Efficient Huffman Coding for Sorted Input | Greedy Algo-4
We recommend to read following post as a prerequisite for this.
Greedy Algorithms | Set 3 (Huffman Coding)
Time complexity of the algorithm discussed in above post is O(nLogn). If we know that the given array is sorted (by non-decreasing order of frequency), we can generate Huffman codes in O(n) time. Following is a O(n) algorithm for sorted input.
1. Create two empty queues.
2. Create a leaf node for each unique character and Enqueue it to the first queue in non-decreasing order of frequency. Initially second queue is empty.
3. Dequeue two nodes with the minimum frequency by examining the front of both queues. Repeat the following steps two times
1. If second queue is empty, dequeue from first queue.
2. If first queue is empty, dequeue from second queue.
3. Else, compare the front of two queues and dequeue the minimum.
4. Create a new internal node with frequency equal to the sum of the two nodes frequencies. Make the first Dequeued node as its left child and the second Dequeued node as right child. Enqueue this node to second queue.
5. Repeat steps#3 and #4 while there is more than one node in the queues. The remaining node is the root node and the tree is complete.
C++
// C++ Program for Efficient Huffman Coding for Sorted input #include <bits/stdc++.h> using namespace std; // This constant can be avoided by explicitly // calculating height of Huffman Tree #define MAX_TREE_HT 100 // A node of huffman tree class QueueNode { public : char data; unsigned freq; QueueNode *left, *right; }; // Structure for Queue: collection // of Huffman Tree nodes (or QueueNodes) class Queue { public : int front, rear; int capacity; QueueNode** array; }; // A utility function to create a new Queuenode QueueNode* newNode( char data, unsigned freq) { QueueNode* temp = new QueueNode[( sizeof (QueueNode))]; temp->left = temp->right = NULL; temp->data = data; temp->freq = freq; return temp; } // A utility function to create a Queue of given capacity Queue* createQueue( int capacity) { Queue* queue = new Queue[( sizeof (Queue))]; queue->front = queue->rear = -1; queue->capacity = capacity; queue->array = new QueueNode*[(queue->capacity * sizeof (QueueNode*))]; return queue; } // A utility function to check if size of given queue is 1 int isSizeOne(Queue* queue) { return queue->front == queue->rear && queue->front != -1; } // A utility function to check if given queue is empty int isEmpty(Queue* queue) { return queue->front == -1; } // A utility function to check if given queue is full int isFull(Queue* queue) { return queue->rear == queue->capacity - 1; } // A utility function to add an item to queue void enQueue(Queue* queue, QueueNode* item) { if (isFull(queue)) return ; queue->array[++queue->rear] = item; if (queue->front == -1) ++queue->front; } // A utility function to remove an item from queue QueueNode* deQueue(Queue* queue) { if (isEmpty(queue)) return NULL; QueueNode* temp = queue->array[queue->front]; if (queue->front == queue ->rear) // If there is only one item in queue queue->front = queue->rear = -1; else ++queue->front; return temp; } // A utility function to get from of queue QueueNode* getFront(Queue* queue) { if (isEmpty(queue)) return NULL; return queue->array[queue->front]; } /* A function to get minimum item from two queues */ QueueNode* findMin(Queue* firstQueue, Queue* secondQueue) { // Step 3.a: If first queue is empty, dequeue from // second queue if (isEmpty(firstQueue)) return deQueue(secondQueue); // Step 3.b: If second queue is empty, dequeue from // first queue if (isEmpty(secondQueue)) return deQueue(firstQueue); // Step 3.c: Else, compare the front of two queues and // dequeue minimum if (getFront(firstQueue)->freq < getFront(secondQueue)->freq) return deQueue(firstQueue); return deQueue(secondQueue); } // Utility function to check if this node is leaf int isLeaf(QueueNode* root) { return !(root->left) && !