# Find the Nth row in Pascal’s Triangle

Given a non-negative integer N, the task is to find the Nth row of Pascal’s Triangle
Note: The row index starts from 0.

Pascal’s Triangle:

1 1
1 2 1
1 3 3 1
1 4 6 4 1

Examples:

Input: N = 3
Output: 1, 3, 3, 1
Explanation:
The elements in the 3rd row are 1 3 3 1.
Input: N = 0
Output:

Naive Approach:
The simplest approach to solve the problem is to use Recursion. Find the row of the previous index first using recursion and then calculate the values of the current row with the help of the previous one. Repeat this process up to the Nth row.
Below is the implementation of the above approach:

## Java

 `// Java Program to find the Nth ` `// index row in Pascal's triangle ` `import` `java.util.ArrayList; ` `public` `class` `geeks { ` ` `  `    ``// Function to find the elements ` `    ``// of rowIndex in Pascal's Triangle ` `    ``public` `static` `ArrayList getRow( ` `        ``int` `rowIndex) ` `    ``{ ` `        ``ArrayList currow ` `            ``= ``new` `ArrayList(); ` `        ``// 1st element of every row is 1 ` `        ``currow.add(``1``); ` ` `  `        ``// Check if the row that has to ` `        ``// be returned is the first row ` `        ``if` `(rowIndex == ``0``) { ` `            ``return` `currow; ` `        ``} ` `        ``// Generate the previous row ` `        ``ArrayList prev = getRow(rowIndex ` `                                         ``- ``1``); ` ` `  `        ``for` `(``int` `i = ``1``; i < prev.size(); i++) { ` `            ``// Generate the elements ` `            ``// of the current row ` `            ``// by the help of the ` `            ``// previous row ` `            ``int` `curr = prev.get(i - ``1``) ` `                       ``+ prev.get(i); ` `            ``currow.add(curr); ` `        ``} ` `        ``currow.add(``1``); ` ` `  `        ``// Return the row ` `        ``return` `currow; ` `    ``} ` ` `  `    ``// Driver Program ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``3``; ` `        ``ArrayList arr = getRow(n); ` ` `  `        ``for` `(``int` `i = ``0``; i < arr.size(); i++) { ` `            ``if` `(i == arr.size() - ``1``) ` `                ``System.out.print(arr.get(i)); ` `            ``else` `                ``System.out.print(arr.get(i) ` `                                 ``+ ``", "``); ` `        ``} ` `    ``} ` `} `

## C#

 `// C# program to find the Nth ` `// index row in Pascal's triangle ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG{ ` ` `  `// Function to find the elements ` `// of rowIndex in Pascal's Triangle ` `public` `static` `List<``int``> getRow(``int` `rowIndex) ` `{ ` `    ``List<``int``> currow = ``new` `List<``int``>(); ` `     `  `    ``// 1st element of every row is 1 ` `    ``currow.Add(1); ` ` `  `    ``// Check if the row that has to ` `    ``// be returned is the first row ` `    ``if` `(rowIndex == 0) ` `    ``{ ` `        ``return` `currow; ` `    ``} ` `    ``// Generate the previous row ` `    ``List<``int``> prev = getRow(rowIndex - 1); ` ` `  `    ``for``(``int` `i = 1; i < prev.Count; i++)  ` `    ``{ ` `         `  `        ``// Generate the elements ` `        ``// of the current row ` `        ``// by the help of the ` `        ``// previous row ` `        ``int` `curr = prev[i - 1] + prev[i]; ` `        ``currow.Add(curr); ` `    ``} ` `    ``currow.Add(1); ` ` `  `    ``// Return the row ` `    ``return` `currow; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `n = 3; ` `    ``List<``int``> arr = getRow(n); ` ` `  `    ``for``(``int` `i = 0; i < arr.Count; i++) ` `    ``{ ` `        ``if` `(i == arr.Count - 1) ` `            ``Console.Write(arr[i]); ` `        ``else` `            ``Console.Write(arr[i] + ``", "``); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```1, 3, 3, 1
```

Efficient Approach:
Follow the steps below to optimize the above approach:

• Unlike the above approach, we will just generate only the numbers of the Nth row.
• We can observe that the Nth row of the Pascals triangle consists of following sequence:

