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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:
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

 




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