C++ Program to Print Pascal's Triangle

Pascal's Triangle

#include<iostream>

 using namespace std;

unsigned int factorial(unsigned int const num){

  if( num <=1 )
    return 1;
  else
    return num*factorial(num-1);
}

int main( void ){

  int row = 0;
    
  cout<<"Enter the number of rows: ";
  cin>>row;
  
  //Method 1

  for(unsigned int n=0;n<row;++n){

    //spacing before row depends on how many rows there are
    //row-i = number of space needed before row starts
    
    for(unsigned int j=1;j<(row-n);++j){
      cout<<"  ";
    }

    unsigned int n_factorial = factorial(n);
    //printing the binomial coefficient using combination formula 
    //Number of binomial coefficient on each line
    // is equivalent to the current row number (i.e., the counter i)
    //n choose k or nCk = n!/(k!(n-k)!)
    
    for(unsigned int k=0;k<=n;++k){
      cout<<"   "<<n_factorial/(factorial(n-k)*factorial(k));
    }

    cout<<endl<<endl;
  }
  

  //Method 2 
 
  for(unsigned int n=0;n<row;++n){

    for(unsigned int j=1;j<(row-n);++j){
      cout<<"  ";
    }
    
    unsigned int coeff = 1;
    
    for(unsigned int k=0;k<=n;++k){
      cout<<"   "<<coeff;
      coeff = coeff*(n-k)/(k+1);
    }
  
    cout<<endl<<endl;
  }
 return 0;

}

Output:

Enter number of rows: 5
Method 1:
           1

         1   1

       1   2   1

     1   3   3   1

   1   4   6   4   1

Method 2:
           1

         1   1

       1   2   1

     1   3   3   1

   1   4   6   4   1

Comments

Non-decreasing Array - LeetCode

Given an array with n integers, your task is to check if it could become non-decreasing by modifying at most 1 element.We define an array is non-decreasing if array[i] <= array[i + 1] holds for every i (1 <= i < n). Example 1: Input: [4,2,3] Output: True Explanation: You could modify the first 4 to 1 to get a non-decreasing array. Example 2: Input: [4,2,1] Output: False Explanation: You can't get a non-decreasing array by modify at most one element. Note: The n belongs to [1, 10,000]. Source: LeetCode At first glance, it looks like an easy and straightforward problem, but believe me it's greedy problem. In the problem, you are allowed to make atmost "one" modification to make a non-decreasing array. Nondecreasing means that every element is greater than or equal to the one before it e.g. [1,3,4,4,5,6,9]. For example, [2,2,3,2,4] can this become non-decreasing? Yes, by replacing 3 with 2 =>[2,2,2,2,4]. What about this:...

Binary Tree Paths

Source: LeetCode Solution: We do a preorder traversal add each node in the string followed by an arrow "->" if the current node is a leaf node (have no right or left child) then we add the string to the list, otherwise, we add an arrow and make a recursive call in a preorder fashion. The previous string is stored in stack and we can use it later when the call is returned to add another possible routes from root to the leaf nodes. Complexity Analysis Time complexity : O ( n ) . n nodes need to be traversed ( n represents the number of nodes in a given tree) . Solution: public List< String > binaryTreePaths( TreeNode root) { ArrayList< String > path = new ArrayList (); binaryPath(root, "" ,path); return path; } public static void binaryPath( TreeNode root, String s, List< String > path){ if (root == null ) return ; ...

Merge Two Binary Trees

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree. Example 1: Input: Tree 1 Tree 2 1 2 / \ / \ 3 2 1 3 / \ \ 5 4 7 Output: Merged tree: 3 / \ 4 5 / \ \ 5 4 7 Note: The merging process must start from the root nodes of both trees. Source: LeetCode Solution: We traverse both trees in a p...

{ Programming }

Search This Blog