Minimum cells traversed to reach corner where every cell represents jumps

Suppose A is at position (0, 0) of a 2-D grid containing ‘m’ rows and ‘n’ columns. His aim is to reach the bottom right point of this grid traveling through as minimum number of cells as possible.

Each cell of the grid contains a positive integer that defines the number of cells A can jump either in the right or the downward direction when he reaches that cell.

Find the minimum no of cells that need to be touched in order to reach bottom right corner.

Examples:

Input : 2 4 2
        5 3 8
        1 1 1
Output :
So following two paths exist to reach (2, 2) from (0, 0)
(0, 0) => (0, 2) => (2, 2) 
(0, 0) => (2, 0) => (2, 1) => (2, 2) 

Hence the output for this test case should be 3

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

Following is a Breadth First Search(BFS) solution of the problem:

Think of this matrix as tree and (0, 0) as root and apply BFS using level order traversal.
Push the coordinates and no of jumps in a queue.
Pop the queue after every level of tree.
Add the value at cell to the coordinates while traversing right and downward direction.
Return no of cells touched while jumping when it reaches bottom right corner.

C++

// C++ program to reach bottom right corner using 
// minimum jumps. 
#include <bits/stdc++.h> 
using namespace std; 
#define R 3 
#define C 3 
  
// function to check coordinates are in valid range. 
bool safe(int x, int y) 
{ 
    if (x < R && y < C && x >= 0 && y >= 0) 
        return true; 
    return false; 
} 
  
// function to return minimum no of cells to reach 
// bottom right cell. 
int matrixJump(int M[R][C], int R1, int C1) 
{ 
    queue<pair<int, pair<int, int> > > q; 
  
    // push the no of cells and coordinates in a queue. 
    q.push(make_pair(1, make_pair(R1, C1))); 
  
    while (!q.empty()) { 
        int x = q.front().second.first; // x coordinate 
        int y = q.front().second.second; // y coordinate 
        int no_of_cells = q.front().first; // no of cells 
  
        q.pop(); 
  
        // when it reaches bottom right return no of cells 
        if (x == R - 1 && y == C - 1)             
            return no_of_cells; 
  
        int v = M[x][y]; 
  
        if (safe(x + v, y)) 
            q.push(make_pair(no_of_cells + 1, make_pair(x + v, y))); 
  
        if (safe(x, y + v)) 
            q.push(make_pair(no_of_cells + 1, make_pair(x, y + v))); 
    } 
  
    // when destination cannot be reached 
    return -1; 
} 
  
// driver function 
int main() 
{ 
    int M[R][C] = { { 2, 4, 2 }, 
                    { 5, 3, 8 }, 
                    { 1, 1, 1 } }; 
    cout << matrixJump(M, 0, 0); 
    return 0; 
} 

Java

// Java program to reach bottom right corner 
// using minimum jumps. 
import java.util.*; 
  
class GFG 
{ 
static int R = 3, C = 3; 
  
// function to check coordinates are in valid range. 
static boolean safe(int x, int y) 
{ 
    if (x < R && y < C && x >= 0 && y >= 0) 
        return true; 
    return false; 
} 
  
// pair class 
static class pair<T, R> 
{ 
    T first; 
    R second; 
    pair(T t, R r) 
    { 
        first = t; 
        second = r; 
    } 
} 
  
// function to return minimum no of cells  
// to reach bottom right cell. 
static int matrixJump(int M[][], int R1, int C1) 
{ 
    Queue<pair<Integer,  
          pair<Integer,  
               Integer>>> q = new LinkedList<>(); 
  
    // push the no of cells and coordinates in a queue. 
    q.add(new pair(1, new pair(R1, C1))); 
  
    while (q.size() > 0) 
    { 
        int x = q.peek().second.first; // x coordinate 
        int y = q.peek().second.second; // y coordinate 
        int no_of_cells = q.peek().first; // no of cells 
  
        q.remove(); 
  
        // when it reaches bottom right return no of cells 
        if (x == R - 1 && y == C - 1)          
            return no_of_cells; 
  
        int v = M[x][y]; 
  
        if (safe(x + v, y)) 
            q.add(new pair(no_of_cells + 1,  
                  new pair(x + v, y))); 
  
        if (safe(x, y + v)) 
            q.add(new pair(no_of_cells + 1,  
                  new pair(x, y + v))); 
    } 
  
