == n - 1)        return 0;     if (i > n - 1 || j > n - 1)        return 9999999;     // if a state has been solved before    // it won't be evaluated again.    if (v[i][j])        return dp[i][j];     v[i][j] = 1;     // recurrence relation    dp[i][j] = 1 + min(minSteps(i + arr[i][j], j, arr),                       minSteps(i, j + arr[i][j], arr));     return dp[i][j];} // Driver Codeint main(){    int arr[n][n] = { { 2, 1, 2 },                      { 1, 1, 1 },                      { 1, 1, 1 } };     int ans = minSteps(0, 0, arr);    if (ans >= 9999999)        cout << -1;    else        cout << ans;     return 0;}
// Java program to implement above approachclass GFG {     static int n = 3;     // 2d array to store    // states of dp    static int dp[][] = new int[n][n];     // array to determine whether    // a state has been solved before    static int[][] v = new int[n][n];     // Function to find the minimum number of    // steps to reach the end of matrix    static int minSteps(int i, int j, int arr[][])    {        // base cases        if (i == n - 1 && j == n - 1) {            return 0;        }         if (i > n - 1 || j > n - 1) {            return 9999999;        }         // if a state has been solved before        // it won't be evaluated again.        if (v[i][j] == 1) {            return dp[i][j];        }         v[i][j] = 1;         // recurrence relation        dp[i][j] = 1 + Math.min(minSteps(i + arr[i][j], j, arr), minSteps(i, j + arr[i][j], arr));         return dp[i][j];    }     // Driver Code    public static void main(String[] args)    {        int arr[][] = { { 2, 1, 2 },                        { 1, 1, 1 },                        { 1, 1, 1 } };         int ans = minSteps(0, 0, arr);        if (ans >= 9999999) {            System.out.println(-1);        }        else {            System.out.println(ans);        }    }} // This code contributed by Rajput-Ji
# Python3 program to implement above approachimport numpy as np; n = 3 # 2d array to store# states of dpdp = np.zeros((n, n)); # array to determine whether# a state has been solved beforev = np.zeros((n, n)); # Function to find the minimum number of# steps to reach the end of matrixdef minSteps(i, j, arr) :     # base cases    if (i == n - 1 and j == n - 1) :        return 0;     if (i > n - 1 or j > n - 1) :        return 9999999;     # if a state has been solved before    # it won't be evaluated again.    if (v[i][j]) :        return dp[i][j];     v[i][j] = 1;     # recurrence relation    dp[i][j] = 1 + min(minSteps(i + arr[i][j], j, arr),                    minSteps(i, j + arr[i][j], arr));     return dp[i][j];  # Driver Codearr = [ [ 2, 1, 2 ],            [ 1, 1, 1 ],            [ 1, 1, 1 ]            ];             ans = minSteps(0, 0, arr);if (ans >= 9999999) :    print(-1);         else :    print(ans); # This code is contributed by AnkitRai01
// C# program to implement above approachusing System; class GFG{         static int n = 3;     // 2d array to store    // states of dp    static int [,]dp = new int[n, n];     // array to determine whether    // a state has been solved before    static int[,] v = new int[n, n];     // Function to find the minimum number of    // steps to reach the end of matrix    static int minSteps(int i, int j, int [,]arr)    {        // base cases        if (i == n - 1 && j == n - 1)        {            return 0;        }         if (i > n - 1 || j > n - 1)        {            return 9999999;        }         // if a state has been solved before        // it won't be evaluated again.        if (v[i, j] == 1)        {            return dp[i, j];        }         v[i, j] = 1;         // recurrence relation        dp[i, j] = 1 + Math.Min(minSteps(i + arr[i,j], j, arr),                            minSteps(i, j + arr[i,j], arr));         return dp[i, j];    }     // Driver Code    static public void Main ()    {        int [,]arr = { { 2, 1, 2 },                        { 1, 1, 1 },                        { 1, 1, 1 } };         int ans = minSteps(0, 0, arr);        if (ans >= 9999999)        {            Console.WriteLine(-1);        }        else        {            Console.WriteLine(ans);        }    }} // This code contributed by ajit.
<script>     // Javascript program to implement    // above approach         let n = 3;       // 2d array to store    // states of dp    let dp = new Array(n);    for(let i = 0; i < n; i++)    {        dp[i] = new Array(n);    }       // array to determine whether    // a state has been solved before    let v = new Array(n);    for(let i = 0; i < n; i++)    {        v[i] = new Array(n);    }       // Function to find the minimum number of    // steps to reach the end of matrix    function minSteps(i, j, arr)    {        // base cases        if (i == n - 1 && j == n - 1)        {            return 0;        }           if (i > n - 1 || j > n - 1)        {            return 9999999;        }           // if a state has been solved before        // it won't be evaluated again.        if (v[i][j] == 1) {            return dp[i][j];        }           v[i][j] = 1;           // recurrence relation        dp[i][j] = 1 + Math.min(minSteps(i +        arr[i][j], j, arr), minSteps(i, j + arr[i][j], arr));           return dp[i][j];    }         let arr = [ [ 2, 1, 2 ],               [ 1, 1, 1 ],               [ 1, 1, 1 ] ];       let ans = minSteps(0, 0, arr);    if (ans >= 9999999) {      document.write(-1);    }    else {      document.write(ans);    }     </script>
2
 
Time Complexity: O(N2). 
Rajput-Ji
jit_t
ankthon
mukesh07
simmytarika5
Divide and Conquer
Dynamic Programming
Matrix
Dynamic Programming
Divide and Conquer
Matrix
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