Skip to content
Related Articles

Related Articles

Improve Article

Count of ways to traverse a Matrix and return to origin in K steps

  • Difficulty Level : Hard
  • Last Updated : 14 Jun, 2021

Given three integers N, M and K, where N and M are the dimensions of the matrix and K is the maximum possible steps, the task is to count the number ways to start from (0, 0) and return traversing the matrix by using K steps only. 
Note: In each step, one can move up, down, left, right, or stay at the current position. The answer could be large, so print the answer modulo 109 + 7
Examples: 
 

Input: N = 2, M = 2, K = 2 
Output: 3 
Explanation: 
Three ways are: 
1)Stay, Stay. 
2)Up, Down 
3)Right, Left
Input: N = 3, M = 3, K = 4 
Output: 23 
 

Approach: The problem can also be solved using Memoization technique. Follow the steps below to solve the problem: 
 

  • Starting from position (0, 0), recursively call every possible position and decrease the value of steps by 1. 
     
  • Create a 3D array of size N * M * K ( DP[N][M][S] ) 
     
Possible ways for each step
can be calculated by:

 DP[i][j][k]
  =  Recursion(i+1, j, k-1)  +
     Recursion(i-1, j, k-1)  +  
     Recursion(i, j-1, k-1)  +  
     Recursion(i, j+1, k-1)  +  
     Recursion(i, j, k-1).

With base condition returning 0,
whenever
i >= l or i = b or j < 0 or k < 0.

Below is the implementation of the above approach: 
 

C++




// C++ program to count total
// number of ways to return
// to origin after completing
// given number of steps.
 
#include <bits/stdc++.h>
using namespace std;
#define MOD 1000000007
 
long long dp[101][101][101];
int N, M, K;
 
// Function Initialize dp[][][]
// array with -1
void Initialize()
{
 
    for (int i = 0; i <= 100; i++)
        for (int j = 0; j <= 100; j++)
            for (int z = 0; z <= 100; z++)
                dp[i][j][z] = -1;
}
 
// Function returns the total count
int CountWays(int i, int j, int k)
{
    if (i >= N || i < 0
        || j >= M || j < 0 || k < 0)
        return 0;
 
    if (i == 0 && j == 0
        && k == 0)
        return 1;
 
    if (dp[i][j][k] != -1)
        return dp[i][j][k];
    else
        dp[i][j][k]
            = (CountWays(i + 1, j, k - 1) % MOD
               + CountWays(i - 1, j, k - 1) % MOD
               + CountWays(i, j - 1, k - 1) % MOD
               + CountWays(i, j + 1, k - 1) % MOD
               + CountWays(i, j, k - 1) % MOD)
              % MOD;
 
    return dp[i][j][k];
}
 
// Driver Program
int main()
{
    N = 3;
    M = 3;
    K = 4;
 
    Initialize();
    cout << CountWays(0, 0, K)
         << "\n";
 
    return 0;
}
Java



// Java program to count total
// number of ways to return
// to origin after completing
// given number of steps.
class GFG{
     
static int [][][] dp = new int[101][101][101];
static int N, M, K;
static int MOD = 1000000007;
     
// Function Initialize dp[][][]
// array with -1
public static void Initialize()
{
    for(int i = 0; i <= 100; i++)
       for(int j = 0; j <= 100; j++)
          for(int z = 0; z <= 100; z++)
             dp[i][j][z] = -1;
}
     
// Function returns the total count
public static int CountWays(int i, int j, int k)
{
    if (i >= N || i < 0 ||
        j >= M || j < 0 || k < 0)
        return 0;
     
    if (i == 0 && j == 0 && k == 0)
        return 1;
     
    if (dp[i][j][k] != -1)
        return dp[i][j][k];
     
    else
        dp[i][j][k] = (CountWays(i + 1, j, k - 1) % MOD +
                       CountWays(i - 1, j, k - 1) % MOD +
                       CountWays(i, j - 1, k - 1) % MOD +
                       CountWays(i, j + 1, k - 1) % MOD +
                       CountWays(i, j, k - 1) % MOD) % MOD;
     
    return dp[i][j][k];
}
 
// Driver code
public static void main(String[] args)
{
    N = 3;
    M = 3;
    K = 4;
     
    Initialize();
    System.out.println(CountWays(0, 0, K));
}
}
 
// This code is contributed by grand_master
Python3



# Python3 program to count total
# number of ways to return
# to origin after completing
# given number of steps.
MOD = 1000000007
 
dp = [[[0 for i in range(101)]
          for i in range(101)]
          for i in range(101)]
N, M, K = 0, 0, 0
 
# Function Initialize dp[][][]
# array with -1
def Initialize():
 
    for i in range(101):
        for j in range(101):
            for z in range(101):
                dp[i][j][z] = -1
 
# Function returns the total count
def CountWays(i, j, k):
 
    if (i >= N or i < 0 or
        j >= M or j < 0 or k < 0):
        return 0
 
    if (i == 0 and j == 0 and k == 0):
        return 1
 
    if (dp[i][j][k] != -1):
        return dp[i][j][k]
    else:
        dp[i][j][k] = (CountWays(i + 1, j, k - 1) % MOD +
                       CountWays(i - 1, j, k - 1) % MOD +
                       CountWays(i, j - 1, k - 1) % MOD +
                       CountWays(i, j + 1, k - 1) % MOD +
                       CountWays(i, j, k - 1) % MOD) % MOD
    return dp[i][j][k]
 