    // when destination cannot be reached 
    return -1; 
} 
  
// Driver Code 
public static void main(String ars[]) 
{ 
    int M[][] = {{ 2, 4, 2 }, 
                 { 5, 3, 8 }, 
                 { 1, 1, 1 }}; 
    System.out.print( matrixJump(M, 0, 0)); 
} 
} 
  
// This code is contributed by Arnab Kundu 

Python3

# Python3 program to reach bottom  
# right corner using minimum jumps.  
  
R, C = 3, 3
  
# Function to check coordinates are in valid range.  
def safe(x, y):  
  
    if x < R and y < C and x >= 0 and y >= 0:  
        return True
    return False
  
# Function to return minimum no of  
# cells to reach bottom right cell.  
def matrixJump(M, R1, C1):  
  
    q = []  
  
    # push the no of cells and coordinates in a queue.  
    q.append((1, (R1, C1)))  
  
    while len(q) != 0:  
        x = q[0][1][0] # x coordinate  
        y = q[0][1][1] # y coordinate  
        no_of_cells = q[0][0] # no of cells  
  
        q.pop(0)  
  
        # when it reaches bottom right return no of cells  
        if x == R - 1 and y == C - 1:          
            return no_of_cells  
  
        v = M[x][y]  
  
        if safe(x + v, y):  
            q.append((no_of_cells + 1, (x + v, y)))  
  
        if safe(x, y + v):  
            q.append((no_of_cells + 1, (x, y + v)))  
      
    # when destination cannot be reached  
    return -1
  
# Driver code  
if __name__ == "__main__":  
  
    M = [[2, 4, 2],  
        [5, 3, 8],  
        [1, 1, 1]] 
          
    print(matrixJump(M, 0, 0)) 
  
# This code is contributed by Rituraj Jain 

C#

// C# program to reach bottom right corner 
// using minimum jumps. 
using System; 
using System.Collections; 
using System.Collections.Generic; 
  
class GFG 
{ 
      
static int R = 3, C = 3; 
  
// function to check coordinates are in valid range. 
static Boolean safe(int x, int y) 
{ 
    if (x < R && y < C && x >= 0 && y >= 0) 
        return true; 
    return false; 
} 
  
// Pair class 
public class Pair<T, U>  
{ 
    public T first; 
    public U second; 
    public Pair() { 
    } 
  
    public Pair(T first, U second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
  
}; 
  
// function to return minimum no of cells  
// to reach bottom right cell. 
static int matrixJump(int [,]M, int R1, int C1) 
{ 
    Queue<Pair<int,Pair<int,int>>> q =  
    new Queue<Pair<int,Pair<int,int>>>(); 
  
    // push the no of cells and coordinates in a queue. 
    q.Enqueue(new Pair<int, Pair<int, int>>( 
                1, new Pair<int, int>(R1, C1))); 
  
    while (q.Count > 0) 
    { 
        int x = q.Peek().second.first; // x coordinate 
        int y = q.Peek().second.second; // y coordinate 
        int no_of_cells = q.Peek().first; // no of cells 
  
        q.Dequeue(); 
  
        // when it reaches bottom right return no of cells 
        if (x == R - 1 && y == C - 1)          
            return no_of_cells; 
  
        int v = M[x, y]; 
  
        if (safe(x + v, y)) 
            q.Enqueue(new Pair<int,Pair<int,int>>(no_of_cells + 1,  
                new Pair<int,int>(x + v, y))); 
  
        if (safe(x, y + v)) 
            q.Enqueue(new Pair<int,Pair<int,int>>(no_of_cells + 1,  
                new Pair<int,int>(x, y + v))); 
    } 
  
    // when destination cannot be reached 
    return -1; 
} 
  
// Driver Code 
public static void Main(String []ars) 
{ 
    int [,]M = {{ 2, 4, 2 }, 
                { 5, 3, 8 }, 
                { 1, 1, 1 }}; 
    Console.Write( matrixJump(M, 0, 0)); 
} 
} 
  
// This code is contributed by Arnab Kundu 

Output:

Time Complexity : O(n)
Auxiliary Space : O(n)

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My Personal Notes

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

C++

Java

Python3

C#

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