Given a two-dimensional matrix, in how way can someone traverse it from top-left to bottom-right?
Condition- At any particular cell the possible moves are either down or right, no other steps possible.
Stop when the end is reached.
Examples:
Input : 3 3 Output : 6 Input : 5 5 Output : 70
If we look closely, we will find that the number of ways a cell can be reached is = Number of ways it can reach the cell above it + number of ways it can reach the cell which is left of it.
So, start filling the 2D array according to it and return the last cell after completely filling the array.
Implementation:
C++
// C++ program using recursive solution to count// number of ways to reach mat[m-1][n-1] from// mat[0][0] in a matrix mat[][]#include <bits/stdc++.h>using namespace std;// Returns The number of way from top-left // to mat[m-1][n-1]int countPaths(int m, int n){ // Return 1 if it is the first row or // first column if (m == 1 || n == 1) return 1; // Recursively find the no of way to // reach the last cell. return countPaths(m - 1, n) + countPaths(m, n - 1);}// Driver codeint main(){ int n = 5; int m = 5; cout << countPaths(n, m); return 0;} |
Java
// Java program using recursive // solution to count number of // ways to reach mat[m-1][n-1] from// mat[0][0] in a matrix mat[][]import java.lang.*;import java.util.*;class GFG{// Returns The number of way // from top-left to mat[m-1][n-1]public int countPaths(int m, int n){ // Return 1 if it is the first // row or first column if (m == 1 || n == 1) return 1; // Recursively find the no of // way to reach the last cell. return countPaths(m - 1, n) + countPaths(m, n - 1);}// Driver Codepublic static void main(String args[]){ GFG g = new GFG(); int n = 5, m = 5; System.out.println(g.countPaths(n, m));}}// This code is contributed // by Akanksha Rai(Abby_akku) |
Python3
# Python3 program using recursive solution to count # number of ways to reach mat[m-1][n-1] from # mat[0][0] in a matrix mat[][]# Returns The number of way from top-left # to mat[m-1][n-1] def countPaths(m, n) : # Return 1 if it is the first row or # first column if m == 1 or n == 1 : return 1 # Recursively find the no of way to # reach the last cell. return (countPaths(m - 1, n) + countPaths(m, n - 1))# Driver code if __name__ == "__main__" : n = 5 m = 5 print(countPaths(n, m))# This code is contributed by ANKITRAI1 |
C#
// C# program using recursive // solution to count number of // ways to reach mat[m-1][n-1] from// mat[0][0] in a matrix mat[][]using System;class GFG{// Returns The number of way // from top-left to mat[m-1][n-1]public int countPaths(int m, int n){ // Return 1 if it is the first // row or first column if (m == 1 || n == 1) return 1; // Recursively find the no of // way to reach the last cell. return countPaths(m - 1, n) + countPaths(m, n - 1);}// Driver Codepublic static void Main(){ GFG g = new GFG(); int n = 5, m = 5; Console.WriteLine(g.countPaths(n, m)); Console.Read();}}// This code is contributed // by SoumikMondal |
PHP
<?php// PHP program using recursive // solution to count number of // ways to reach mat[m-1][n-1] from // mat[0][0] in a matrix mat[][] // Returns The number of way // from top-left to mat[m-1][n-1] function countPaths($m, $n) { // Return 1 if it is the // first row or first column if ($m == 1 || $n == 1) return 1; // Recursively find the no of // way to reach the last cell. return countPaths($m - 1, $n) + countPaths($m, $n - 1); } // Driver code $n = 5; $m = 5; echo countPaths($n, $m); // This code is contributed by jit_t?> |
Javascript
<script> // Javascript program using recursive // solution to count number of // ways to reach mat[m-1][n-1] from // mat[0][0] in a matrix mat[][] // Returns The number of way // from top-left to mat[m-1][n-1] function countPaths(m, n) { // Return 1 if it is the first // row or first column if (m == 1 || n == 1) return 1; // Recursively find the no of // way to reach the last cell. return countPaths(m - 1, n) + countPaths(m, n - 1); } let n = 5, m = 5; document.write(countPaths(n, m));// This code is contributed by suresh07.</script> |
