Given three integers M, N and K, the task is to count all the possible paths from the cell (0, 0, 0) to cell (M-1, N-1, K-1) in a matrix of size (M, N, K). Movement is allowed only in three directions, which are along the positive direction of the three axes i.e. from any cell (i, j, k) movement is allowed to cells (i+1, j, k), (i, j+1, k) and (i, j, k+1).
Examples:
Input: M = 1, N = 1, K = 1
Output: 1
Explanation: As the source and destination both are same, there is only one possible way to stay there.Input: M = 2, N = 2, K = 2
Output: 6
Explanation: The possible paths are :
- (0, 0, 0) -> (0, 1, 0) -> (0, 1, 1) -> (1, 1, 1)
- (0, 0, 0) -> (0, 1, 0) -> (1, 1, 0) -> (1, 1, 1)
- (0, 0, 0) -> (1, 0, 0) -> (1, 1, 0) -> (1, 1, 1)
- (0, 0, 0) -> (1, 0, 0) -> (1, 0, 1) -> (1, 1, 1)
- (0, 0, 0) -> (0, 0, 1) -> (0, 1, 1) -> (1, 1, 1)
- (0, 0, 0) -> (0, 0, 1) -> (1, 0, 1) -> (1, 1, 1)
Approach: This problem can be solved by using dynamic programming. Follow the below steps, to solve this problem:
- Create a 3D array of size M*N*K, say dp and initialise dp[0][0][0] to 1.
- Use three nested loop where the outermost iterates from i = 0 to i < M, the innermost iterates from k = 0 to k < K and the middle one from j = 0 to j < N.
- Here, dp[i][j][k] indicates the number of paths from cell (0, 0, 0) to the cell (i, j, k).
- So, any cell(i, j, k), can be reached from only 3 other cells which are (i-1, j, k), (i, j-1, k) and (i, j, k-1).
- Therefore, the relation among the cells here is:
dp[i][j][k] = dp[i – 1][j][k] + dp[i][j – 1][k] + dp[i][j][k – 1]
- Fill the whole dp matrix using the above relation and then return dp[M-1][N-1][K-1] as the answer.
Below is the implementation of the above approach:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;
// Function to find to the number of ways// from the cell(0, 0, 0)// to cell(M-1, N-1, K-1)int numberOfWays(int M, int N, int K)
{ vector<vector<vector<int> > > dp(
M, vector<vector<int> >(
N, vector<int>(K)));
// Initialising dp
dp[0][0][0] = 1;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
for (int k = 0; k < K; ++k) {
if (i == 0 and j == 0
and k == 0) {
continue;
}
dp[i][j][k] = 0;
if (i - 1 >= 0) {
dp[i][j][k] += dp[i - 1][j][k];
}
if (j - 1 >= 0) {
dp[i][j][k] += dp[i][j - 1][k];
}
if (k - 1 >= 0) {
dp[i][j][k] += dp[i][j][k - 1];
}
}
}
}
return dp[M - 1][N - 1][K - 1];
}// Driver Codeint main()
{ int M = 2, N = 2, K = 2;
cout << numberOfWays(M, N, K);
return 0;
} |
Java
/*package whatever //do not write package name here */import java.io.*;
class GFG {
// Function to find to the number of ways
// from the cell(0, 0, 0)
// to cell(M-1, N-1, K-1)
static int numberOfWays(int M, int N, int K)
{
int [][][] dp = new int[M][N][K];
// Initialising dp
dp[0][0][0] = 1;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
for (int k = 0; k < K; ++k) {
if (i == 0 && j == 0
&& k == 0) {
continue;
}
dp[i][j][k] = 0;
if (i - 1 >= 0) {
dp[i][j][k] += dp[i - 1][j][k];
}
if (j - 1 >= 0) {
dp[i][j][k] += dp[i][j - 1][k];
}
if (k - 1 >= 0) {
dp[i][j][k] += dp[i][j][k - 1];
}
