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

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// 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<Integer> getRow(
        int rowIndex)
    {
        ArrayList<Integer> currow
            = new ArrayList<Integer>();
        // 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<Integer> 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<Integer> 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)
                                 + ", ");
        }
    }
}

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C#

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// 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

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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++

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// C++ program to implement
// the above approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

1, 5, 10, 10, 5, 1

 

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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