Given a string str of length L, the task is to find the first occurrence of a non-repeating character in the string.
Examples:
Input: str = “neveropen”
Output: fInput: str = “programmer”
Output: p
Note: Please refer this article for HashMap approach and this article as space-optimized approach.
Linked List Approach: The idea is to use Linked List to keep track of the unique elements in the string. Below is the illustration of the approach:
- Iterate over the string for each character in the string and add the character in the Linked List on basis of the below conditions:
- If the character is already present in the Linked List, then remove the existing character node from the linked list.
- Otherwise, Add the character into the linked list.
- Finally, the character at the first node of the linked list is the first non-repeating character of the string.
Below is the implementation of the above approach:
C++14
// C++ implementation to find the// first non-repeating element of the// string using Linked List#include <bits/stdc++.h>using namespace std;// Function to find the first// non-repeating element of the// given string using Linked Listvoid firstNonRepElement(string str){ map<char, int> mpp; for(auto i:str) { mpp[i]++; } for(auto i:str) { if (mpp[i] == 1) { cout << i << endl; return; } } return;}// Driver Codeint main(){ string str = "neveropen"; // Function Call firstNonRepElement(str);}// This code is contributed by mohit kumar 29 |
Java
// Java implementation to find the // first non-repeating element // of the string using Linked List import java.util.LinkedList; public class FirstNonRepeatingElement { // Function to find the first // non-repeating element of the // given string using Linked List static void firstNonRepElement(String str) { LinkedList<Character> list = new LinkedList<Character>(); list.add(str.charAt(0)); for (int i = 1; i < str.length(); i++) { if (list.contains(str.charAt(i))) list.remove(list.indexOf( str.charAt(i))); else list.add(str.charAt(i)); } System.out.println(list.get(0)); } // Driver Code public static void main(String[] args) { String str = "neveropen"; // Function Call firstNonRepElement(str); } } |
Python3
# Python3 implementation to find the# first non-repeating element of the# string using Linked Listimport collections# Function to find the first# non-repeating element of the# given string using Linked Listdef firstNonRepElement(str): # Make list as a linkedlist list = collections.deque()# linkedlist list.append(str[0]) for i in range(len(str)): if str[i] in list: list.remove(str[i]) else: list.append(str[i]) print(list[0])# Driver Codeif __name__=='__main__': str = "neveropen"; # Function Call firstNonRepElement(str); # This code is contributed by pratham76 |
C#
// C# implementation to find the// first non-repeating element// of the string using Linked Listusing System; using System.Collections; using System.Collections.Generic;class FirstNonRepeatingElement{// Function to find the first// non-repeating element of the// given string using Linked Liststatic void firstNonRepElement(string str){ LinkedList<char> list = new LinkedList<char>(); list.AddLast(str[0]); for(int i = 1; i < str.Length; i++) { if (list.Contains(str[i])) list.Remove(str[i]); else list.AddLast(str[i]); } Console.Write(list.First.Value);}// Driver Codepublic static void Main(string[] args){ string str = "neveropen"; // Function call firstNonRepElement(str);}}// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript implementation to find the// first non-repeating element// of the string using Linked List // Function to find the first // non-repeating element of the // given string using Linked Listfunction firstNonRepElement(str){ let list = []; list.push(str[0]); for (let i = 1; i < str.length; i++) { if (list.includes(str[i])) list.splice(list.indexOf(str[i]),1); else list.push(str[i]); } document.write(list[0]);}// Driver Codelet str = "neveropen";// Function CallfirstNonRepElement(str);// This code is contributed by patel2127</script> |
Output
f
Performance Analysis:
- Time Complexity: O(N * 26)
- Auxiliary Space: O(N)
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