Pascal’s Triangle

Pascal’s triangle is a triangular array of binomial coefficients. Write a function that takes an integer value n as input and prints first n lines of Pascal’s triangle. Following are the first 6 rows of Pascal’s Triangle.

1  
1 1 
1 2 1 
1 3 3 1 
1 4 6 4 1 
1 5 10 10 5 1 

Method 1 ( O(n^3) time complexity )
The number of entries in every line is equal to line number. For example, the first line has “1”, the second line has “1 1”, the third line has “1 2 1”,.. and so on. Every entry in a line is value of a Binomial Coefficient. The value of ith entry in line number line is C(line, i). The value can be calculated using following formula. 

C(line, i)   = line! / ( (line-i)! * i! ) 

A simple method is to run two loops and calculate the value of Binomial Coefficient in inner loop. 

C++

//  C++ code for Pascal's Triangle
#include <iostream>
using namespace std;

// See https://www.geeksforgeeks.org/space-and-time-efficient-binomial-coefficient/
// for details of this function
int binomialCoeff(int n, int k);

// Function to print first
// n lines of Pascal's
// Triangle
void printPascal(int n)
{
    // Iterate through every line and
    // print entries in it
    for (int line = 0; line < n; line++)
    {
        // Every line has number of
        // integers equal to line
        // number
        for (int i = 0; i <= line; i++)
            cout <<" "<< binomialCoeff(line, i);
        cout <<"\n";
    }
}

// See https://www.geeksforgeeks.org/space-and-time-efficient-binomial-coefficient/
// for details of this function
int binomialCoeff(int n, int k)
{
    int res = 1;
    if (k > n - k)
    k = n - k;
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }

    return res;
}

// Driver program
int main()
{
    int n = 7;
    printPascal(n);
    return 0;
}

// This code is contributed by shivanisinghss2110

C

//  C++ code for Pascal's Triangle
#include <stdio.h>

// See https://www.geeksforgeeks.org/space-and-time-efficient-binomial-coefficient/
// for details of this function
int binomialCoeff(int n, int k);

// Function to print first
// n lines of Pascal's
// Triangle
void printPascal(int n)
{
    // Iterate through every line and
    // print entries in it
    for (int line = 0; line < n; line++)
    {
        // Every line has number of
        // integers equal to line
        // number
        for (int i = 0; i <= line; i++)
            printf("%d ",
                    binomialCoeff(line, i));
        printf("\n");
    }
}

// See https://www.geeksforgeeks.org/space-and-time-efficient-binomial-coefficient/
// for details of this function
int binomialCoeff(int n, int k)
{
    int res = 1;
    if (k > n - k)
    k = n - k;
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }

    return res;
}

// Driver program
int main()
{
    int n = 7;
    printPascal(n);
    return 0;
}

Java

// Java code for Pascal's Triangle
import java.io.*;

class GFG {

    // Function to print first
    // n lines of Pascal's Triangle
    static void printPascal(int n)
    {

    // Iterate through every line
    // and print entries in it
    for (int line = 0; line < n; line++)
    {
        // Every line has number of
        // integers equal to line number
        for (int i = 0; i <= line; i++)
        System.out.print(binomialCoeff
                        (line, i)+" ");

        System.out.println();
    }
    }

    // Link for details of this function
    // https://www.geeksforgeeks.org/space-and-time-efficient-binomial-coefficient/
    static int binomialCoeff(int n, int k)
    {
        int res = 1;

        if (k > n - k)
        k = n - k;

        for (int i = 0; i < k; ++i)
        {
            res *= (n - i);
            res /= (i + 1);
        }
        return res;
    }

    // Driver code
    public static void main(String args[])
    {
    int n = 7;
    printPascal(n);
    }
}

/*This code is contributed by Nikita Tiwari.*/

Python3

# Python 3 code for Pascal's Triangle
# A simple O(n^3)
# program for
# Pascal's Triangle

# Function to print
# first n lines of
# Pascal's Triangle
def printPascal(n) :

    # Iterate through every line
    # and print entries in it
    for line in range(0, n) :

        # Every line has number of
        # integers equal to line
        # number
        for i in range(0, line + 1) :
            print(binomialCoeff(line, i),
                " ", end = "")
        print()