# Driver code
if __name__ == '__main__':
     
    N = 3
    M = 3
    K = 4
 
    Initialize()
    print(CountWays(0, 0, K))
 
# This code is contributed by mohit kumar 29
C#



// C# program to count total
// number of ways to return
// to origin after completing
// given number of steps.
using System;
 
class GFG{
     
static int [,,] dp = new int[101, 101, 101];
static int N, M, K;
static int MOD = 1000000007;
     
// Function Initialize [,]dp[]
// array with -1
public static void Initialize()
{
    for(int i = 0; i <= 100; i++)
        for(int j = 0; j <= 100; j++)
            for(int z = 0; z <= 100; z++)
                dp[i, j, z] = -1;
}
     
// Function returns the total count
public static int CountWays(int i, int j, int k)
{
    if (i >= N || i < 0 ||
        j >= M || j < 0 || k < 0)
        return 0;
     
    if (i == 0 && j == 0 && k == 0)
        return 1;
     
    if (dp[i, j, k] != -1)
        return dp[i, j, k];
     
    else
        dp[i, j, k] = (CountWays(i + 1, j, k - 1) % MOD +
                       CountWays(i - 1, j, k - 1) % MOD +
                       CountWays(i, j - 1, k - 1) % MOD +
                       CountWays(i, j + 1, k - 1) % MOD +
                       CountWays(i, j, k - 1) % MOD) % MOD;
     
    return dp[i, j, k];
}
 
// Driver code
public static void Main(String[] args)
{
    N = 3;
    M = 3;
    K = 4;
     
    Initialize();
    Console.WriteLine(CountWays(0, 0, K));
}
}
 
// This code is contributed by Amit Katiyar
Javascript



<script>
 
// JavaScript program to count total
// number of ways to return
// to origin after completing
// given number of steps.
 
var MOD =  1000000007;
 
var dp = Array.from(Array(101), ()=>Array(101));
for(var i =0; i<101; i++)
        for(var j =0; j<101; j++)
            dp[i][j] = new Array(101).fill(-1);
var N, M, K;
 
// Function returns the total count
function CountWays(i, j, k)
{
    if (i >= N || i < 0
        || j >= M || j < 0 || k < 0)
        return 0;
 
    if (i == 0 && j == 0
        && k == 0)
        return 1;
 
    if (dp[i][j][k] != -1)
        return dp[i][j][k];
    else
        dp[i][j][k]
            = (CountWays(i + 1, j, k - 1) % MOD
               + CountWays(i - 1, j, k - 1) % MOD
               + CountWays(i, j - 1, k - 1) % MOD
               + CountWays(i, j + 1, k - 1) % MOD
               + CountWays(i, j, k - 1) % MOD)
              % MOD;
 
    return dp[i][j][k];
}
 
// Driver Program
N = 3;
M = 3;
K = 4;
document.write( CountWays(0, 0, K))
 
 
</script>
Output: 
23
 
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.
In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.



My Personal Notes
arrow_drop_up
Next

Sort an array by swapping elements of different type specified by another array
Recommended Articles
Page :
Count the number of ways to traverse a Matrix
09, Oct 18
Count of ways to traverse a Matrix according to given conditions
21, Mar 20
Number of ways to reach (X, Y) in a matrix starting from the origin
18, Nov 19
Number of ways to reach (M, N) in a matrix starting from the origin without visiting (X, Y)
25, Jun 20
Traverse matrix in L shape
20, Oct 18
Traverse a given Matrix using Recursion
03, Apr 20
Traverse the matrix in Diagonally Bottum-Up fashion using Recursion
07, May 20
Min number of moves to traverse entire Matrix through connected cells with equal values
29, Jun 20
Count ways to reach the Nth stair using multiple 1 or 2 steps and a single step 3
22, Mar 20
Count ways to reach Nth Stairs by taking 1 and 2 steps with exactly one 3 step
18, Feb 20
Count of ways to travel a cyclic path in N steps in a Triangular Pyramid
21, May 20
Minimum number of steps to convert a given matrix into Upper Hessenberg matrix
27, Jun 19
Minimum steps required to convert the matrix into lower hessenberg matrix
27, Jun 19
Minimum number of steps to convert a given matrix into Diagonally Dominant Matrix
04, Jul 19
Maximum points from top left of matrix to bottom right and return back
01, Oct 17
Return previous element in an expanding matrix
18, May 16
Return an array of anti-diagonals of given N*N square matrix
26, Apr 20
Counts paths from a point to reach Origin
13, Jun 17
Counts paths from a point to reach Origin : Memory Optimized
08, Apr 20
Count of ways to generate a Matrix with product of each row and column as 1 or -1
08, Jul 20
Generate matrix from given Sparse Matrix using Linked List and reconstruct the Sparse Matrix
19, Apr 21
Generate a Matrix such that given Matrix elements are equal to Bitwise OR of all corresponding row and column elements of generated Matrix
15, Jan 21
Queries to add, remove and return the difference of maximum and minimum.
18, Oct 18
Minimum steps to reach any of the boundary edges of a matrix | Set-2
23, Dec 18
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
Saif Haider
@Saif Haider
Vote for difficulty
Current difficulty :
Hard





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox