Given two number N and M. The task is to find the number of shortest paths to reach the cell(i, j) in the grid of size N × M when the moves started from the bottom-left corner
Note: cell(i, j) represents the ith row and jth column in the grid
Below image shows some of the shortest paths to reach cell(1, 4) in 4 × 4 grid
Examples :
Input : N = 3, M = 4
Output : 1 3 6 10
1 2 3 4
1 1 1 1
Input : N = 5, M = 2
Output : 1 5
1 4
1 3
1 2
1 1
Approach : An efficient approach is to compute the grid starting from the bottom-left corner.
- The number of shortest paths to reach cell(n, i) is 1, where, 1 < = i < = M
- The number of shortest paths to reach cell(i, 1) is 1, where, 1 < = i < = N
- The number of shortest paths to reach cell(i, j) are the sum the number of shortest paths of cell(i-1, j) and (i, j+1), where, 1 < = j < = M and 1 < = i < = N
Below is the implementation of the above approach :
C++
// CPP program to find number of shortest paths#include <bits/stdc++.h>using namespace std;// Function to find number of shortest pathsvoid NumberOfShortestPaths(int n, int m){ int a[n][m]; for (int i = 0; i < n; i++) memset(a[i], 0, sizeof(a[i])); // Compute the grid starting from // the bottom-left corner for (int i = n - 1; i >= 0; i--) { for (int j = 0; j < m; j++) { if (j == 0 or i == n - 1) a[i][j] = 1; else a[i][j] = a[i][j - 1] + a[i + 1][j]; } } // Print the grid for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { cout << a[i][j] << " "; } cout << endl; }}// Driver codeint main(){ int n = 5, m = 2; // Function call NumberOfShortestPaths(n, m); return 0;} |
Java
// Java program to find number of shortest pathsimport java.io.*;class GFG{// Function to find number of shortest pathsstatic void NumberOfShortestPaths(int n, int m){ int [][]a = new int[n][m]; // Compute the grid starting from // the bottom-left corner for (int i = n - 1; i >= 0; i--) { for (int j = 0; j < m; j++) { if (j == 0 || i == n - 1) a[i][j] = 1; else a[i][j] = a[i][j - 1] + a[i + 1][j]; } } // Print the grid for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { System.out.print(a[i][j] + " "); } System.out.println(); }}// Driver codepublic static void main(String[] args){ int n = 5, m = 2; // Function call NumberOfShortestPaths(n, m);}}// This code is contributed by Princi Singh |
Python3
# Python 3 program to find # number of shortest paths# Function to find number of shortest pathsdef NumberOfShortestPaths(n, m): a = [[0 for i in range(m)] for j in range(n)] for i in range(n): for j in range(m): a[i][j] = 0 # Compute the grid starting from # the bottom-left corner i = n - 1 while(i >= 0): for j in range(m): if (j == 0 or i == n - 1): a[i][j] = 1 else: a[i][j] = a[i][j - 1] + \ a[i + 1][j] i -= 1 # Print the grid for i in range(n): for j in range(m): print(a[i][j], end = " ") print("\n", end = "")# Driver codeif __name__ == '__main__': n = 5 m = 2 # Function call NumberOfShortestPaths(n, m) # This code is contributed by# Surendra_Gangwar |
C#
// C# program to find number of shortest pathsusing System;class GFG{// Function to find number of shortest pathsstatic void NumberOfShortestPaths(int n, int m){ int [,]a = new int[n, m]; // Compute the grid starting from // the bottom-left corner for (int i = n - 1; i >= 0; i--) { for (int j = 0; j < m; j++) { if (j == 0 || i == n - 1) a[i, j] = 1; else a[i, j] = a[i, j - 1] + a[i + 1, j]; } } // Print the grid for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { Console.Write(a[i, j] + " "); } Console.Write("\n"); }}// Driver codepublic static void Main(String[] args){ int n = 5, m = 2; // Function call NumberOfShortestPaths(n, m);}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// javascript program to find number of shortest paths // Function to find number of shortest paths function NumberOfShortestPaths(n , m) { var a = Array(n).fill().map(() => Array(m).fill(0)); // Compute the grid starting from // the bottom-left corner for (var i = n - 1; i >= 0; i--) { for (j = 0; j < m; j++) { if (j == 0 || i == n - 1) a[i][j] = 1; else a[i][j] = a[i][j - 1] + a[i + 1][j]; } } // Print the grid for (var i = 0; i < n; i++) { for (j = 0; j < m; j++) { document.write(a[i][j] + " "); } document.write("<br/>"); } } // Driver code var n = 5, m = 2; // Function call NumberOfShortestPaths(n, m);// This code is contributed by gauravrajput1 </script> |
Output
1 5 1 4 1 3 1 2 1 1
Time complexity: O(N × M), where N is number of rows and M is number of columns of the grid.
Auxiliary Space: O(N × M), where N is number of rows and M is number of columns of the grid.
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