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

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 are the implementation of the above idea:

## C++

 `// C++ program to impplement above approach ` ` `  `#include ` `#define n 3 ` `using` `namespace` `std; ` ` `  `// 2d array to store ` `// states of dp ` `int` `dp[n][n]; ` ` `  `// array to determine whether ` `// a state has been solved before ` `int` `v[n][n]; ` ` `  `// Function to find the minimum number of ` `// steps to reach the end of matrix ` `int` `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 Code ` `int` `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

 `// Java program to impplement above approach ` `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 ` `    ``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

 `# Python3 program to implement above approach  ` `import` `numpy as np; ` ` `  `n ``=` `3` ` `  `# 2d array to store  ` `# states of dp  ` `dp ``=` `np.zeros((n, n));  ` ` `  `# array to determine whether  ` `# a state has been solved before  ` `v ``=` `np.zeros((n, n));  ` ` `  `# Function to find the minimum number of  ` `# steps to reach the end of matrix  ` `def` `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 Code  ` `arr ``=` `[ [ ``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#

 `// C# program to impplement above approach ` `using` `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. `

Output:

```2
```

Time Complexity: O(N2).

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : Rajput-Ji, jit_t, AnkitRai01