First Non Repeating Character Using Hash in DSA Go - 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.

func firstNonRepeatingChar(s string) byte {
    freq := make(map[byte]int)
    for i := 0; i < len(s); i++ {
        freq[s[i]]++
    }
    for i := 0; i < len(s); i++ {
        if freq[s[i]] == 1 {
            return s[i]
        }
    }
    return 0
}

This code counts how many times each character appears, then finds the first character that appears only once.

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 does two full passes over the characters.

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, so the growth is linear.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows directly in proportion to the string length.

Common Mistake

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

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

Interview Connect

Understanding how to analyze loops like this helps you explain your code clearly and shows you know how to judge efficiency in real problems.

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 is the main purpose of using a hash map in finding the first non-repeating character in a string?

easy

A. To count the frequency of each character efficiently

B. To sort the characters alphabetically

C. To reverse the string quickly

D. To store characters in a linked list

Solution

1. Step 1: Understand the role of hash map

   A hash map stores each character as a key and counts how many times it appears in the string.

2. Step 2: Use frequency to find non-repeating character

   By checking the counts, we can find the first character with count 1, which is the first non-repeating character.

3. Final Answer:

   To count the frequency of each character efficiently -> Option A

4. Quick Check:

   Hash map counts characters = frequency count [OK]

Hint: Hash maps count characters fast to find unique ones [OK]

Common Mistakes:

• Thinking hash map sorts characters
• Using hash map to reverse string
• Confusing hash map with linked list

2. Which of the following Go code snippets correctly declares a hash map to store character counts for a string?

easy

A. counts := make(map[string]int)

B. counts := make([]int, len(s))

C. counts := make(map[int]string)

D. counts := make(map[rune]int)

Solution

1. Step 1: Identify correct map key type for characters

   In Go, characters from strings are best represented as 'rune' type, so map[rune]int is correct for counting characters.

2. Step 2: Check other options

   counts := make(map[string]int) uses string as key which is incorrect for single characters; counts := make(map[int]string) reverses key and value types; counts := make([]int, len(s)) is a slice, not a map.

3. Final Answer:

   counts := make(map[rune]int) -> Option D

4. Quick Check:

   Use map[rune]int for character counts in Go [OK]

Hint: Use map[rune]int to count characters in Go strings [OK]

Common Mistakes:

• Using map[string]int for single characters
• Confusing slices with maps
• Using wrong key-value types in map

3. What will be the output of the following Go code snippet?

package main
import "fmt"
func main() {
    s := "swiss"
    counts := make(map[rune]int)
    for _, ch := range s {
        counts[ch]++
    }
    for _, ch := range s {
        if counts[ch] == 1 {
            fmt.Println(string(ch))
            break
        }
    }
}

medium

A. w

B. s

C. i

D. No output

Solution

1. Step 1: Count character frequencies in "swiss"

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

2. Step 2: Find first character with count 1 in order

   Order in string: 's'(3), 'w'(1), 'i'(1), 's'(3), 's'(3). First with count 1 is 'w'.

3. Final Answer:

   w -> Option A

4. Quick Check:

   First unique char in "swiss" is 'w' [OK]

Hint: Check counts in string order; first count 1 is answer [OK]

Common Mistakes:

• Picking 's' which repeats
• Choosing 'i' ignoring order
• Expecting no output

4. Identify the bug in this Go code that tries to find the first non-repeating character:

package main
import "fmt"
func main() {
    s := "hello"
    counts := make(map[rune]int)
    for i := 0; i < len(s); i++ {
        counts[rune(s[i])] = counts[rune(s[i])] + 1
    }
    for i := 0; i < len(s); i++ {
        if counts[rune(s[i])] == 1 {
            fmt.Println(s[i])
            break
        }
    }
}

medium

A. Loop should use range instead of index

B. Map key should be string, not rune

C. Printing s[i] prints byte, not character

D. Counts map is not initialized

Solution

1. Step 1: Check how character is printed

   Printing s[i] prints a byte as integer, not the character itself.

2. Step 2: Fix printing to convert byte to string

   Use fmt.Println(string(s[i])) to print the character properly.

3. Final Answer:

   Printing s[i] prints byte, not character -> Option C

4. Quick Check:

   Print string(s[i]) to show character, not byte [OK]

Hint: Convert byte to string when printing characters [OK]

Common Mistakes:

• Ignoring printing type mismatch
• Changing map key type unnecessarily
• Not initializing map (already done)

5. Given the string "aabbccdde", which code modification will correctly find the first non-repeating character using a hash map in Go, considering the string may contain Unicode characters?

hard

A. Use map[byte]int and iterate with 'for i := 0; i < len(s); i++'

B. Use map[rune]int and iterate with 'for _, ch := range s' to count frequencies

C. Use map[string]int and split string into substrings of length 2

D. Use map[int]int and convert characters to ASCII codes manually

Solution

1. Step 1: Choose correct map key type for Unicode

   Using map[rune]int allows counting Unicode characters correctly, unlike byte which only handles ASCII.

2. Step 2: Use range loop to iterate runes

   Using 'for _, ch := range s' iterates over runes, correctly handling multi-byte Unicode characters.

3. Final Answer:

   Use map[rune]int and iterate with 'for _, ch := range s' to count frequencies -> Option B

4. Quick Check:

   map[rune]int + range loop handles Unicode correctly [OK]

Hint: Use rune keys and range loop for Unicode-safe counting [OK]

Common Mistakes:

• Using byte keys breaks Unicode characters
• Splitting string incorrectly into substrings
• Using int keys without clear mapping