Given a string str containing only lowercase characters. The problem is to print the characters along with their frequency in the order of their occurrence using Binary Tree
Examples:
Input: str = “aaaabbnnccccz”
Output: “a4b2n2c4z”
Explanation:
Input: str = “neveropen”
Output: g2e4k2s2for
Approach:
- Start with the first character in the string.
- Perform a level order insertion of the character in the Binary Tree
- Pick the next character:
- If the character has been seen and we encounter it during level order insertion increase the count of the node.
- If the character has not been seen so far, go to step number 2.
- Repeat the process for all the characters of the string.
- Print the level order traversal of the tree which should output the desired output.
Below is the implementation of the above approach:
C++
// C++ implementation of// the above approach#include <bits/stdc++.h>using namespace std;// Node in the tree where// data holds the character// of the string and cnt// holds the frequencystruct node { char data; int cnt; node *left, *right;};// Function to add a new// node to the Binary Treenode* add(char data){ // Create a new node and // populate its data part, // set cnt as 1 and left // and right children as NULL node* newnode = new node; newnode->data = data; newnode->cnt = 1; newnode->left = newnode->right = NULL; return newnode;}// Function to add a node// to the Binary Tree in// level ordernode* addinlvlorder(node* root, char data){ if (root == NULL) { return add(data); } // Use the queue data structure // for level order insertion // and push the root of tree to Queue queue<node*> Q; Q.push(root); while (!Q.empty()) { node* temp = Q.front(); Q.pop(); // If the character to be // inserted is present, // update the cnt if (temp->data == data) { temp->cnt++; break; } // If the left child is // empty add a new node // as the left child if (temp->left == NULL) { temp->left = add(data); break; } else { // If the character is present // as a left child, update the // cnt and exit the loop if (temp->left->data == data) { temp->left->cnt++; break; } // Add the left child to // the queue for further // processing Q.push(temp->left); } // If the right child is empty, // add a new node to the right if (temp->right == NULL) { temp->right = add(data); break; } else { // If the character is present // as a right child, update the // cnt and exit the loop if (temp->right->data == data) { temp->right->cnt++; break; } // Add the right child to // the queue for further // processing Q.push(temp->right); } } return root;}// Function to print the// level order traversal of// the Binary Treevoid printlvlorder(node* root){ // Add the root to the queue queue<node*> Q; Q.push(root); while (!Q.empty()) { node* temp = Q.front(); // If the cnt of the character // is more than one, display cnt if (temp->cnt > 1) { cout << temp->data << temp->cnt; } // If the cnt of character // is one, display character only else { cout << temp->data; } Q.pop(); // Add the left child to // the queue for further // processing if (temp->left != NULL) { Q.push(temp->left); } // Add the right child to // the queue for further // processing if (temp->right != NULL) { Q.push(temp->right); } }}// Driver codeint main(){ string s = "neveropen"; node* root = NULL; // Add individual characters // to the string one by one // in level order for (int i = 0; i < s.size(); i++) { root = addinlvlorder(root, s[i]); } // Print the level order // of the constructed // binary tree printlvlorder(root); return 0;} |
Java
// Java implementation of// the above approachimport java.util.*;class GFG{// Node in the tree where// data holds the character// of the String and cnt// holds the frequencystatic class node{ char data; int cnt; node left, right;};// Function to add a new// node to the Binary Treestatic node add(char data){ // Create a new node and // populate its data part, // set cnt as 1 and left // and right children as null node newnode = new node(); newnode.data = data; newnode.cnt = 1; newnode.left = newnode.right = null; return newnode;}// Function to add a node// to the Binary Tree in// level orderstatic node addinlvlorder(node root, char data){ if (root == null) { return add(data); } // Use the queue data structure // for level order insertion // and push the root of tree to Queue Queue<node> Q = new LinkedList<node>(); Q.add(root); while (!Q.isEmpty()) { node temp = Q.peek(); Q.remove(); // If the character to be // inserted is present, // update the cnt if (temp.data == data) { temp.cnt++; break; } // If the left child is // empty add a new node // as the left child if (temp.left == null) { temp.left = add(data); break; } else { // If the character is present // as a left child, update the // cnt and exit the loop if (temp.left.data == data) { temp.left.cnt++; break; } // Add