# 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 ```

## We strongly recommend that you click here and practice it, before moving on to the solution.

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.

```// A simple O(n^3) program for Pascal's Triangle
#include <stdio.h>

// See http://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 http://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 to test above function
int main()
{
int n = 7;
printPascal(n);
return 0;
}
```

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 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
```
```// 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.

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