Number of shortest paths to reach every cell from bottom-left cell in the grid

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP progarm to find number of shortest paths
#include <bits/stdc++.h>
using namespace std;
  
// Function to find number of shortest paths
void 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 code
int main()
{
    int n = 5, m = 2;
  
    // Function call
    NumberOfShortestPaths(n, m);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java progarm to find number of shortest paths
class GFG
{
  
// Function to find number of shortest paths
static 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 code
public static void main(String[] args)
{
    int n = 5, m = 2;
  
    // Function call
    NumberOfShortestPaths(n, m);
}
}
  
// This code is contributed by Princi Singh

chevron_right


C#

// C# progarm to find number of shortest paths
using System;

class GFG
{

// Function to find number of shortest paths
static 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 code public static void Main(String[] args) { int n = 5, m = 2; // Function call NumberOfShortestPaths(n, m); } } // This code is contributed by PrinciRaj1992 [tabbyending] Output :

1 5 
1 4 
1 3 
1 2 
1 1 

Time complexity: O(N × M)



My Personal Notes arrow_drop_up

Student of BS computer science

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : princi singh