Find minimum steps required to reach the end of a matrix

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

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

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++

// 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; 
} 

Java

// 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 

Python3

# 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 

C#

// 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. 

Output:

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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My Personal Notes

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

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

C++

Java

Python3

C#

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