Find minimum steps required to reach the end of a matrix | Set 2

Given a 2d-matrix consisting of positive integers, the task is to find the minimum number of steps required to reach the end of the matrix. If we are at cell (i, j) then we can go to all the cells represented by (i + X, j + Y) such that X ≥ 0, Y ≥ 0 and X + Y = arr[i][j]. If no path exists then print -1.

Examples:

Input: arr[][] = {
{4, 1, 1},
{1, 1, 1},
{1, 1, 1}}
Output: 1
The path will be from {0, 0} -> {2, 2} as manhattan distance
between two is 4.
Thus, we are reaching there in 1 step.



Input: arr[][] = {
{1, 1, 2},
{1, 1, 1},
{2, 1, 1}}
Output: 3

A simple solution will be to explore all possible solutions which will take exponential time.

An efficient solution is to use dynamic programming to solve this problem in polynomial time. Lets decide the states of 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 have arr[i][j] + 1 possible paths.

The recurrence relation will be

dp[i][j] = 1 + min(dp[i][j + arr[i][j]], dp[i + 1][j + arr[i][j] – 1], …., dp[i + arr[i][j]][j])

To reduce the number of terms in recurrence relation, we can put an upper bound on the values of X and Y. How?
We know that i + X < N. Thus, X < N – i otherwise they would go out of bounds.
Similarly, Y < N – j

0 ≤ Y < N – j …(1)
X + Y = arr[i][j] …(2)
Substituting value of Y from second into first, we get
X ≥ arr[i][j] + j – N + 1

From above we get another lower bound on constraint of X i.e. X ≥ arr[i][j] + j – N + 1.
So, new lower bound on X becomes X ≥ max(0, arr[i][j] + j – N + 1).
Also X ≤ min(arr[i][j], N – i – 1).

Our recurrence relation optimises to

dp[i][j] = 1 + min(dp[i + max(0, arr[i][j] + j – N + 1)][j + arr[i][j] – max(0, arr[i][j] + j – N + 1)], …., dp[i + min(arr[i][j], N – i – 1)][j + arr[i][j] – min(arr[i][j], N – i – 1)])

Below is the implementation of the above approach:

C++

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// C++ implementation of the 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 return the minimum steps required
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;
    dp[i][j] = 9999999;
  
    // Recurrence relation
    for (int k = max(0, arr[i][j] + j - n + 1);
         k <= min(n - i - 1, arr[i][j]); k++) {
        dp[i][j] = min(dp[i][j], minSteps(i + k, j + arr[i][j] - k, arr));
    }
  
    dp[i][j]++;
  
    return dp[i][j];
}
  
// Driver code
int main()
{
    int arr[n][n] = { { 4, 1, 2 },
                      { 1, 1, 1 },
                      { 2, 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 implementation of the 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 return the minimum steps required
    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;
        dp[i][j] = 9999999;
  
        // Recurrence relation
        for (int k = Math.max(0, arr[i][j] + j - n + 1);
             k <= Math.min(n - i - 1, arr[i][j]); k++) {
            dp[i][j] = Math.min(dp[i][j],
                                minSteps(i + k, j + arr[i][j] - k, arr));
        }
  
        dp[i][j]++;
  
        return dp[i][j];
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int arr[][] = { { 4, 1, 2 },
                        { 1, 1, 1 },
                        { 2, 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 implementation of the 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 return the minimum steps required 
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
    dp[i][j] = 9999999
  
    # Recurrence relation 
    for k in range(max(0, arr[i][j] + j - n + 1),min(n - i - 1, arr[i][j]) + 1) :
        dp[i][j] = min(dp[i][j], minSteps(i + k, j + arr[i][j] - k, arr)); 
      
  
    dp[i][j] += 1
  
    return dp[i][j]; 
  
  
# Driver code 
if __name__ == "__main__"
  
    arr =
            [ 4, 1, 2 ], 
            [ 1, 1, 1 ], 
            [ 2, 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# implementation of the 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 return the minimum steps required
    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;
        dp[i, j] = 9999999;
  
        // Recurrence relation
        for (int k = Math.Max(0, arr[i,j] + j - n + 1);
            k <= Math.Min(n - i - 1, arr[i,j]); k++)
        {
            dp[i,j] = Math.Min(dp[i,j],
                                minSteps(i + k, j + arr[i,j] - k, arr));
        }
  
        dp[i,j]++;
  
        return dp[i,j];
    }
  
    // Driver code
    static public void Main ()
    {
        int [,]arr = { { 4, 1, 2 },
                        { 1, 1, 1 },
                        { 2, 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:

1

The time complexity of the above approach will be O(n3). Each state takes O(n) time in worst case to solve.



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Improved By : Rajput-Ji, jit_t, AnkitRai01