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

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**Method 1 ( O(n^3) time complexity )**

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 ** i**th 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++

// Java code for Pascal's Triangle #include <stdio.h> // See https://www.geeksforgeeks.org/archives/25621 // 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/archives/25621 // 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/archives/25621 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.*/

## Python

# 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/archives/25621 # 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.

Time complexity of this method is O(n^3). Following are optimized methods.

**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.

// A O(n^2) time and O(n^2) extra space method for Pascal's Triangle void printPascal(int n) { int arr[n][n]; // An auxiliary array to store generated pscal triangle values // 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]; printf("%d ", arr[line][i]); } printf("\n"); } }

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 ** i**th 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.

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

// 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"); } }

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.

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.