}
}
}
return dp[M - 1][N - 1][K - 1];
}
// Driver Code
public static void main (String[] args)
{
int M = 2, N = 2, K = 2;
System.out.println(numberOfWays(M, N, K));
}
}// This code is contributed by Potta Lokesh |
Python3
# Python code for the above approach# Function to find to the number of ways# from the cell(0, 0, 0)# to cell(M-1, N-1, K-1)def numberOfWays(M, N, K):
dp = [[[0 for i in range(K)] for j in range(N)] for k in range(M)]
# Initialising dp
dp[0][0][0] = 1;
for i in range(M):
for j in range(N):
for k in range(K):
if (i == 0 and j == 0 and k == 0):
continue;
dp[i][j][k] = 0;
if (i - 1 >= 0):
dp[i][j][k] += dp[i - 1][j][k];
if (j - 1 >= 0):
dp[i][j][k] += dp[i][j - 1][k];
if (k - 1 >= 0):
dp[i][j][k] += dp[i][j][k - 1];
return dp[M - 1][N - 1][K - 1];
# Driver CodeM = 2
N = 2
K = 2;
print(numberOfWays(M, N, K));
# This code is contributed by gfgking |
C#
/*package whatever //do not write package name here */using System;
class GFG {
// Function to find to the number of ways
// from the cell(0, 0, 0)
// to cell(M-1, N-1, K-1)
static int numberOfWays(int M, int N, int K)
{
int[, , ] dp = new int[M, N, K];
// Initialising dp
dp[0, 0, 0] = 1;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
for (int k = 0; k < K; ++k) {
if (i == 0 && j == 0 && k == 0) {
continue;
}
dp[i, j, k] = 0;
if (i - 1 >= 0) {
dp[i, j, k] += dp[i - 1, j, k];
}
if (j - 1 >= 0) {
dp[i, j, k] += dp[i, j - 1, k];
}
if (k - 1 >= 0) {
dp[i, j, k] += dp[i, j, k - 1];
}
}
}
}
return dp[M - 1, N - 1, K - 1];
}
// Driver Code
public static void Main(string[] args)
{
int M = 2, N = 2, K = 2;
Console.WriteLine(numberOfWays(M, N, K));
}
}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript code for the above approach
// Function to find to the number of ways
// from the cell(0, 0, 0)
// to cell(M-1, N-1, K-1)
const numberOfWays = (M, N, K) => {
let dp = new Array(M).fill(0).map(() => new Array(N).fill(0).map(() => new Array(K).fill(0)));
// Initialising dp
dp[0][0][0] = 1;
for (let i = 0; i < M; ++i) {
for (let j = 0; j < N; ++j) {
for (let k = 0; k < K; ++k) {
if (i == 0 && j == 0
&& k == 0) {
continue;
}
dp[i][j][k] = 0;
if (i - 1 >= 0) {
dp[i][j][k] += dp[i - 1][j][k];
}
if (j - 1 >= 0) {
dp[i][j][k] += dp[i][j - 1][k];
}
if (k - 1 >= 0) {
dp[i][j][k] += dp[i][j][k - 1];
}
}
}
}
return dp[M - 1][N - 1][K - 1];
}
// Driver Code
let M = 2, N = 2, K = 2;
document.write(numberOfWays(M, N, K));
// This code is contributed by rakeshsahni</script> |
Output
6
Time Complexity: O(M*N*K)
Space Complexity: O(M*N*K)
Efficient approach : Optimize space complexity
In previous approach we can see that the dp[i][j][k] is depend upon dp[i – 1][j][k], dp[i][j – 1][k] and dp[i][j][k – 1] so we can convert 3d matric into current and previous 2d matrix where i-1 refer to previous and i refer to current.
Implantation steps :
- Create two 2d matrix refer to previous and current matrix.
- Initialize both matrix starting point with 1.
- Now the approach is same as the previous code but we have to only convert dp[i] to current and dp[i-1] to previous.
