Introduction
A good programmer is the one who can write the most optimized codes. To develop this ability, the knowledge of data structures and algorithms becomes essential. Due to this reason, the knowledge of Data Structures and Algorithms is frequently tested in interviews for SDE(Software Development Engineering) roles. The importance of DSA cannot be emphasized enough, especially if you aim to get placed in product-based companies like Google, Amazon, Microsoft, etc.
For questions related to each data structure, refer to the blog Must-Do Coding Interview Questions for Product Based Companies.
This blog will discuss the interview problem: find the first non-repeating character in a stream previously asked in companies like Amazon, Adobe, Microsoft, Yahoo, etc.
Problem Statement
Given an input stream of characters consisting only of lowercase alphabets. Find the first non-repeating character in the input string each time a new character is inserted into the stream. If there is no non-repeating character, then append '-1' to the answer.
Example
Input:
aabcbc
Output:
a -1 b b c -1
Explanation:
- When the input stream is "a," the first non-repeating character, "a," is appended.
- When the input stream is "aa," there is no first non-repeating character, so "-1" is appended.
- When the input stream is "aab," the first non-repeating character, "b," is appended.
- When the input stream is "aabc," the first non-repeating character, "b," is appended.
- When the input stream is "aabcb," the first non-repeating character, "c," is appended.
- When the input stream is "aabcbc," there is no first non-repeating character, so "-1" is appended.
Explanation
Now let's see various approaches to finding the first non-repeating character in a stream.
Approach 1: Using Queue
In this approach, a queue data structure and a character array are used. The frequency of the characters is stored in the frequencyCount array, and the characters are stored in the queue. If the frequencyCount is greater than 1, the character is removed from the queue as it is repeating. Else, the front element of the queue is displayed. If the queue is empty, i.e., no more non-repeating characters are present, -1 is printed.
Steps
- Create a character array to store the frequency count of the characters.
- Create a queue to store the characters.
- Traverse the string.
- And the character and increment the frequency count.
- If the queue contains repeating characters, remove the character from the queue.
- If the queue contains repeating non-repeating characters, display the front element of the queue and break the while loop.
- If the queue is empty, i.e., no more non-repeating characters are present, -1 is printed.
Working
Code
import java.util.LinkedList;
import java.util.Queue;
public class Main {
public static String firstNonRepeatingCharacter(String str) {
StringBuilder resultantString = new StringBuilder();
int[] characterFrequency = new int[26];
// queue to store the characters
Queue<Character> queue = new LinkedList<Character>();
// traverse whole stream of characters
for (int i = 0; i < str.length(); i++) {
char ch = str.charAt(i);
// push the character into the queue
queue.add(ch);
// increment the frequency count
characterFrequency[ch - 'a']++;
// check for the non repeating character
while (!queue.isEmpty()) {
// when the character is repeated
if (characterFrequency[queue.peek() - 'a'] > 1)
queue.remove();
// when the character is not repeating
else {
resultantString.append(queue.peek() + " ");
break;
}
}
// if there is no non-repeating character
if (queue.isEmpty())
resultantString.append("-1 ");
}
return resultantString.toString();
}
public static void main(String[] args) {
String str = "aabcbc";
String result = firstNonRepeatingCharacter(str);
System.out.println(result);
}
}
Output
a -1 b b c -1
Complexity Analysis
Time Complexity: O(n) as the string is traversed once.
Space Complexity: O(n) as extra space is required to store the characters in the queue.
Where "n" is the number of characters in the string.
Approach 2: Using Doubly Linked List
In this approach, a Doubly Linked List and a hashmap are used. The Doubly Linked List stores the characters, and the hashmap stores the character as the key and the node( storing the character) as the value. If the hashmap does not contain the character, the character is stored in the hashmap and doubly linked list. Else, the value of the character is set to null in the hashmap.
Steps
- Create a hashmap to store the character and node.
- Create a doubly linked list to store the characters.
- Traverse the string.
- If the hashmap does not contain the character:
- Add character to the doubly linked list.
- Add the node containing character to the hashmap with character as the key.
- Add the character stored in the head of the doubly linked list to the resultant string.