Output
70
The above solution has exponential time complexity. It can be optimized using Dynamic Programming as there are overlapping subproblems (highlighted below in partial recursion tree for m=3, n=3)
Space Complexity: 2(m+n)
CP(3, 3)
/ \
CP(2, 3) CP(3, 2)
/ \ / \
CP(1,3) CP(2,2) CP(2,2) CP(3,1)
Implementation:
C++
// A simple recursive solution to count// number of ways to reach mat[m-1][n-1] from// mat[0][0] in a matrix mat[][]#include <bits/stdc++.h>using namespace std;// Returns The number of way from top-left // to mat[m-1][n-1]int countPaths(int m, int n){ int dp[m+1][n+1]; for (int i=1; i<=m; i++) { for (int j=1; j<=n; j++) { if (i==1 || j == 1) dp[i][j] = 1; else dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[m][n];}// Driver codeint main(){ int n = 5; int m = 5; cout << countPaths(n, m); return 0;} |
Java
// A simple recursive solution to count// number of ways to reach mat[m-1][n-1] from// mat[0][0] in a matrix mat[][]class GFG{ // Returns The number of way from top-left // to mat[m-1][n-1] static int countPaths(int m, int n) { int [][]dp=new int[m+1][n+1]; for (int i=1; i<=m; i++) { for (int j=1; j<=n; j++) { if (i==1 || j == 1) dp[i][j] = 1; else dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[m][n]; } // Driver code public static void main(String []args) { int n = 5; int m = 5; System.out.println(countPaths(n, m)); }}// This code is contributed // by ihritik (Hritik Raj) |
Python3
# A simple recursive solution to # count number of ways to reach # mat[m-1][n-1] from mat[0][0] # in a matrix mat[][]# Returns The number of way # from top-left to mat[m-1][n-1]def countPaths(m, n): dp = [[0 for i in range(m + 1)] for j in range(n + 1)] for i in range(1, m + 1): for j in range(1, n + 1): if (i == 1 or j == 1): dp[i][j] = 1 else: dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) return dp[m][n]# Driver codeif __name__ =="__main__": n = 5 m = 5 print(countPaths(n, m))# This code is contributed # by ChitraNayal |
C#
// A simple recursive solution to count// number of ways to reach mat[m-1][n-1] from// mat[0][0] in a matrix mat[][]using System;class GFG{ // Returns The number of way from top-left // to mat[m-1][n-1] static int countPaths(int m, int n) { int [,]dp=new int[m+1,n+1]; for (int i=1; i<=m; i++) { for (int j=1; j<=n; j++) { if (i==1 || j == 1) dp[i,j] = 1; else dp[i,j] = dp[i-1,j] + dp[i,j-1]; } } return dp[m,n]; } // Driver code public static void Main() { int n = 5; int m = 5; Console.WriteLine(countPaths(n, m)); }}// This code is contributed // by ihritik (Hritik Raj) |
PHP
<?php// A simple recursive solution to count// number of ways to reach mat[m-1][n-1] // from mat[0][0] in a matrix mat[][]// Returns The number of way from top-left // to mat[m-1][n-1]function countPaths($m, $n){ $dp; for ($i = 1; $i <= $m; $i++) { for ($j = 1; $j <= $n; $j++) { if ($i == 1 || $j == 1) $dp[$i][$j] = 1; else $dp[$i][$j] = $dp[$i - 1][$j] + $dp[$i][$j - 1]; } } return $dp[$m][$n];}// Driver code$n = 5;$m = 5; echo countPaths($n, $m);// This code is contributed by Rajput-Ji?> |
Javascript
<script> // A simple recursive solution to count // number of ways to reach mat[m-1][n-1] from // mat[0][0] in a matrix mat[][] // Returns The number of way from top-left // to mat[m-1][n-1] function countPaths(m, n) { let dp = new Array(m+1); for (let i=0; i<=m; i++) { dp[i] = new Array(n + 1); for (let j=0; j<=n; j++) { dp[i][j] = 0; } } for (let i=1; i<=m; i++) { for (let j=1; j<=n; j++) { if (i==1 || j == 1) dp[i][j] = 1; else dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[m][n]; } let n = 5; let m = 5; document.write(countPaths(n, m)); </script> |
Output
70
Time Complexity: O(m * n)
Auxiliary Space: O(m * n)
Another Method(Efficient):
There is one more efficient way of reaching the solution in O(m) or O(n) whichever is greater.
