Count Words with Given Prefix in DSA Go - Time & Space Complexity

Time Complexity: Count Words with Given Prefix

O(n * m)

Understanding Time Complexity

We want to understand how long it takes to count words that start with a certain prefix in a list.

The question is: how does the time grow when the list or prefix size changes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func countPrefix(words []string, prefix string) int {
    count := 0
    for _, word := range words {
        if len(word) >= len(prefix) && word[:len(prefix)] == prefix {
            count++
        }
    }
    return count
}
    

This code counts how many words in the list start with the given prefix.

Identify Repeating Operations

• Primary operation: Looping through each word in the list.
• How many times: Once for each word, so as many times as the list size (n).
• Secondary operation: Checking prefix by comparing characters up to prefix length (m).
• How many times: Up to prefix length for each word.

How Execution Grows With Input

As the number of words (n) grows, the loop runs more times. Also, longer prefixes (m) mean more characters to check per word.

Input Size (n)	Prefix Length (m)	Approx. Operations
10	3	About 10 * 3 = 30 checks
100	3	About 100 * 3 = 300 checks
1000	3	About 1000 * 3 = 3000 checks

Pattern observation: Operations grow linearly with both the number of words and the prefix length.

Final Time Complexity

Time Complexity: O(n * m)

This means the time grows proportionally to the number of words times the length of the prefix.

Common Mistake

[X] Wrong: "Checking prefix is always O(1) because prefix is small."

[OK] Correct: Even if prefix is small, it still requires checking each character for every word, so it depends on prefix length and number of words.

Interview Connect

Understanding how loops and string checks combine helps you explain your code clearly and shows you know how to analyze performance in real tasks.

Self-Check

"What if we stored words in a trie data structure? How would the time complexity change when counting words with a prefix?"

Practice

1. What does the problem "Count Words with Given Prefix" ask you to do?

easy

A. Count the total number of characters in all words

B. Find the longest word in a list

C. Count how many words in a list start with a specific prefix

D. Sort the words alphabetically

Solution

1. Step 1: Understand the problem statement

   The problem asks to find how many words begin with a given prefix string.

2. Step 2: Identify the key operation

   We need to check each word and count those that start with the prefix.

3. Final Answer:

   Count how many words in a list start with a specific prefix -> Option C

4. Quick Check:

   Counting words by prefix = Count how many words in a list start with a specific prefix [OK]

Hint: Look for words starting exactly with the prefix [OK]

Common Mistakes:

• Confusing prefix with substring anywhere in word
• Counting total words instead of filtered ones
• Ignoring case sensitivity if required

2. Which of the following Go code snippets correctly checks if a word starts with a given prefix pref?

easy

A. strings.HasPrefix(word, pref)

B. strings.Contains(word, pref)

C. word[:len(pref)] == pref

D. word == pref

Solution

1. Step 1: Recall Go standard library functions

   strings.HasPrefix(word, pref) returns true if word starts with pref.

2. Step 2: Compare other options

   strings.Contains checks anywhere, slicing may panic if prefix longer, equality checks full word.

3. Final Answer:

   strings.HasPrefix(word, pref) -> Option A

4. Quick Check:

   Use HasPrefix for prefix check [OK]

Hint: Use strings.HasPrefix for safe prefix check [OK]

Common Mistakes:

• Using strings.Contains which checks anywhere in word
• Slicing without length check causing runtime panic
• Comparing whole word instead of prefix

3. What is the output of the following Go code?

package main
import (
  "fmt"
  "strings"
)
func countPrefix(words []string, prefix string) int {
  count := 0
  for _, w := range words {
    if strings.HasPrefix(w, prefix) {
      count++
    }
  }
  return count
}
func main() {
  words := []string{"apple", "app", "application", "banana", "apply"}
  fmt.Println(countPrefix(words, "app"))
}

medium

A. 4

B. 2

C. 3

D. 5

Solution

1. Step 1: Identify words starting with "app"

   Words: "apple", "app", "application", "banana", "apply". "banana" does not start with "app".

2. Step 2: Count matching words

   Matching words: "apple", "app", "application", "apply" -> 4 words.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Count of words starting with "app" = 4 [OK]

Hint: Count words starting exactly with prefix using HasPrefix [OK]

Common Mistakes:

• Counting "banana" incorrectly
• Off-by-one error in counting
• Misreading prefix length

4. The following Go code is intended to count words starting with a prefix but causes a runtime error. What is the error?

func countPrefix(words []string, prefix string) int {
  count := 0
  for _, w := range words {
    if w[:len(prefix)] == prefix {
      count++
    }
  }
  return count
}

medium

A. No error, code works fine

B. Index out of range panic if word shorter than prefix

C. Logical error: counts words containing prefix anywhere

D. Syntax error due to missing import

Solution

1. Step 1: Analyze slicing operation

   The code slices w[:len(prefix)] without checking if w is long enough.

2. Step 2: Identify runtime error cause

   If w is shorter than prefix, slicing causes index out of range panic.

3. Final Answer:

   Index out of range panic if word shorter than prefix -> Option B

4. Quick Check:

   Slicing without length check causes panic [OK]

Hint: Check word length before slicing prefix [OK]

Common Mistakes:

• Ignoring length check before slicing
• Assuming code compiles without imports
• Confusing prefix check with substring check

5. You have a list of words and want to count how many start with each prefix from a list of prefixes. Which approach is best for efficient counting in Go?

hard

A. Use strings.Contains to check prefix in each word

B. For each prefix, loop over all words and count matches

C. Sort words and use binary search for each prefix

D. Build a trie from words, then query counts for each prefix

Solution

1. Step 1: Understand the problem scale

   Counting for many prefixes over many words needs efficiency.

2. Step 2: Compare approaches

   Looping over all words for each prefix is slow (O(P*W)). Sorting + binary search helps but trie allows prefix counting efficiently.

3. Step 3: Choose best data structure

   Trie stores words by prefix, enabling fast prefix count queries.

4. Final Answer:

   Build a trie from words, then query counts for each prefix -> Option D

5. Quick Check:

   Trie enables efficient prefix counting [OK]

Hint: Use trie for fast prefix count queries [OK]

Common Mistakes:

• Using nested loops causing slow performance
• Using strings.Contains which is incorrect for prefix
• Ignoring data structure benefits