```NC0, NC1, ......, NCN - 1, NCN
```
• Since, NC0 = 1, the following values of the sequence can be generated by the following equation:

```NCr = (NCr - 1 * (N - r + 1)) / r where 1 ≤ r ≤ N
```

Below is the implementation of the above approach:

## C++

 `// C++ program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Print the N-th row of the ` `// Pascal's Triangle ` `void` `generateNthrow(``int` `N) ` `{ ` `    ``// nC0 = 1 ` `    ``int` `prev = 1; ` `    ``cout << prev; ` ` `  `    ``for` `(``int` `i = 1; i <= N; i++) { ` `        ``// nCr = (nCr-1 * (n - r + 1))/r ` `        ``int` `curr = (prev * (N - i + 1)) / i; ` `        ``cout << ``", "` `<< curr; ` `        ``prev = curr; ` `    ``} ` `} ` ` `  `// Driver Program ` `int` `main() ` `{ ` ` `  `    ``int` `N = 5; ` `    ``generateNthrow(N); ` `    ``return` `0; ` `} `

## Java

 `// Java program to implement the above approach ` `import` `java.io.*; ` ` `  `class` `GFG{ ` ` `  `// Print the N-th row of the ` `// Pascal's Triangle ` `static` `void` `generateNthrow(``int` `N) ` `{ ` ` `  `    ``// nC0 = 1 ` `    ``int` `prev = ``1``; ` `    ``System.out.print(prev); ` `     `  `    ``for``(``int` `i = ``1``; i <= N; i++) ` `    ``{ ` `         `  `       ``// nCr = (nCr-1 * (n - r + 1))/r ` `       ``int` `curr = (prev * (N - i + ``1``)) / i; ` `       ``System.out.print(``", "` `+ curr); ` `       ``prev = curr; ` `    ``} ` `} ` `     `  `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `N = ``5``; ` `    ``generateNthrow(N); ` `} ` `} ` ` `  `// This code is contributed by shubhamsingh10 `

## Python3

 `# Python3 program to implement the above approach  ` ` `  `# Print the N-th row of the  ` `# Pascal's Triangle ` `def` `generateNthRow (N): ` ` `  `    ``# nC0 = 1 ` `    ``prev ``=` `1` `    ``print``(prev, end ``=` `'') ` ` `  `    ``for` `i ``in` `range``(``1``, N ``+` `1``): ` ` `  `        ``# nCr = (nCr-1 * (n - r + 1))/r ` `        ``curr ``=` `(prev ``*` `(N ``-` `i ``+` `1``)) ``/``/` `i ` `        ``print``(``","``, curr, end ``=` `'') ` `        ``prev ``=` `curr ` ` `  `# Driver code ` `N ``=` `5` ` `  `# Function calling ` `generateNthRow(N)  ` ` `  `# This code is contributed by himanshu77 `

## C#

 `// C# program to implement the above approach ` `using` `System; ` `using` `System.Collections.Generic; ` `class` `GFG{ ` ` `  `// Print the N-th row of the ` `// Pascal's Triangle ` `static` `void` `generateNthrow(``int` `N) ` `{ ` ` `  `    ``// nC0 = 1 ` `    ``int` `prev = 1; ` `    ``Console.Write(prev); ` `     `  `    ``for``(``int` `i = 1; i <= N; i++) ` `    ``{ ` `         `  `        ``// nCr = (nCr-1 * (n - r + 1))/r ` `        ``int` `curr = (prev * (N - i + 1)) / i; ` `        ``Console.Write(``", "` `+ curr); ` `        ``prev = curr; ` `    ``} ` `} ` `     `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `N = 5; ` `    ``generateNthrow(N); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```1, 5, 10, 10, 5, 1

```

Time Complexity: O(N)
Auxiliary Space: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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