(root->right); } // A utility function to print an array of size n void printArr( int arr[], int n) { int i; for (i = 0; i < n; ++i) cout << arr[i]; cout << endl; } // The main function that builds Huffman tree QueueNode* buildHuffmanTree( char data[], int freq[], int size) { QueueNode *left, *right, *top; // Step 1: Create two empty queues Queue* firstQueue = createQueue(size); Queue* secondQueue = createQueue(size); // Step 2:Create a leaf node for each unique character // and Enqueue it to the first queue in non-decreasing // order of frequency. Initially second queue is empty for ( int i = 0; i < size; ++i) enQueue(firstQueue, newNode(data[i], freq[i])); // Run while Queues contain more than one node. Finally, // first queue will be empty and second queue will // contain only one node while ( !(isEmpty(firstQueue) && isSizeOne(secondQueue))) { // Step 3: Dequeue two nodes with the minimum // frequency by examining the front of both queues left = findMin(firstQueue, secondQueue); right = findMin(firstQueue, secondQueue); // Step 4: Create a new internal node with frequency // equal to the sum of the two nodes frequencies. // Enqueue this node to second queue. top = newNode( '$' , left->freq + right->freq); top->left = left; top->right = right; enQueue(secondQueue, top); } return deQueue(secondQueue); } // Prints huffman codes from the root of Huffman Tree. It // uses arr[] to store codes void printCodes(QueueNode* root, int arr[], int top) { // Assign 0 to left edge and recur if (root->left) { arr[top] = 0; printCodes(root->left, arr, top + 1); } // Assign 1 to right edge and recur if (root->right) { arr[top] = 1; printCodes(root->right, arr, top + 1); } // If this is a leaf node, then it contains one of the // input characters, print the character and its code // from arr[] if (isLeaf(root)) { cout << root->data << ": " ; printArr(arr, top); } } // The main function that builds a Huffman Tree and print // codes by traversing the built Huffman Tree void HuffmanCodes( char data[], int freq[], int size) { // Construct Huffman Tree QueueNode* root = buildHuffmanTree(data, freq, size); // Print Huffman codes using the Huffman tree built // above int arr[MAX_TREE_HT], top = 0; printCodes(root, arr, top); } // Driver code int main() { char arr[] = { 'a' , 'b' , 'c' , 'd' , 'e' , 'f' }; int freq[] = { 5, 9, 12, 13, 16, 45 }; int size = sizeof (arr) / sizeof (arr[0]); HuffmanCodes(arr, freq, size); return 0; } // This code is contributed by rathbhupendra |
C
// C Program for Efficient Huffman Coding for Sorted input #include <stdio.h> #include <stdlib.h> // This constant can be avoided by explicitly calculating // height of Huffman Tree #define MAX_TREE_HT 100 // A node of huffman tree struct QueueNode { char data; unsigned freq; struct QueueNode *left, *right; }; // Structure for Queue: collection of Huffman Tree nodes (or // QueueNodes) struct Queue { int front, rear; int capacity; struct QueueNode** array; }; // A utility function to create a new Queuenode struct QueueNode* newNode( char data, unsigned freq) { struct QueueNode* temp = ( struct QueueNode*) malloc ( sizeof ( struct QueueNode)); temp->left = temp->right = NULL; temp->data = data; temp->freq = freq; return temp; } // A utility function to create a Queue of given capacity struct Queue* createQueue( int capacity) { struct Queue* queue = ( struct Queue*) malloc ( sizeof ( struct Queue)); queue->front = queue->rear = -1; queue->capacity = capacity; queue->array = ( struct QueueNode**) malloc ( queue->capacity * sizeof ( struct QueueNode*)); return queue; } // A utility function to check if size of given queue is 1 int isSizeOne( struct Queue* queue) { return queue->front == queue->rear && queue->front != -1; } // A utility function to check if given queue is empty int isEmpty( struct Queue* queue) { return queue->front == -1; } // A utility function to check if given queue is full int isFull( struct Queue* queue) { return queue->rear == queue->capacity - 1; } // A utility function to add an item to queue void enQueue( struct Queue* queue, struct QueueNode* item) { if (isFull(queue)) return ; queue->array[++queue->rear] = item; if (queue->front == -1) ++queue->front; } // A utility function to remove an item from queue struct QueueNode* deQueue( struct Queue* queue) { if (isEmpty(queue)) return NULL; struct QueueNode* temp = queue->array[queue->front]; if (queue->front == queue ->rear) // If there is only one item in queue queue->front = queue->rear = -1; else ++queue->front; return temp; } // A utility function to get from of queue struct QueueNode* getFront( struct Queue* queue) { if (isEmpty(queue)) return NULL; return queue->array[queue->front]; } /* A function to get minimum item from two queues */ struct QueueNode* findMin( struct Queue* firstQueue, struct Queue* secondQueue) { // Step 3.a: If first queue is empty, dequeue from // second queue if (isEmpty(firstQueue)) return deQueue(secondQueue); // Step 3.b: If second queue is empty, dequeue from // first queue if (isEmpty(secondQueue)) return deQueue(firstQueue); // Step 3.c: Else, compare the front of two queues and // dequeue minimum if (getFront(firstQueue)->freq < getFront(secondQueue)->freq) return deQueue(firstQueue); return deQueue(secondQueue); } // Utility function to check if this node is leaf int isLeaf( struct QueueNode* root) { return !(root->left) && !(root->right); } // A utility function to print an array of size n void printArr( int arr[], int n) { int i; for (i = 0; i < n; ++i) printf ( "%d" , arr[i]); printf ( "\n" ); } // The main function that builds Huffman tree struct QueueNode* buildHuffmanTree( char data[], int freq[], int size) { struct QueueNode *left, *right, *top; // Step 1: Create two empty queues struct Queue* firstQueue = createQueue(size); struct Queue* secondQueue = createQueue(size); // Step 2:Create a leaf node for each unique character // and Enqueue it to the first queue in non-decreasing // order of frequency. Initially second queue is empty for ( int i = 0; i < size; ++i) enQueue(firstQueue, newNode(data[i], freq[i])); // Run while Queues contain more than one node. Finally, // first queue will be empty and second queue will // contain only one node while ( !(isEmpty(firstQueue) && isSizeOne(secondQueue))) { // Step 3: Dequeue two nodes with the minimum // frequency by examining the front of both queues left = findMin(firstQueue, secondQueue); right = findMin(firstQueue, secondQueue); // Step 4: Create a new internal node with frequency // equal to the sum of the two nodes frequencies. // Enqueue this node to second queue. top = newNode( '$' , left->freq + right->freq); top->left = left; top->right = right; enQueue(secondQueue, top); } return deQueue(secondQueue); } // Prints huffman codes from the root of Huffman Tree. It // uses arr[] to store codes void printCodes( struct QueueNode* root, int arr[], int top) { // Assign 0 to left edge and recur if (root->left) { arr[top] = 0; printCodes(root->left, arr, top + 1); } // Assign 1 to right edge and recur if (root->right) { arr[top] = 1; printCodes(root->right, arr, top + 1); } // If this is a leaf node, then it contains one of the // input characters, print the character and its code // from arr[] if (isLeaf(root)) { printf ( "%c: " , root->data); printArr(arr, top); } } // The main function that builds a Huffman Tree and print // codes by traversing the built Huffman Tree void HuffmanCodes( char data[], int freq[], int size) { // Construct Huffman Tree struct QueueNode* root = buildHuffmanTree(data, freq, size); // Print Huffman codes using the Huffman tree built // above int arr[MAX_TREE_HT], top = 0; printCodes(root, arr, top); } // Driver program to test above functions int main() { char arr[] = { 'a' , 'b' , 'c' , 'd' , 'e' , 'f' }; int freq[] = { 5, 9, 12, 13, 16, 45 }; int size = sizeof (arr) / sizeof (arr[0]); HuffmanCodes(arr, freq, size); return 0; } |
Python3
# Python3 program for Efficient Huffman Coding # for Sorted input # Class for the nodes of the Huffman tree class QueueNode: def __init__( self , data = None , freq = None , left = None , right = None ): self .data = data self .freq = freq self .left = left self .right = right # Function to check if the following # node is a leaf node def isLeaf( self ): return ( self .left = = None and self .right = = None ) # Class for the two Queues class Queue: def __init__( self ): self .queue = [] # Function for checking if the # queue has only 1 node def isSizeOne( self ): return len ( self .queue) = = 1 # Function for checking if # the queue is empty def isEmpty( self ): return self .queue = = [] # Function to add item to the queue def enqueue( self , x): self .queue.append(x) # Function to remove