C++ Program For Pascal’s Triangle

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 

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 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;
}

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 

Auxiliary Space: O(1)

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. 

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;
}

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. 

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;
}

Output:

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