# See https://www.geeksforgeeks.org/space-and-time-efficient-binomial-coefficient/
# for details of this function
def binomialCoeff(n, k) :
    res = 1
    if (k > n - k) :
        k = n - k
    for i in range(0 , k) :
        res = res * (n - i)
        res = res // (i + 1)

    return res

# Driver program
n = 7
printPascal(n)


# This code is contributed by Nikita Tiwari.

C#

// C# code for Pascal's Triangle
using System;

class GFG {

    // Function to print first
    // n lines of Pascal's Triangle
    static void printPascal(int n)
    {

    // Iterate through every line
    // and print entries in it
    for (int line = 0; line < n; line++)
    {
        // Every line has number of
        // integers equal to line number
        for (int i = 0; i <= line; i++)
        Console.Write(binomialCoeff
                        (line, i)+" ");

        Console.WriteLine();
    }
    }

    // Link for details of this function
    // https://www.geeksforgeeks.org/space-and-time-efficient-binomial-coefficient/
    static int binomialCoeff(int n, int k)
    {
        int res = 1;

        if (k > n - k)
        k = n - k;

        for (int i = 0; i < k; ++i)
        {
            res *= (n - i);
            res /= (i + 1);
        }
        return res;
    }

    // Driver code
    public static void Main()
    {
    int n = 7;
    printPascal(n);
    }
}

/*This code is contributed by vt_m.*/

PHP

<?php
// PHP implementation for
// Pascal's Triangle

// for details of this function
function binomialCoeff($n, $k)
{
    $res = 1;
    if ($k > $n - $k)
    $k = $n - $k;
    for ($i = 0; $i < $k; ++$i)
    {
        $res *= ($n - $i);
        $res /= ($i + 1);
    }
return $res;
}

// Function to print first
// n lines of Pascal's
// Triangle
function printPascal($n)
{
    // Iterate through every line and
    // print entries in it
    for ($line = 0; $line < $n; $line++)
    {
        // Every line has number of
        // integers equal to line
        // number
        for ($i = 0; $i <= $line; $i++)
                echo "".binomialCoeff($line, $i)." ";

        echo "\n";
    }
}

// Driver Code
$n=7;
printPascal($n);

// This code is contributed by Mithun Kumar
?>

Javascript

<script>

// Javascript code for Pascal's Triangle

    // Function to print first
    // n lines of Pascal's Triangle
    function printPascal(n)
    {

    // Iterate through every line
    // and print entries in it
    for (let line = 0; line < n; line++)
    {
        // Every line has number of
        // integers equal to line number
        for (let i = 0; i <= line; i++)
        document.write(binomialCoeff
                        (line, i)+" ");

        document.write("<br />");
    }
    }

    // Link for details of this function
    // https://www.geeksforgeeks.org/space-and-time-efficient-binomial-coefficient/
    function binomialCoeff(n, k)
    {
        let res = 1;

        if (k > n - k)
        k = n - k;

        for (let i = 0; i < k; ++i)
        {
            res *= (n - i);
            res /= (i + 1);
        }
        return res;
    }


// Driver Code

    let n = 7;
    printPascal(n);

</script>

Output

 1
 1 1
 1 2 1
 1 3 3 1
 1 4 6 4 1
 1 5 10 10 5 1
 1 6 15 20 15 6 1

Time complexity: O(n^3)
Auxiliary Space: O(1)

Method 2( O(n^2) time and O(n^2) extra space )
If we take a closer at the triangle, we observe that every entry is sum of the two values above it. So we can create a 2D array that stores previously generated values. To generate a value in a line, we can use the previously stored values from array. 

Steps to solve the problem:-

step1- Declare an 2-D array array of size n*n.

step2- Iterate through line 0 to line n:

           *Iterate through i=0 to present the  line:

                  *check if present line is equal to i or i=0 than arr[line][i]=1 .

