First Non Repeating Character Using Hash in DSA Java - Time & Space Complexity

Time Complexity: First Non Repeating Character Using Hash

O(n)

Understanding Time Complexity

We want to find how long it takes to find the first character that appears only once in a string using a hash map.

How does the time needed grow as the string gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

import java.util.HashMap;

public class Solution {
  public static char firstNonRepeatingChar(String s) {
    HashMap countMap = new HashMap<>();
    for (char c : s.toCharArray()) {
      countMap.put(c, countMap.getOrDefault(c, 0) + 1);
    }
    for (char c : s.toCharArray()) {
      if (countMap.get(c) == 1) return c;
    }
    return '_';
  }
}
    

This code counts how many times each character appears, then finds the first character with count one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two loops over the string characters.
• How many times: Each loop runs once over all characters, so twice in total.

How Execution Grows With Input

As the string length grows, the code goes through all characters twice.

Input Size (n)	Approx. Operations
10	About 20 steps (2 x 10)
100	About 200 steps (2 x 100)
1000	About 2000 steps (2 x 1000)

Pattern observation: The operations grow roughly twice the input size, which means it grows linearly.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the first non-repeating character grows directly with the length of the string.

Common Mistake

[X] Wrong: "Because there are two loops, the time complexity is O(n²)."

[OK] Correct: The loops run one after another, not nested, so their times add up, not multiply.

Interview Connect

Understanding this linear time approach shows you can use simple data structures like hash maps to solve problems efficiently, a skill valued in many coding challenges.

Self-Check

"What if we used a nested loop to check each character against all others instead of a hash map? How would the time complexity change?"

Practice

1. What does the "First Non Repeating Character Using Hash" algorithm primarily do?

easy

A. Reverses the string using a hash map

B. Counts the total number of characters in a string

C. Sorts characters of a string in alphabetical order

D. Finds the first character in a string that appears only once

Solution

1. Step 1: Understand the problem goal

   The algorithm aims to find the first character in the string that does not repeat anywhere else.

2. Step 2: Role of hash map

   A hash map stores character counts to quickly check which characters appear once.

3. Final Answer:

   Finds the first character in a string that appears only once -> Option D

4. Quick Check:

   First non repeating character = Finds the first character in a string that appears only once [OK]

Hint: Look for the first unique character using counts [OK]

Common Mistakes:

• Thinking it counts total characters only
• Confusing sorting with finding unique character
• Assuming it reverses the string

2. Which Java collection is best suited to store character counts for this problem?

easy

A. ArrayList<Character>

B. HashMap

C. LinkedList<Character>

D. TreeSet<Character>

Solution

1. Step 1: Identify data structure needs

   We need to map each character to its count, so a key-value structure is required.

2. Step 2: Choose appropriate Java collection

   HashMap stores characters as keys and their counts as values efficiently.

3. Final Answer:

   HashMap -> Option B

4. Quick Check:

   Use HashMap for character counts [OK]

Hint: Use HashMap for key-value pairs of char and count [OK]

Common Mistakes:

• Using list which doesn't map keys to counts
• Using TreeSet which stores unique sorted elements only
• Using LinkedList which is not key-value based

3. What is the output of the following Java code snippet?

String s = "swiss";
HashMap map = new HashMap<>();
for(char c : s.toCharArray()) {
  map.put(c, map.getOrDefault(c, 0) + 1);
}
for(char c : s.toCharArray()) {
  if(map.get(c) == 1) {
    System.out.println(c);
    break;
  }
}

medium

A. w

B. s

C. i

D. No output

Solution

1. Step 1: Count character frequencies

   Characters and counts: s=3, w=1, i=1.

2. Step 2: Find first character with count 1

   Iterate string: 's'(3), 'w'(1) -> first non repeating is 'w'.

3. Final Answer:

   w -> Option A

4. Quick Check:

   First unique char in "swiss" = w [OK]

Hint: Check counts then first unique char in order [OK]

Common Mistakes:

• Choosing 's' which repeats
• Choosing 'i' which is unique but appears after 'w'
• Assuming no output if not careful

4. Identify the error in this code snippet for finding the first non repeating character:

String s = "hello";
HashMap map = new HashMap<>();
for(int i = 0; i < s.length(); i++) {
  map.put(s.charAt(i), 1);
}
for(int i = 0; i < s.length(); i++) {
  if(map.get(s.charAt(i)) == 1) {
    System.out.println(s.charAt(i));
    break;
  }
}

medium

A. The map.put always sets count to 1, ignoring repeats

B. The loop should start from 1, not 0

C. The map should use Character keys, not int

D. The break statement is missing

Solution

1. Step 1: Analyze map population logic

   The code sets count to 1 for every character without incrementing, so repeats are not counted.

2. Step 2: Effect on finding non repeating character

   Since all counts are 1, the code incorrectly treats all characters as unique.

3. Final Answer:

   The map.put always sets count to 1, ignoring repeats -> Option A

4. Quick Check:

   Count update missing causes wrong results [OK]

Hint: Increment count instead of setting 1 every time [OK]

Common Mistakes:

• Not incrementing counts in map
• Wrong loop indices
• Confusing key types in map

5. Given the string "aabbccddeeffg", which character will be the first non repeating character using the hash method? What if the string is "aabbccddeeffggh"?

hard

A. First: g, Second: g

B. First: a, Second: h

C. First: g, Second: h

D. First: h, Second: h

Solution

1. Step 1: Analyze first string "aabbccddeeffg"

   All characters except 'g' appear twice; 'g' appears once, so first non repeating is 'g'.

2. Step 2: Analyze second string "aabbccddeeffggh"

   Now 'g' appears twice, 'h' appears once, so first non repeating is 'h'.

3. Final Answer:

   First: g, Second: h -> Option C

4. Quick Check:

   Check counts carefully for each string [OK]

Hint: Check counts carefully; last unique changes with input [OK]

Common Mistakes:

• Assuming first character is always unique
• Ignoring updated counts in second string
• Confusing repeated characters as unique