- If the hashmap contains the character:
- If the character is repeating and the node is not null, delete the node and store the value as null in the hashmap.
- If the head is null, add -1 to the resultant string. Else add the character value of the head node.
Working
Code
import java.util.HashMap;
// node of the doubly linked list
class Node {
char ch;
Node previous;
Node next;
Node(char ch) {
this.ch = ch;
}
}
public class Main {
// head and tail of the doubly linked list
Node head = null, tail = null;
// add node at the end of the doubly linked list
public void addNode(char ch) {
Node newNode = new Node(ch);
// if doubly linked list is empty
if (head == null) {
head = tail = newNode;
return;
}
// if doubly linked list is not empty
tail.next = newNode;
newNode.previous = tail;
tail = newNode;
}
// delete the node of the doubly linked list
void deleteNode(Node del) {
// if doubly linked list is empty or the node to be deleted is empty
if (head == null || del == null) {
return;
}
// delete head node
if (head == del) {
head = del.next;
}
// delete tail node
if (tail == del)
tail = tail.previous;
// change next pointer only if node to be deleted is not the tail node
if (del.next != null) {
del.next.previous = del.previous;
}
// change previous pointer only if node to be deleted is not the first node
if (del.previous != null) {
del.previous.next = del.next;
}
return;
}
// function that returns the string of first non repeating characters
public static String firstNonRepeatingCharacter(String str) {
StringBuilder resultantString = new StringBuilder();
// hashmap to store the character and node
HashMap<Character, Node> hashmap = new HashMap<Character, Node>();
// doubly linked list
Main dll = new Main();
for (int i = 0; i < str.length(); i++) {
char ch = str.charAt(i);
// if the hashmap does not contain the character
if (!hashmap.containsKey(ch)) {
// add the node to the doubly linked list
dll.addNode(ch);
// add the character to hashmap
hashmap.put(ch, dll.tail);
// add the character to the resultant string
resultantString.append(dll.head.ch + " ");
} else {
// if the character is encountered for the second time
if (hashmap.get(ch) != null) {
dll.deleteNode(hashmap.get(ch));
// replace the node of the respective character with null
hashmap.replace(ch, null);
}
if (dll.head == null) {
resultantString.append("-1 ");
} else {
resultantString.append(dll.head.ch + " ");
}
}
}
return resultantString.toString();
}
// driver Code
public static void main(String[] args) {
String str = "aabcbc";
String result = firstNonRepeatingCharacter(str);
System.out.println(result);
}
}
Output:
a -1 b b c -1
Complexity Analysis
Time Complexity: O(n) as the string is traversed once.
Space Complexity: O(n) as extra space is required to store the doubly linked list and the hashmap.
Where "n" is the number of characters in the string.
Frequently Asked Questions
What is the first non-repeating character in a stream stack?
If the character given is aabccdd. Here the occurrence of a is 2, b is 1, c is 2, and d is 2. The character which is not repeated in the first time is b. So, the first non-repeating character is b.
How to find first non repeated character in a string using stream?
Scan the string, then save the number of characters in a hashmap. Get a character count for each character in String by traversing the Map. When we are reading a string from start to last, if any character's count is 1, return that character.
What is the program to find the first non repeated character in a word?
In Java, we may locate the first non-repeating character in a string using the indexOf() and lastIndexOf() methods.
public class Main {
public static void main(String args[]) {
String str ="abbbcccda";
for(char i :str.toCharArray()){
if ( str.indexOf(i) == str.lastIndexOf(i)) {
System.out.println("First non-repeating character is: "+i);
break;
}
}
}
}Conclusion
This blog discussed the various methods to find the first non-repeating character in a stream along with their implementation using data structures such as queue, array, hashmaps, and doubly-linked lists.
Recommended Problems
- Top Google Coding Interview Questions
- Top Amazon Coding Interview Questions
- Top Microsoft Coding Interview Questions
Also check out some of the Guided Paths on topics such as Data Structure and Algorithms, Competitive Programming, etc. as well as some Contests and some more Interview Experiences curated by top Industry Experts only on Coding Ninjas Studio.
Cheers!
