Longest Common Prefix in DSA Java - Time & Space Complexity

Time Complexity: Longest Common Prefix

O(n * m)

Understanding Time Complexity

We want to know how the time to find the longest common prefix grows as we add more words or longer words.

How does the work increase when input size changes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0) return "";
    String prefix = strs[0];
    for (int i = 1; i < strs.length; i++) {
        while (strs[i].indexOf(prefix) != 0) {
            prefix = prefix.substring(0, prefix.length() - 1);
            if (prefix.isEmpty()) return "";
        }
    }
    return prefix;
}
    

This code finds the longest common prefix among an array of strings by comparing prefixes and shortening them until they match.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Comparing prefixes using indexOf and shortening the prefix inside a nested loop.
• How many times: Outer loop runs once per string (n-1 times), inner loop can run up to length of prefix times (m times).

How Execution Grows With Input

As the number of strings (n) or the length of the prefix (m) grows, the work increases roughly by multiplying these two.

Input Size (n)	Approx. Operations
10 strings, prefix length 5	About 10 * 5 = 50
100 strings, prefix length 10	About 100 * 10 = 1000
1000 strings, prefix length 20	About 1000 * 20 = 20,000

Pattern observation: The operations grow linearly with both the number of strings and the prefix length.

Final Time Complexity

Time Complexity: O(n * m)

This means the time grows proportionally with the number of strings and the length of the shortest prefix.

Common Mistake

[X] Wrong: "The time depends only on the number of strings, not their length."

[OK] Correct: The prefix length affects how many characters we compare, so longer prefixes mean more work per string.

Interview Connect

Understanding this complexity helps you explain how your solution scales and shows you can analyze nested loops clearly.

Self-Check

"What if we sorted the strings first and then compared only the first and last strings? How would the time complexity change?"

Practice

1. What does the Longest Common Prefix (LCP) of a list of strings represent?

easy

A. The longest substring found anywhere in all strings

B. The longest ending substring common to all strings

C. The longest starting substring common to all strings

D. The shortest substring common to any two strings

Solution

1. Step 1: Understand the prefix concept

   A prefix is a substring starting at the beginning of a string.

2. Step 2: Define longest common prefix

   The longest common prefix is the longest substring starting at the beginning that all strings share.

3. Final Answer:

   The longest starting substring common to all strings -> Option C

4. Quick Check:

   Prefix means start of string = longest common start [OK]

Hint: Longest common prefix starts at string beginnings [OK]

Common Mistakes:

• Confusing prefix with substring anywhere
• Thinking it is the longest suffix
• Mixing prefix with shortest substring

2. Which of the following Java method signatures is correct for a function that finds the longest common prefix of an array of strings named strs?

easy

A. public String longestCommonPrefix(String[] strs)

B. public void longestCommonPrefix(String[] strs)

C. public int longestCommonPrefix(String[] strs)

D. public String longestCommonPrefix(List<String> strs)

Solution

1. Step 1: Identify return type for prefix

   The longest common prefix is a string, so return type must be String.

2. Step 2: Check parameter type

   The input is an array of strings, so parameter type is String[].

3. Final Answer:

   public String longestCommonPrefix(String[] strs) -> Option A

4. Quick Check:

   Return String, input String[] = public String longestCommonPrefix(String[] strs) [OK]

Hint: Return type matches prefix type; input is String array [OK]

Common Mistakes:

• Using void return type instead of String
• Returning int instead of String
• Using List<String> instead of String[]

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

public class Test {
  public static String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0) return "";
    String prefix = strs[0];
    for (int i = 1; i < strs.length; i++) {
      while (!strs[i].startsWith(prefix)) {
        prefix = prefix.substring(0, prefix.length() - 1);
        if (prefix.isEmpty()) return "";
      }
    }
    return prefix;
  }
  public static void main(String[] args) {
    String[] input = {"flower", "flow", "flight"};
    System.out.println(longestCommonPrefix(input));
  }
}

medium

A. "flow"

B. "fl"

C. "f"

D. ""

Solution

1. Step 1: Initialize prefix and check each string

   Start with prefix = "flower". Check "flow": "flow" starts with "flower"? No, shorten prefix to "flowe", then "flow", which matches.

2. Step 2: Check next string "flight"

   "flight" starts with "flow"? No, shorten prefix to "flo", then "fl", which matches.

3. Final Answer:

   "fl" -> Option B

4. Quick Check:

   Common start of all strings = "fl" [OK]

Hint: Shorten prefix until all strings start with it [OK]

Common Mistakes:

• Returning first string without checking others
• Not shortening prefix correctly
• Returning empty string too early

4. Identify the error in the following Java method for finding the longest common prefix:

public String longestCommonPrefix(String[] strs) {
  if (strs.length == 0) return null;
  String prefix = strs[0];
  for (int i = 1; i <= strs.length; i++) {
    while (!strs[i].startsWith(prefix)) {
      prefix = prefix.substring(0, prefix.length() - 1);
      if (prefix.isEmpty()) return "";
    }
  }
  return prefix;
}

medium

A. The substring method is used incorrectly

B. The method should return null instead of empty string when no prefix

C. The prefix should be initialized as an empty string

D. The loop condition should be i < strs.length, not i <= strs.length

Solution

1. Step 1: Check loop bounds

   The for loop uses i <= strs.length, which causes ArrayIndexOutOfBoundsException because valid indices are 0 to strs.length - 1.

2. Step 2: Correct loop condition

   Change to i < strs.length to avoid accessing invalid index.

3. Final Answer:

   The loop condition should be i < strs.length, not i <= strs.length -> Option D

4. Quick Check:

   Array indices go from 0 to length-1 [OK]

Hint: Loop index must be less than array length [OK]

Common Mistakes:

• Using i <= length causing runtime error
• Returning null instead of empty string
• Incorrect substring usage

5. Given the array String[] words = {"interview", "interrupt", "integrate", "integral"};, which of the following modifications to the longest common prefix algorithm will correctly find the prefix "int" even if some strings are empty or null?

hard

A. Add checks to skip null or empty strings before comparing prefixes

B. Return the first string immediately if any string is null or empty

C. Ignore the first string and start prefix from the second string

D. Use the longest string as the initial prefix instead of the first string

Solution

1. Step 1: Understand problem with null or empty strings

   Null or empty strings cause errors or incorrect prefix results if not handled.

2. Step 2: Modify algorithm to skip or handle null/empty

   Check each string before comparing; skip or handle null/empty to avoid errors and find correct prefix.

3. Final Answer:

   Add checks to skip null or empty strings before comparing prefixes -> Option A

4. Quick Check:

   Handle null/empty strings safely to find correct prefix [OK]

Hint: Always check for null or empty strings before prefix checks [OK]

Common Mistakes:

• Not handling null causing exceptions
• Returning first string without checks
• Starting prefix from wrong string