the left child to // the queue for further // processing Q.add(temp.left); } // If the right child is empty, // add a new node to the right if (temp.right == null) { temp.right = add(data); break; } else { // If the character is present // as a right child, update the // cnt and exit the loop if (temp.right.data == data) { temp.right.cnt++; break; } // Add the right child to // the queue for further // processing Q.add(temp.right); } } return root;}// Function to print the// level order traversal of// the Binary Treestatic void printlvlorder(node root){ // Add the root to the queue Queue<node> Q = new LinkedList<node>(); Q.add(root); while (!Q.isEmpty()) { node temp = Q.peek(); // If the cnt of the character // is more than one, display cnt if (temp.cnt > 1) { System.out.print((temp.data +""+ temp.cnt)); } // If the cnt of character // is one, display character only else { System.out.print((char)temp.data); } Q.remove(); // Add the left child to // the queue for further // processing if (temp.left != null) { Q.add(temp.left); } // Add the right child to // the queue for further // processing if (temp.right != null) { Q.add(temp.right); } }}// Driver codepublic static void main(String[] args){ String s = "neveropen"; node root = null; // Add individual characters // to the String one by one // in level order for (int i = 0; i < s.length(); i++) { root = addinlvlorder(root, s.charAt(i)); } // Print the level order // of the constructed // binary tree printlvlorder(root);}}// This code is contributed by Rajput-Ji |
Python3
# Python implementation of# the above approach# Node in the tree where# data holds the character# of the String and cnt# holds the frequencyclass node: def __init__(self): self.data = '' self.cnt = 0 self.left = None self.right = None# Function to add a new# node to the Binary Treedef add(data): # Create a new node and # populate its data part, # set cnt as 1 and left # and right children as None newnode = node() newnode.data = data newnode.cnt = 1 newnode.left = newnode.right = None return newnode# Function to add a node# to the Binary Tree in# level orderdef addinlvlorder(root, data): if (root == None): return add(data) # Use the queue data structure # for level order insertion # and push the root of tree to Queue Q = [] Q.append(root) while (len(Q) != 0): temp = Q[0] Q = Q[1:] # If the character to be # inserted is present, # update the cnt if (temp.data == data): temp.cnt += 1 break # If the left child is # empty add a new node # as the left child if (temp.left == None): temp.left = add(data) break else: # If the character is present # as a left child, update the # cnt and exit the loop if (temp.left.data == data): temp.left.cnt += 1 break # push the left child to # the queue for further # processing Q.append(temp.left) # If the right child is empty, # add a new node to the right if (temp.right == None): temp.right = add(data) break else: # If the character is present # as a right child, update the # cnt and exit the loop if (temp.right.data == data): temp.right.cnt += 1 break # push the right child to # the queue for further # processing Q.append(temp.right) return root# Function to print the# level order traversal of# the Binary Treedef printlvlorder(root): # push the root to the queue Q = [] Q.append(root) while (len(Q) != 0): temp = Q[0] # If the cnt of the character # is more than one, display cnt if (temp.cnt > 1): print(f"{temp.data}{temp.cnt}",end="") # If the cnt of character # is one, display character only else: print(temp.data,end="") Q = Q[1:] # push the left child to # the queue for further # processing if (temp.left != None): Q.append(temp.left) # push the right child to # the queue for further # processing if (temp.right != None): Q.append(temp.right)# Driver codes = "neveropen"root = None# push individual characters# to the String one by one# in level orderfor i in range(len(s)): root = addinlvlorder(root, s[i]) # Print the level order# of the constructed# binary treeprintlvlorder(root)# This code is contributed by shinjanpatra |
C#
// C# implementation of// the above approachusing System;using System.Collections.Generic;class GFG{// Node in the tree where// data holds the character// of the String and cnt// holds the frequencypublic class node{ public char data; public int cnt; public node left, right;};// Function to add a new// node to the Binary Treestatic node add(char data){ // Create a new node and // populate its data part, // set cnt as 1 and left // and right children as null node newnode = new node(); newnode.data = data; newnode.cnt = 1; newnode.left = newnode.right = null; return newnode;}// Function to add a node// to the Binary Tree in// level orderstatic node addinlvlorder(node root, char data){ if (root == null) { return add(data); } // Use the queue data structure // for level order insertion // and push the root of tree to Queue List<node> Q = new List<node>(); Q.Add(root); while (Q.Count != 0) { node temp = Q[0]; Q.RemoveAt(0); // If the character to be // inserted is present, // update the cnt if (temp.data == data) { temp.cnt++; break; } // If the left child is // empty add a new node // as the left child if (temp.left == null) { temp.left = add(data); break; } else { // If the character is present // as a left child, update the // cnt and exit the loop if (temp.left.data == data) { temp.left.cnt++; break; } // Add the left child to // the queue for further // processing Q.Add(temp.left); } // If the right child is empty, // add a new node to the right if (temp.right == null) { temp.right = add(data); break; } else { // If the character is present // as a right child, update the // cnt and exit the loop if (temp.right.data == data) { temp.right.cnt++; break; } // Add the right child to // the queue for further // processing Q.Add(temp.right); } } return root;}// Function to print the// level order traversal of// the Binary Treestatic void printlvlorder(node root){ // Add the root to the queue List<node> Q = new List<node>(); Q.Add(root); while (Q.Count != 0) { node temp = Q[0]; // If the cnt of the character // is more than one, display cnt if (temp.cnt > 1) { Console.Write((temp.data +""+ temp.cnt)); } // If the cnt of character // is one, display character only else { Console.Write((char)temp.data); } Q.RemoveAt(0); // Add the left child to // the queue for further // processing if (temp.left != null) { Q.Add(temp.left); } // Add the right child to // the queue for further // processing if (temp.right != null) { Q.Add(temp.right); } }}// Driver codepublic static void Main(String[] args){ String s = "neveropen"; node root = null; // Add individual characters // to the String one by one // in level order for (int i = 0; i < s.Length; i++) { root = addinlvlorder(root, s[i]); } // Print the level order // of the constructed // binary tree printlvlorder(root);}}// This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript implementation of// the above approach// Node in the tree where// data holds the character// of the String and cnt// holds the frequencyclass node{ constructor() { this.data = ''; this.cnt = 0; this.left = null; this.right = null; }};// Function to add a new// node to the Binary Treefunction add(data){ // Create a new node and // populate its data part, // set cnt as 1 and left // and right children as null var newnode = new node(); newnode.data = data; newnode.cnt = 1; newnode.left = newnode.right = null; return newnode;}// Function to add a node// to the Binary Tree in// level orderfunction addinlvlorder(root, data){ if (root == null) { return add(data); } // Use the queue data structure // for level order insertion // and push the root of tree to Queue var Q = []; Q.push(root); while (Q.length != 0) { var temp = Q[0]; Q.shift(); // If the character to be // inserted is present, // update the cnt if (temp.data == data) { temp.cnt++; break; } // If the left child is // empty add a new node // as the left child if (temp.left == null) { temp.left = add(data); break; } else { // If the character is present // as a left child, update the // cnt and exit the loop if (temp.left.data == data) { temp.left.cnt++; break; } // push the left child to // the queue for further // processing Q.push(temp.left); } // If the right child is empty, // add a new node to the right if (temp.right == null) { temp.right = add(data); break; } else { // If the character is present // as a right child, update the // cnt and exit the loop if (temp.right.data == data) { temp.right.cnt++; break; } // push the right child to // the queue for further // processing Q.push(temp.right); } } return root;}// Function to print the// level order traversal of// the Binary Treefunction printlvlorder(root){ // push the root to the queue var Q = []; Q.push(root); while (Q.length != 0) { var temp = Q[0]; // If the cnt of the character // is more than one, display cnt if (temp.cnt > 1) { document.write((temp.data +""+ temp.cnt)); } // If the cnt of character // is one, display character only else { document.write(temp.data); } Q.shift(); // push the left child to // the queue for further // processing if (temp.left != null) { Q.push(temp.left); } // push the right child to // the queue for further // processing if (temp.right != null) { Q.push(temp.right); } }}// Driver codevar s = "neveropen";var root = null;// push individual characters// to the String one by one// in level orderfor(var i = 0; i < s.length; i++){ root = addinlvlorder(root, s[i]);}// Print the level order// of the constructed// binary treeprintlvlorder(root);</script> |
Output:
g2e4k2s2for
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