item from the queue def dequeue( self ): return self .queue.pop( 0 ) # Function to get minimum item from two queues def findMin(firstQueue, secondQueue): # Step 3.1: If second queue is empty, # dequeue from first queue if secondQueue.isEmpty(): return firstQueue.dequeue() # Step 3.2: If first queue is empty, # dequeue from second queue if firstQueue.isEmpty(): return secondQueue.dequeue() # Step 3.3: Else, compare the front of # two queues and dequeue minimum if (firstQueue.queue[ 0 ].freq < secondQueue.queue[ 0 ].freq): return firstQueue.dequeue() return secondQueue.dequeue() # The main function that builds Huffman tree def buildHuffmanTree(data, freq, size): # Step 1: Create two empty queues firstQueue = Queue() secondQueue = Queue() # Step 2: Create a leaf node for each unique # character and Enqueue it to the first queue # in non-decreasing order of frequency. # Initially second queue is empty. for i in range (size): firstQueue.enqueue(QueueNode(data[i], freq[i])) # Run while Queues contain more than one node. # Finally, first queue will be empty and # second queue will contain only one node while not (firstQueue.isEmpty() and secondQueue.isSizeOne()): # Step 3: Dequeue two nodes with the minimum # frequency by examining the front of both queues left = findMin(firstQueue, secondQueue) right = findMin(firstQueue, secondQueue) # Step 4: Create a new internal node with # frequency equal to the sum of the two # nodes frequencies. Enqueue this node # to second queue. top = QueueNode( "$" , left.freq + right.freq, left, right) secondQueue.enqueue(top) return secondQueue.dequeue() # Prints huffman codes from the root of # Huffman tree. It uses arr[] to store codes def printCodes(root, arr): # Assign 0 to left edge and recur if root.left: arr.append( 0 ) printCodes(root.left, arr) arr.pop( - 1 ) # Assign 1 to right edge and recur if root.right: arr.append( 1 ) printCodes(root.right, arr) arr.pop( - 1 ) # If this is a leaf node, then it contains # one of the input characters, print the # character and its code from arr[] if root.isLeaf(): print (f "{root.data}: " , end = "") for i in arr: print (i, end = "") print () # The main function that builds a Huffman # tree and print codes by traversing the # built Huffman tree def HuffmanCodes(data, freq, size): # Construct Huffman Tree root = buildHuffmanTree(data, freq, size) # Print Huffman codes using the Huffman # tree built above arr = [] printCodes(root, arr) # Driver code arr = [ "a" , "b" , "c" , "d" , "e" , "f" ] freq = [ 5 , 9 , 12 , 13 , 16 , 45 ] size = len (arr) HuffmanCodes(arr, freq, size) # This code is contributed by Kevin Joshi |
C++14
// Clean c++ stl code to generate huffman codes if the array // is sorted in non-decreasing order #include <bits/stdc++.h> using namespace std; // Node structure for creating a binary tree struct Node { char ch; int freq; Node* left; Node* right; Node( char c, int f, Node* l = nullptr, Node* r = nullptr) : ch(c) , freq(f) , left(l) , right(r){}; }; // Find the min freq node between q1 and q2 Node* minNode(queue<Node*>& q1, queue<Node*>& q2) { Node* temp; if (q1.empty()) { temp = q2.front(); q2.pop(); return temp; } if (q2.empty()) { temp = q1.front(); q1.pop(); return temp; } if (q1.front()->freq < q2.front()->freq) { temp = q1.front(); q1.pop(); return temp; } else { temp = q2.front(); q2.pop(); return temp; } } // Function to print the generated huffman codes void printHuffmanCodes(Node* root, string str = "" ) { if (!root) return ; if (root->ch != '$' ) { cout << root->ch << ": " << str << '\n' ; return ; } printHuffmanCodes(root->left, str + "0" ); printHuffmanCodes(root->right, str + "1" ); return ; } // Function to generate huffman codes void generateHuffmanCode(vector<pair< char , int > > v) { if (!v.size()) return ; queue<Node*> q1; queue<Node*> q2; for ( auto it = v.begin(); it != v.end(); ++it) q1.push( new Node(it->first, it->second)); while (!q1.empty() or q2.size() > 1) { Node* l = minNode(q1, q2); Node* r = minNode(q1, q2); Node* node = new Node( '$' , l->freq + r->freq, l, r); q2.push(node); } printHuffmanCodes(q2.front()); return ; } int main() { vector<pair< char , int > > v = { { 'a' , 5 }, { 'b' , 9 }, { 'c' , 12 }, { 'd' , 13 }, { 'e' , 16 }, { 'f' , 45 } }; generateHuffmanCode(v); return 0; } |
