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

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++

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// C++ program to impplement above approach
  
#include <bits/stdc++.h>
#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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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.

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Output:

2

Time Complexity: O(N2).



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Improved By : Rajput-Ji, jit_t, AnkitRai01