*else update arr[line][i] to arr[line-1][i-1] + arr[line-1][i] .

                  *print arr[line][i].

          *shift to next line.

C++

// C++ program for Pascal’s Triangle
// A O(n^2) time and O(n^2) extra space
// method for Pascal's Triangle
#include <bits/stdc++.h>
using namespace std;

void printPascal(int n)
{

    // An auxiliary array to store
    // generated pascal triangle values
    int arr[n][n];

    // Iterate through every line and
    // print integer(s) in it
    for (int line = 0; line < n; line++)
    {
        // Every line has number of integers
        // equal to line number
        for (int i = 0; i <= line; i++)
        {
        // First and last values in every row are 1
        if (line == i || i == 0)
            arr[line][i] = 1;
        // Other values are sum of values just
        // above and left of above
        else
            arr[line][i] = arr[line - 1][i - 1] +
                            arr[line - 1][i];
        cout << arr[line][i] << " ";
        }
        cout << "\n";
    }
}

// Driver code
int main()
{
    int n = 5;
    printPascal(n);
    return 0;
}

// This code is Contributed by Code_Mech.

C

// C program for Pascal’s Triangle
// A O(n^2) time and O(n^2) extra space
// method for Pascal's Triangle
void printPascal(int n)
{
// An auxiliary array to store
// generated pascal triangle values
int arr[n][n];

// Iterate through every line and print integer(s) in it
for (int line = 0; line < n; line++)
{
    // Every line has number of integers
    // equal to line number
    for (int i = 0; i <= line; i++)
    {
    // First and last values in every row are 1
    if (line == i || i == 0)
        arr[line][i] = 1;
    // Other values are sum of values just
    // above and left of above
    else
        arr[line][i] = arr[line-1][i-1] + arr[line-1][i];
    printf("%d ", arr[line][i]);
    }
    printf("\n");
}
}
// Driver code
int main()
{
int n = 5;
    printPascal(n);
    return 0;
}

Java

// java program for Pascal's Triangle
// A O(n^2) time and O(n^2) extra
// space method for Pascal's Triangle
import java.io.*;

class GFG {
    public static void main (String[] args) {
        int n = 5;
        printPascal(n);
    }

public static void printPascal(int n)
{
// An auxiliary array to store generated pascal triangle values
int[][] arr = new int[n][n];

// Iterate through every line and print integer(s) in it
for (int line = 0; line < n; line++)
{
    // Every line has number of integers equal to line number
    for (int i = 0; i <= line; i++)
    {
    // First and last values in every row are 1
    if (line == i || i == 0)
        arr[line][i] = 1;
    else // Other values are sum of values just above and left of above
        arr[line][i] = arr[line-1][i-1] + arr[line-1][i];
    System.out.print(arr[line][i]);
    }
    System.out.println("");
}
}
}

Python3

# Python3 program for Pascal's Triangle

# A O(n^2) time and O(n^2) extra
# space method for Pascal's Triangle
def printPascal(n:int):

    # An auxiliary array to store
    # generated pascal triangle values
    arr = [[0 for x in range(n)]
              for y in range(n)]

    # Iterate through every line
    # and print integer(s) in it
    for line in range (0, n):

        # Every line has number of
        # integers equal to line number
        for i in range (0, line + 1):

            # First and last values
            # in every row are 1
            if(i is 0 or i is line):
                arr[line][i] = 1
                print(arr[line][i], end = " ")

            # Other values are sum of values
            # just above and left of above
            else:
                arr[line][i] = (arr[line - 1][i - 1] +
                                arr[line - 1][i])
                print(arr[line][i], end = " ")
        print("\n", end = "")

# Driver Code
n = 5
printPascal(n)

# This code is contributed
# by Sanju Maderna

C#

// C# program for Pascal's Triangle
// A O(n^2) time and O(n^2) extra
// space method for Pascal's Triangle
using System;

class GFG
{
public static void printPascal(int n)
{

// An auxiliary array to store
// generated pascal triangle values
int[,] arr = new int[n, n];