Javascript
// JavaScript program for the above approach // Class for the nodes of the Huffman tree class QueueNode { constructor(data = null , freq = null , left = null , right = null ) { this .data = data; this .freq = freq; this .left = left; this .right = right; } // Function to check if the following // node is a leaf node isLeaf() { return ( this .left == null && this .right == null ); } } // Class for the two Queues class Queue { constructor() { this .queue = []; } // Function for checking if the // queue has only 1 node isSizeOne() { return this .queue.length == 1; } // Function for checking if // the queue is empty isEmpty() { return this .queue.length == 0; } // Function to add item to the queue enqueue(x) { this .queue.push(x); } // Function to remove item from the queue dequeue() { return this .queue.shift(); } } // Function to get minimum item from two queues function findMin(firstQueue, secondQueue) { // Step 3.1: If second queue is empty, // dequeue from first queue if (secondQueue.isEmpty()) { return firstQueue.dequeue(); } // Step 3.2: If first queue is empty, // dequeue from second queue if (firstQueue.isEmpty()) { return secondQueue.dequeue(); } // Step 3.3: Else, compare the front of // two queues and dequeue minimum if (firstQueue.queue[0].freq < secondQueue.queue[0].freq) { return firstQueue.dequeue(); } return secondQueue.dequeue(); } // The main function that builds Huffman tree function buildHuffmanTree(data, freq, size) { // Step 1: Create two empty queues let firstQueue = new Queue(); let secondQueue = new Queue(); // Step 2: Create a leaf node for each unique // character and Enqueue it to the first queue // in non-decreasing order of frequency. // Initially second queue is empty. for (let i = 0; i < size; i++) { firstQueue.enqueue( new QueueNode(data[i], freq[i])); } // Run while Queues contain more than one node. // Finally, first queue will be empty and // second queue will contain only one node while (!(firstQueue.isEmpty() && secondQueue.isSizeOne())) { // Step 3: Dequeue two nodes with the minimum // frequency by examining the front of both queues let left = findMin(firstQueue, secondQueue); let right = findMin(firstQueue, secondQueue); // Step 4: Create a new internal node with // frequency equal to the sum of the two // nodes frequencies. Enqueue this node // to second queue. let top = new QueueNode( "$" , left.freq + right.freq, left, right); secondQueue.enqueue(top); } return secondQueue.dequeue(); } // Prints huffman codes from the root of // Huffman tree. It uses arr[] to store codes function printCodes(root, arr) { // Assign 0 to left edge and recur if (root.left) { arr.push(0); printCodes(root.left, arr); arr.pop(); } // Assign 1 to right edge and recur if (root.right) { arr.push(1); printCodes(root.right, arr); arr.pop(); } // If this is a leaf node, then it contains // one of the input characters, print the // character and its code from arr[] if (root.isLeaf()) { let output = root.data + ": " ; for (let i = 0; i < arr.length; i++) { output += arr[i]; } console.log(output); } } // The main function that builds a Huffman // tree and print codes by traversing the // built Huffman tree function HuffmanCodes(data, freq, size) { // Construct Huffman Tree let root = buildHuffmanTree(data, freq, size); // Print Huffman codes using the Huffman // tree built above let arr = []; printCodes(root, arr); } // Driver code let arr = [ "a" , "b" , "c" , "d" , "e" , "f" ]; let freq = [5, 9, 12, 13, 16, 45]; let size = arr.length; HuffmanCodes(arr, freq, size); // This code is contributed by Prince Kumar |
Output:
f: 0 c: 100 d: 101 a: 1100 b: 1101 e: 111
Time complexity: O(n)
If the input is not sorted, it needs to be sorted first before it can be processed by the above algorithm. Sorting can be done using heap sort or merge-sort both of which run in Theta(nlogn). So, the overall time complexity becomes O(nlogn) for unsorted input.
Auxiliary Space: O(n)
Reference:
http://en.wikipedia.org/wiki/Huffman_coding
This article is compiled by Aashish Barnwal and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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