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

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

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

• Create a 3D DP to store the solution of the subproblems and initialize it with 0.
• Initialize the DP  with base cases dp[0][0][0] = 1.
• Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
• Return the final solution stored in dp[0][0][k].

Implementation :

C++

// C++ program for above approach

#include <bits/stdc++.h>
using namespace std;
#define MOD 1000000007


// Function returns the total count
int CountWays(int n, int m, int k)
{
    // Initialize dp to store
    // computations of subproblems
      int dp[k+1][n+1][m+1];
    memset(dp, 0, sizeof(dp));

  // Base Case
    dp[0][0][0] = 1;

     // iterate over subproblems and get the current
    // solution for previous computations
    for (int l = 1; l <= k; l++) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {

                  // update DP current value
                dp[i][j][l] = ((i > 0 ? dp[i - 1][j][l - 1] : 0)
                    + (i < n - 1 ? dp[i + 1][j][l - 1] : 0)
                    + (j > 0 ? dp[i][j - 1][l - 1] : 0)
                    + (j < m - 1 ? dp[i][j + 1][l - 1] : 0)
                    + dp[i][j][l - 1]) % MOD;
            }
        }
    }

      // return final answer
    return dp[0][0][k];
}

// Driver Program
int main()
{
    int N = 3;
    int M = 3;
    int K = 4;

      // function call
    cout << CountWays(N, M, K) << "\n";

    return 0;
}

// This code is contributed by bhardwajji.

Python3

# Python 3 program for above approach

MOD = 1000000007

# Function returns the total count
def CountWays(n, m, k):
    # Initialize dp to store
    # computations of subproblems
    dp = [[[0 for i in range(m+1)] for j in range(n+1)] for k in range(k+1)]

    # Base Case
    dp[0][0][0] = 1

    # iterate over subproblems and get the current
    # solution for previous computations
    for l in range(1, k+1):
        for i in range(n):
            for j in range(m):
                # update DP current value
                dp[l][i][j] = ((dp[l-1][i-1][j] if i > 0 else 0)
                               + (dp[l-1][i+1][j] if i < n-1 else 0)
                               + (dp[l-1][i][j-1] if j > 0 else 0)
                               + (dp[l-1][i][j+1] if j < m-1 else 0)
                               + dp[l-1][i][j]) % MOD

    # return final answer
    return dp[k][0][0]

# Driver Program
if __name__ == "__main__":
    N = 3
    M = 3
    K = 4

    # function call
    print(CountWays(N, M, K))

C#

using System;

class GFG {
  const int MOD = 1000000007;

  // Function returns the total count
  static int CountWays(int n, int m, int k)
  {
    // Initialize dp to store
    // computations of subproblems
    int[][][] dp = new int[k + 1][][];
    for (int i = 0; i <= k; i++) {
      dp[i] = new int[n + 1][];
      for (int j = 0; j <= n; j++) {
        dp[i][j] = new int[m + 1];
      }
    }

    // Base Case
    dp[0][0][0] = 1;

    // iterate over subproblems and get the current
    // solution for previous computations
    for (int l = 1; l <= k; l++) {
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
          // update DP current value
          dp[l][i][j]
            = ((i > 0 ? dp[l - 1][i - 1][j] : 0)
               + (i < n - 1
                  ? dp[l - 1][i + 1][j]
                  : 0)
               + (j > 0 ? dp[l - 1][i][j - 1]
                  : 0)
               + (j < m - 1
                  ? dp[l - 1][i][j + 1]
                  : 0)
               + dp[l - 1][i][j])
            % MOD;
        }
      }
    }

    // return final answer
    return dp[k][0][0];
  }

  // Driver Program
  static void Main(string[] args)
  {
    int N = 3;
    int M = 3;
    int K = 4;

    // function call
    Console.WriteLine(CountWays(N, M, K));
  }
}

Javascript

const MOD = 1000000007;

function CountWays(n, m, k) {
    const dp = new Array(k + 1)
        .fill(null)
        .map(() =>
            new Array(n + 1)
                .fill(null)
                .map(() =>
                    new Array(m + 1).fill(0)
                )
        );

    // Base Case
    dp[0][0][0] = 1;

    // Iterate over subproblems and get the current solution for previous computations
    for (let l = 1; l <= k; l++) {
        for (let i = 0; i < n; i++) {
            for (let j = 0; j < m; j++) {
                // Update DP current value
                dp[l][i][j] = (
                    (i > 0 ? dp[l - 1][i - 1][j] : 0) +
                    (i < n - 1 ? dp[l - 1][i + 1][j] : 0) +
                    (j > 0 ? dp[l - 1][i][j - 1] : 0) +
                    (j < m - 1 ? dp[l - 1][i][j + 1] : 0) +
                    dp[l - 1][i][j]
                ) % MOD;
            }
        }
    }

    // Return final answer
    return dp[k][0][0];
}

// Driver Program
const N = 3;
const M = 3;
const K = 4;

// Function call
console.log(CountWays(N, M, K));

Output

23

Time Complexity : O(N*K*M)
Auxiliary Space : O(N*K*M)