From ac24f7d6f4d97899794120b2f9035a5372f6246c Mon Sep 17 00:00:00 2001
From: Mayank Kumar Jha <mk9440@users.noreply.github.com>
Date: Fri, 22 Sep 2017 01:21:56 +0530
Subject: [PATCH] Add files via upload

Huffman Encoding with Proper comments
---
 Huffman.java | 158 +++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 158 insertions(+)
 create mode 100644 Huffman.java

diff --git a/Huffman.java b/Huffman.java
new file mode 100644
index 00000000..b93b7e7c
--- /dev/null
+++ b/Huffman.java
@@ -0,0 +1,158 @@
+
+import java.util.Comparator;
+import java.util.Iterator;
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Scanner;
+import java.util.Stack;
+/**
+ *
+ * @author Mayank Kumar (mk9440)
+ */
+/*
+Output :
+
+Enter number of distinct letters 
+6
+Enter letters with its frequncy to encode
+Enter letter : a
+Enter frequncy : 45
+
+Enter letter : b
+Enter frequncy : 13
+
+Enter letter : c
+Enter frequncy : 12
+
+Enter letter : d
+Enter frequncy : 16
+
+Enter letter : e
+Enter frequncy : 9
+
+Enter letter : f
+Enter frequncy : 5
+
+Letter		Encoded Form
+a		0 
+b		1 0 1 
+c		1 0 0 
+d		1 1 1 
+e		1 1 0 1 
+f		1 1 0 0 
+
+*/
+
+class Node{
+String letr="";    
+int freq=0,data=0;
+Node left=null,right=null;
+}
+
+//A comparator class to sort list on the basis of their frequency
+class comp implements Comparator<Node>{
+ @Override
+    public int compare(Node o1, Node o2) {
+        if(o1.freq>o2.freq){return 1;}
+        else if(o1.freq<o2.freq){return -1;}
+        else{return 0;}
+    }
+}
+
+
+public class Huffman {
+    
+    // A simple function to print a given list
+    //I just made it for debugging
+    public static void print_list(List li){
+    Iterator<Node> it=li.iterator();
+    while(it.hasNext()){Node n=it.next();System.out.print(n.freq+" ");}System.out.println();
+    }
+    
+    //Function for making tree (Huffman Tree)
+    public static Node make_huffmann_tree(List li){
+    //Sorting list in increasing order of its letter frequency    
+    li.sort(new comp());
+    Node temp=null;
+    Iterator it=li.iterator();
+    //System.out.println(li.size());
+    //Loop for making huffman tree till only single node remains in list
+    while(true){
+    temp=new Node();
+    //a and b are  Node which are to be combine to make its parent
+    Node a=new Node(),b=new Node();
+    a=null;b=null;
+    //checking if list is eligible for combining or not
+    //here first assignment of it.next in a will always be true as list till end will
+    //must have atleast one node
+    a=(Node)it.next();
+    //Below condition is to check either list has 2nd node or not to combine 
+    //If this condition will be false, then it means construction of huffman tree is completed
+    if(it.hasNext()){b=(Node)it.next();}
+    //Combining first two smallest nodes in list to make its parent whose frequncy 
+    //will be equals to sum of frequency of these two nodes 
+    if(b!=null){
+    temp.freq=a.freq+b.freq;a.data=0;b.data=1;//assigining 0 and 1 to left and right nodes
+    temp.left=a;temp.right=b;
+    //after combing, removing first two nodes in list which are already combined
+    li.remove(0);//removes first element which is now combined  -step1
+    li.remove(0);//removes 2nd element which comes on 1st position after deleting first in step1
+    li.add(temp);//adding new combined node to list
+    //print_list(li);  //For visualizing each combination step
+    }
+    //Sorting after combining to again repeat above on sorted frequency list
+    li.sort(new comp());    
+    it=li.iterator();//resetting list pointer to first node (head/root of tree)
+    if(li.size()==1){return (Node)it.next();} //base condition ,returning root of huffman tree    
+    }
+}
+    
+    //Function for finding path between root and given letter ch
+    public static void dfs(Node n,String ch){
+        Stack<Node> st=new Stack(); // stack for storing path
+        int freq=n.freq; // recording root freq to avoid it adding in path encoding
+        find_path_and_encode(st,n,ch,freq);
+    }
+    
+    //A simple utility function to print stack (Used for printing path)
+    public static void print_path(Stack<Node> st){
+    for(int i=0;i<st.size();i++){
+    System.out.print(st.elementAt(i).data+" ");
+    }
+    }
+    
+    public static void find_path_and_encode(Stack<Node> st,Node root,String s,int f){
+    //Base condition
+    if(root!= null){
+    if(root.freq!=f){st.push(root);} // avoiding root to add in path/encoding bits
+    if(root.letr.equals(s)){print_path(st);return;} // Recursion stopping condition when path gets founded
+    find_path_and_encode(st,root.left,s,f);
+    find_path_and_encode(st,root.right,s,f); 
+    //Popping if path not found in right or left of this node,because we previously 
+    //pushed this node in taking a mindset that it might be in path
+    st.pop();    
+    }    
+    }
+    
+    public static void main(String args[]){
+    List <Node> li=new LinkedList<>();
+    Scanner in=new Scanner(System.in);
+    System.out.println("Enter number of distinct letters ");
+    int n=in.nextInt();
+    String s[]=new String[n];
+    System.out.print("Enter letters with its frequncy to encode\n");
+    for(int  i=0;i<n;i++){
+    Node a=new Node();
+    System.out.print("Enter letter : ");
+    a.letr=in.next();s[i]=a.letr;
+    System.out.print("Enter frequncy : ");
+    a.freq=in.nextInt();System.out.println();
+    li.add(a);
+    }
+    Node root=new Node();
+    root=make_huffmann_tree(li);
+    System.out.println("Letter\t\tEncoded Form");
+    for(int i=0;i<n;i++){
+    System.out.print(s[i]+"\t\t");dfs(root,s[i]);System.out.println();}
+     }
+}