We can permute the number of right operations and down operations.
Explanation:
In the given figure, we can see that for a matrix of 3×3, we need 2 right operations and 2 down operations.
Thus, we can permute these operations in any order, still we can reach the bottom-right.
“RRDD”,”RDRD”,”RDDR”,”DRDR”,”DDRR”,”DRRD”
Let the number of columns be m, and the number of rows is n, then no of permutations = (m+n)!/ (m!*n!)
Implementation:
C++
// C++ program for above approach#include<bits/stdc++.h>using namespace std;// Find factorialint factorial(int n){ int res = 1, i; for(i = 2; i <= n; i++) res *= i; return res;}// Find number of ways to reach // mat[m-1][n-1] from mat[0][0]// in a matrix mat[][]] int countWays(int m, int n){ m = m - 1; n = n - 1; return factorial(m + n) / (factorial(m) * factorial(n));}// Driver Codeint main(){ int m = 5; int n = 5; // Function call int result = countWays(m, n); cout << result;}// This code is contributed by chahattekwani71 |
Java
// Java Program for above approachimport java.io.*;class GFG { // Find factorial static int factorial(int n) { int res = 1, i; for (i = 2; i <= n; i++) res *= i; return res; } // Find number of ways to reach // mat[m-1][n-1] from mat[0][0] // in a matrix mat[][]] static int countWays(int m, int n) { m = m - 1; n = n - 1; return factorial(m + n) / (factorial(m) * factorial(n)); } // Driver Code public static void main(String[] args) { int m = 5; int n = 5; // Function Call int result = countWays(m, n); System.out.println(result); }} |
Python3
# Python3 program for # the above approach# Find factorialdef factorial(n): res = 1 for i in range(2, n + 1): res *= i return res# Find number of ways to reach # mat[m-1][n-1] from mat[0][0]# in a matrix mat[][]]def countWays(m, n): m = m - 1 n = n - 1 return (factorial(m + n) // (factorial(m) * factorial(n)))# Driver code m = 5n = 5# Function callresult = countWays(m, n)print(result)# This code is contributed by divyeshrabadiya07 |
C#
// C# Program for above approachusing System;class GFG{ // Find factorialstatic int factorial(int n){ int res = 1, i; for(i = 2; i <= n; i++) res *= i; return res;}// Find number of ways to reach // mat[m-1][n-1] from mat[0][0]// in a matrix mat[][]]static int countWays(int m, int n){ m = m - 1; n = n - 1; return factorial(m + n) / (factorial(m) * factorial(n));}// Driver codestatic void Main() { int m = 5; int n = 5; // Function Call int result = countWays(m, n); Console.WriteLine(result);}}// This code is contributed by divyesh072019 |
Javascript
<script>// JavaScript Program for above approach // Find factorialfunction factorial( n){ var res = 1, i; for(i = 2; i <= n; i++) res *= i; return res;} // Find number of ways to reach// mat[m-1][n-1] from mat[0][0]// in a matrix mat[][]]function countWays( m, n){ m = m - 1; n = n - 1; return factorial(m + n) / (factorial(m) * factorial(n));} // Driver code var m = 5; var n = 5; // Function Call var result = countWays(m, n); document.write(result); </script> |
Output
70
Time Complexity: O(n)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