- At last return the answer
Implementation :
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;
// Function to find to the number of ways// from the cell(0, 0, 0)// to cell(M-1, N-1, K-1)int numberOfWays(int M, int N, int K)
{ vector<vector<int> > curr(N, vector<int>(K, 0));
vector<vector<int> > prev(N, vector<int>(K, 0));
// Initialising dp
prev[0][0] = 1;
curr[0][0] = 1;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
for (int k = 0; k < K; ++k) {
if (i == 0 and j == 0 and k == 0) {
continue;
}
curr[j][k] = 0;
if (i - 1 >= 0) {
curr[j][k] += prev[j][k];
}
if (j - 1 >= 0) {
curr[j][k] += curr[j - 1][k];
}
if (k - 1 >= 0) {
curr[j][k] += curr[j][k - 1];
}
}
}
prev = curr;
}
return curr[N - 1][K - 1];
}// Driver Codeint main()
{ int M = 2, N = 2, K = 2;
cout << numberOfWays(M, N, K);
return 0;
}// this code is contributed by bhardwajji |
Java
public class Main {
// Function to find to the number of ways
// from the cell(0, 0, 0)
// to cell(M-1, N-1, K-1)
public static int numberOfWays(int M, int N, int K)
{
int[][] curr = new int[N][K];
int[][] prev = new int[N][K];
prev[0][0] = 1;
curr[0][0] = 1;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
for (int k = 0; k < K; ++k) {
if (i == 0 && j == 0 && k == 0) {
continue;
}
curr[j][k] = 0;
if (i - 1 >= 0) {
curr[j][k] += prev[j][k];
}
if (j - 1 >= 0) {
curr[j][k] += curr[j - 1][k];
}
if (k - 1 >= 0) {
curr[j][k] += curr[j][k - 1];
}
}
}
for (int j = 0; j < N; j++) {
prev[j] = curr[j].clone();
}
}
return curr[N - 1][K - 1];
}
public static void main(String[] args)
{
int M = 2, N = 2, K = 2;
System.out.println(numberOfWays(M, N, K));
}
} |
Python3
# Function to find the number of ways# from the cell(0, 0, 0)# to cell(M-1, N-1, K-1)def numberOfWays(M, N, K):
curr = [[0 for _ in range(K)] for _ in range(N)]
prev = [[0 for _ in range(K)] for _ in range(N)]
# Initializing dp
prev[0][0] = 1
curr[0][0] = 1
for i in range(M):
for j in range(N):
for k in range(K):
if i == 0 and j == 0 and k == 0:
continue
curr[j][k] = 0
if i - 1 >= 0:
curr[j][k] += prev[j][k]
if j - 1 >= 0:
curr[j][k] += curr[j - 1][k]
if k - 1 >= 0:
curr[j][k] += curr[j][k - 1]
prev[j] = curr[j].copy() # Create a new list for prev[j]
return curr[N - 1][K - 1]
# Driver codeM, N, K = 2, 2, 2
print(numberOfWays(M, N, K))
|
Javascript
// Javascript code for the above approach// Function to find to the number of ways// from the cell(0, 0, 0)// to cell(M-1, N-1, K-1)function numberOfWays(M, N, K) {
let curr = new Array(N);
let prev = new Array(N);
for (let i = 0; i < N; i++) {
curr[i] = new Array(K).fill(0);
prev[i] = new Array(K).fill(0);
}// Initialising dpprev[0][0] = 1;curr[0][0] = 1;for (let i = 0; i < M; i++) {
for (let j = 0; j < N; j++) {
for (let k = 0; k < K; k++) {
if (i === 0 && j === 0 && k === 0) {
continue;
}curr[j][k] = 0;if (i - 1 >= 0) {
curr[j][k] += prev[j][k];}if (j - 1 >= 0) {
curr[j][k] += curr[j - 1][k];}if (k - 1 >= 0) {
curr[j][k] += curr[j][k - 1];}}}for (let j = 0; j < N; j++) {
prev[j] = [...curr[j]];}}return curr[N - 1][K - 1];
}const M = 2;const N = 2;const K = 2;console.log(numberOfWays(M, N, K)); |
C#
// C# code for the above approachusing System;
public class MainClass {
// Function to find to the number of ways
// from the cell(0, 0, 0)
// to cell(M-1, N-1, K-1)
public static int NumberOfWays(int M, int N, int K)
{
int[, ] curr = new int[N, K];
int[, ] prev = new int[N, K];
// Initialising dp
prev[0, 0] = 1;
curr[0, 0] = 1;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
for (int k = 0; k < K; ++k) {
if (i == 0 && j == 0 && k == 0) {
continue;
}
curr[j, k] = 0;
if (i - 1 >= 0) {
curr[j, k] += prev[j, k];
}
if (j - 1 >= 0) {
curr[j, k] += curr[j - 1, k];
}
if (k - 1 >= 0) {
curr[j, k] += curr[j, k - 1];
}
}
}
prev = curr.Clone() as int[, ];
}
return curr[N - 1, K - 1];
}
public static void Main(string[] args)
{
int M = 2, N = 2, K = 2;
Console.WriteLine(NumberOfWays(M, N, K));
}
} |
Output :
6
Time Complexity: O(M*N*K)
Auxiliary Space: O(N*K)
Article Tags :
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