# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 :

 `// CPP program to find number of shortest paths ` `#include ` `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; ` `} `

 `// Java program 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 `

 `# Python 3 program to find  ` `# number of shortest paths ` ` `  `# Function to find number of shortest paths ` `def` `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 code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `5` `    ``m ``=` `2` ` `  `    ``# Function call ` `    ``NumberOfShortestPaths(n, m) ` `     `  `# This code is contributed by ` `# Surendra_Gangwar `

 `// C# program 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 `

Output :

```1 5
1 4
1 3
1 2
1 1
```

Time complexity: O(N × M)

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