// Iterate through every line
// and print integer(s) in it
for (int line = 0; line < n; line++)
{
    // Every line has number of
    // integers equal to line number
    for (int i = 0; i <= line; i++)
    {

    // First and last values
    // in every row are 1
    if (line == i || i == 0)
        arr[line, i] = 1;
    else // Other values are sum of values
         // just above and left of above
        arr[line, i] = arr[line - 1, i - 1] +
                       arr[line - 1, i];
    Console.Write(arr[line, i]);
    }
Console.WriteLine("");
}
}

// Driver Code
public static void Main ()
{
    int n = 5;
    printPascal(n);
}
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

PHP

<?php
// PHP program for Pascal’s Triangle
// A O(n^2) time and O(n^2) extra space
// method for Pascal's Triangle
function printPascal($n)
{
    // An auxiliary array to store
    // generated pascal triangle values
    $arr = array(array());

    // Iterate through every line and
    // print integer(s) in it
    for ($line = 0; $line < $n; $line++)
    {
        // Every line has number of integers
        // equal to line number
        for ($i = 0; $i <= $line; $i++)
        {
            // First and last values in every row are 1
            if ($line == $i || $i == 0)
                $arr[$line][$i] = 1;

            // Other values are sum of values just
            // above and left of above
            else
                $arr[$line][$i] = $arr[$line - 1][$i - 1] +
                                $arr[$line - 1][$i];
            echo $arr[$line][$i] . " ";
        }
        echo "\n";
    }
}

// Driver code
$n = 5;
printPascal($n);

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>

// javascript program for Pascal's Triangle
// A O(n^2) time and O(n^2) extra
// space method for Pascal's Triangle
var n = 5;
printPascal(n);

function printPascal(n)
{
// An auxiliary array to store generated pascal triangle values
arr = a = Array(n).fill(0).map(x => Array(n).fill(0));

// Iterate through every line and print integer(s) in it
for (line = 0; line < n; line++)
{
    // Every line has number of integers equal to line number
    for (i = 0; i <= line; i++)
    {
    // First and last values in every row are 1
    if (line == i || i == 0)
        arr[line][i] = 1;
    else
    // Other values are sum of values just above and left of above
        arr[line][i] = arr[line-1][i-1] + arr[line-1][i];
    document.write(arr[line][i]);
    }
    document.write("<br>");
}
}

// This code is contributed by 29AjayKumar

</script>

Output

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

This method can be optimized to use O(n) extra space as we need values only from previous row. So we can create an auxiliary array of size n and overwrite values. Following is another method uses only O(1) extra space.
Method 3 ( O(n^2) time and O(1) extra space )
This method is based on method 1. We know that ith entry in a line number line is Binomial Coefficient C(line, i) and all lines start with value 1. The idea is to calculate C(line, i) using C(line, i-1). It can be calculated in O(1) time using the following. 
Steps to solve the problem:

1. iterate through line 1 to line n:

          *declare c variable and initialize it to 1.

          *iterate through i=1 till present line:

                    *print c.

                    *update c to c*(line-i)/i.

*shift to next line.

C(line, i)   = line! / ( (line-i)! * i! )
C(line, i-1) = line! / ( (line - i + 1)! * (i-1)! )
We can derive following expression from above two expressions.
C(line, i) = C(line, i-1) * (line - i + 1) / i

So C(line, i) can be calculated from C(line, i-1) in O(1) time

C++

// C++ program for Pascal’s Triangle
// A O(n^2) time and O(1) extra space
// function for Pascal's Triangle
#include <bits/stdc++.h>

using namespace std;
void printPascal(int n)
{

for (int line = 1; line <= n; line++)
{
    int C = 1; // used to represent C(line, i)
    for (int i = 1; i <= line; i++)
    {

        // The first value in a line is always 1
        cout<< C<<" ";
        C = C * (line - i) / i;
    }
    cout<<"\n";
}
}

