Tree: Huffman Decoding
Problem Statement :
Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only the leaves will contain a letter and its frequency count. All other nodes will contain a null instead of a character, and the count of the frequency of all of it and its descendant characters. For instance, consider the string ABRACADABRA. There are a total of 11 characters in the string. This number should match the count in the ultimately determined root of the tree. Our frequencies are A = 5 , B =2, R = 2, C = 1 and D = 1 . The two smallest frequencies are for C and D , both equal to 1 , so we'll create a tree with them. The root node will contain the sum of the counts of its descendants, in this case 1 + 1 = 2. The left node will be the first character encountered, C , and the right will contain D. Next we have 3 items with a character count of 2: the tree we just created, the character B and the character R. The tree came first, so it will go on the left of our new root node. B will go on the right. Repeat until the tree is complete, then fill in the 1's and 0's for the edges. The finished graph looks like: Input characters are only present in the leaves. Internal nodes have a character value of ϕ (NULL). We can determine that our values for characters are: A - 0 B - 111 C - 1100 D - 1101 R - 10 Our Huffman encoded string is: A B R A C A D A B R A 0 111 10 0 1100 0 1101 0 111 10 0 or 01111001100011010111100 To avoid ambiguity, Huffman encoding is a prefix free encoding technique. No codeword appears as a prefix of any other codeword. To decode the encoded string, follow the zeros and ones to a leaf and return the character there. You are given pointer to the root of the Huffman tree and a binary coded string to decode. You need to print the decoded string. Function Description Complete the function decode_huff in the editor below. It must return the decoded string. decode_huff has the following parameters: root: a reference to the root node of the Huffman tree s: a Huffman encoded string Input Format There is one line of input containing the plain string, . Background code creates the Huffman tree then passes the head node and the encoded string to the function. Constraints 1 <= |s| <= 25 Output Format Output the decoded string on a single line.
Solution :
Solution in C :
In C++ :
/*
The structure of the node is
typedef struct node
{
int freq;
char data;
node * left;
node * right;
}node;
*/
void decode_huff(node * root,string s)
{
node *treeroot = root;
// char *result = (char*)malloc(sizeof(char)*len);
int i=0,k=0;
while (s[i]!='\0'){
// printf("s[%d]%c\n", i, s[i]);
if (s[i]== '1' ){
root=root->right;
// printf("i was here");
if (root->data != '\0'){
// printf("i was here");
printf("%c", root->data);
root = treeroot;
}
i++;
}
else {
root=root->left;
if (root->data != '\0'){
printf("%c", root->data);
root = treeroot;
}
i++;
}
}
}
In Java :
/*
class Node
public int frequency; // the frequency of this tree
public char data;
public Node left, right;
*/
void decode(String S ,Node root)
{
Node temp=root;
String ans="";
for(int i=0;i<S.length();i++){
// System.out.println("er1");
if(S.charAt(i)=='0')
temp=temp.left;
else
temp=temp.right;
if(temp.right==null && temp.left==null)
{
ans+=(temp.data);
temp=root;
}
}
System.out.println(ans);
}
In python3 :
"""class Node:
def __init__(self, freq,data):
self.freq= freq
self.data=data
self.left = None
self.right = None
"""
def decodeHuff(root, s):
s=list(s)
s=[int(i) for i in s]
while len(s)!=0 :
decodeHuf(root,s,root)
# Enter your code here. Read input from STDIN. Print output to STDOUT
def decodeHuf(root, s,parent):
if not (root.left or root.right) :
print(root.data,end='')
return
if len(s)==0 :
return
x=s[0]
del s[0]
if x == 1 :
decodeHuf(root.right,s,parent)
else :
decodeHuf(root.left,s,parent)
#Enter Your Code Here
View More Similar Problems
Queue using Two Stacks
A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que
View Solution →Castle on the Grid
You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):
View Solution →Down to Zero II
You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.
View Solution →Truck Tour
Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr
View Solution →Queries with Fixed Length
Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon
View Solution →QHEAP1
This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element
View Solution →