C++ Program For Pascal's Triangle

Pascal's triangle is a triangular arrangement of binomial coefficients where every row begins and ends with 1, and every other element is the sum of the two elements directly above it.

Each row represents the coefficients of a binomial expansion.
It can be generated using several approaches with different time and space complexities.

Example

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

Methods to Generate Pascal's Triangle

This article discusses three approaches to generate Pascal's Triangle, each with different time and space complexities.

Method 1: Using Binomial Coefficients

In this approach, every element is computed independently using the binomial coefficient formula.

Each row contains row + 1 elements.
The iᵗʰ element in row row is C(row, i).
Every coefficient is calculated separately.

Formula

C(n, k) = n! / (k! × (n − k)!)

Algorithm

Iterate through each row.
For every position in the row, calculate the corresponding binomial coefficient.
Print the computed value.

C++

#include <iostream>
using namespace std;

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

Complexity Analysis

Time Complexity: O(n³)
Auxiliary Space: O(1)

Method 2: Using Dynamic Programming

Instead of calculating every coefficient independently, use the property that every element is the sum of the two elements directly above it.

The first and last element of every row is always 1.
Remaining elements are computed using previously generated rows.
Stores the triangle in a 2D array.

Algorithm

Create a 2D array.
Initialize the first and last element of every row as 1.
Compute the remaining elements using previous rows.
Print the generated triangle.

C++

#include <bits/stdc++.h>
using namespace std;

void printPascal(int n)
{
    
    // An auxiliary array to store 
    // generated pascal triangle values
    long long 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

Complexity Analysis

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

Space Optimization

The previous method stores the entire Pascal's Triangle even though only the previous row is required to generate the next row.

The extra space can be reduced to O(n) by storing only one row.
It can be optimized even further by computing each coefficient directly without storing previous rows, as shown in the next method.

Method 3: Using Previous Binomial Coefficient

This is the most efficient approach. Instead of recomputing every coefficient, each value is calculated from the previous coefficient in constant time.

Every row starts with 1.
Each subsequent coefficient is derived from the previous one.
No extra array is required.

Formula

C(line, i) = C(line, i − 1) × (line − i + 1) / i

Algorithm

Start every row with coefficient 1.
Print the current coefficient.
Compute the next coefficient using the recurrence relation.
Repeat until the row is complete.

C++

#include <bits/stdc++.h>

using namespace std;
void printPascal(int n)
{
    
for (int line = 1; line <= n; line++)
{
    long long 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

Complexity Analysis

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

Integer Overflow

Although the optimized approach is the most space-efficient, Pascal's Triangle grows very quickly.

Using long long increases the range compared to int.
However, even long long overflows for sufficiently large values of n.
For very large rows, arbitrary-precision arithmetic is required.

C++ Program For Pascal's Triangle

Methods to Generate Pascal's Triangle

Method 1: Using Binomial Coefficients

Method 2: Using Dynamic Programming

Method 3: Using Previous Binomial Coefficient

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