// Driver code
int main()
{
    int n = 5;
    printPascal(n);
    return 0;
}

// This code is contributed by Code_Mech

C

// C program for Pascal’s Triangle
// A O(n^2) time and O(1) extra space
// function for Pascal's Triangle
void printPascal(int n)
{
for (int line = 1; line <= n; line++)
{
    int C = 1; // used to represent C(line, i)
    for (int i = 1; i <= line; i++)
    {
    printf("%d ", C); // The first value in a line is always 1
    C = C * (line - i) / i;
    }
    printf("\n");
}
}
// Driver code
int main()
{
int n = 5;
    printPascal(n);
    return 0;
}

Java

// Java program for Pascal's Triangle
// A O(n^2) time and O(1) extra
// space method for Pascal's Triangle
import java.io.*;
class GFG {

//Pascal function
public static void printPascal(int n)
{
    for(int line = 1; line <= n; line++)
    {

    int C=1;// used to represent C(line, i)
    for(int i = 1; i <= line; i++)
    {
        // The first value in a line is always 1
        System.out.print(C+" ");
        C = C * (line - i) / i;
    }
    System.out.println();
    }
}

// Driver code
public static void main (String[] args) {
    int n = 5;
    printPascal(n);
}
}
// This code is contributed
// by Archit Puri

Python3

# Python3 program for Pascal's Triangle
# A O(n^2) time and O(1) extra
# space method for Pascal's Triangle

# Pascal function
def printPascal(n):

    for line in range(1, n + 1):
        C = 1; # used to represent C(line, i)
        for i in range(1, line + 1):

            # The first value in a
            # line is always 1
            print(C, end = " ");
            C = int(C * (line - i) / i);
        print("");

# Driver code
n = 5;
printPascal(n);

# This code is contributed by mits

C#

// C# program for Pascal's Triangle
// A O(n^2) time and O(1) extra
// space method for Pascal's Triangle
using System;
class GFG
{

// Pascal function
public static void printPascal(int n)
{
    for(int line = 1;
            line <= n; line++)
    {

    int C = 1;// used to represent C(line, i)
    for(int i = 1; i <= line; i++)
    {
        // The first value in a
        // line is always 1
        Console.Write(C + " ");
        C = C * (line - i) / i;
    }
    Console.Write("\n") ;
    }
}

// Driver code
public static void Main ()
{
    int n = 5;
    printPascal(n);
}
}

// This code is contributed
// by ChitraNayal

PHP

<?php
// PHP program for Pascal's Triangle
// A O(n^2) time and O(1) extra
// space method for Pascal's Triangle

// Pascal function
function printPascal($n)
{
    for($line = 1; $line <= $n; $line++)
    {
        $C = 1;// used to represent C(line, i)
        for($i = 1; $i <= $line; $i++)
        {
            // The first value in a
            // line is always 1
            print($C . " ");
            $C = $C * ($line - $i) / $i;
        }
        print("\n");
    }
}

// Driver code
$n = 5;
printPascal($n);

// This code is contributed by mits
?>

Javascript

<script>

// JavaScript program for Pascal's Triangle
// A O(n^2) time and O(1) extra
// space method for Pascal's Triangle

//Pascal function
function printPascal(n)
{
    for(line = 1; line <= n; line++)
    {

    var C=1;// used to represent C(line, i)
    for(i = 1; i <= line; i++)
    {
        // The first value in a line is always 1
        document.write(C+" ");
        C = C * (line - i) / i;
    }
    document.write("<br>");
    }
}

// Driver code
var n = 5;
printPascal(n);

// This code is contributed by 29AjayKumar

</script>

Output

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

Time Complexity: O(n²)
Auxiliary Space: O(1)

So method 3 is the best method among all, but it may cause integer overflow for large values of n as it multiplies two integers to obtain values. 

Variations of the problem that may be asked in interviews:

i) Find the whole pascal triangle as shown above.

ii) Find just the one element of a pascal’s triangle given row number and column number in O(n) time.

iii) Find a particular row of pascal’s triangle given a row number in O(n) time.

This article is compiled by Rahul and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.