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, LeftInput: 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; } |
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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 |
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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 |
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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 |
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Output:
23
