0 Program: 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.

// A simple O(n^3) program for Pascal's Triangle #include <stdio.h> 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 = 1; line < n; line++)   {     // Every line has number of integers equal to line number     for (int i = 1; i <= line; i++)       printf("%d ", binomialCoeff(line, i));     printf("\n");   } } 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 CoefficientC(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) / 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;  // C(line, i) = C(line, i-1) * (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.