Longest Common Prefix in DSA C++ - Time & Space Complexity

Time Complexity: Longest Common Prefix

O(n * m)

Understanding Time Complexity

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

How does the work change when the list of words or their length increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

string longestCommonPrefix(vector<string>& strs) {
    if (strs.empty()) return "";
    string prefix = strs[0];
    for (int i = 1; i < strs.size(); i++) {
        while (strs[i].find(prefix) != 0) {
            prefix = prefix.substr(0, prefix.size() - 1);
            if (prefix.empty()) return "";
        }
    }
    return prefix;
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer loop goes through each string, and the inner while loop shortens the prefix until it matches the start of the current string.
• How many times: The outer loop runs once per string (n times), and the inner loop can run up to the length of the prefix (m times) in the worst case.

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 (m=5)
10	About 50 checks
100	About 500 checks
1000	About 5000 checks

Pattern observation: Doubling the number of strings roughly doubles the work, and longer prefixes increase the work linearly.

Final Time Complexity

Time Complexity: O(n * m)

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

Common Mistake

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

[OK] Correct: The prefix length matters because we compare characters one by one, so longer prefixes mean more checks per string.

Interview Connect

Understanding this helps you explain how your solution scales with input size, a key skill interviewers look for.

Self-Check

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

Practice

1. What does the Longest Common Prefix (LCP) problem ask you to find given a list of strings?

easy

A. The total count of common characters in all strings

B. The longest starting substring common to all strings

C. The longest ending substring common to all strings

D. The longest substring common to any two strings

Solution

1. Step 1: Understand the problem definition

   The Longest Common Prefix means the longest substring at the start of all strings that is the same.

2. Step 2: Compare options with the definition

   Only The longest starting substring common to all strings matches the definition exactly: longest starting substring common to all strings.

3. Final Answer:

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

4. Quick Check:

   Longest Common Prefix = longest starting substring [OK]

Hint: Focus on the start of strings, not anywhere else [OK]

Common Mistakes:

• Confusing prefix with substring anywhere
• Thinking about suffix instead of prefix
• Counting characters instead of substring

2. Which of the following C++ code snippets correctly initializes a vector of strings named strs with values "flower", "flow", "flight"?

easy

A. std::vector strs = {"flower", "flow", "flight"};

B. std::vector strs = {"flower", "flow", "flight"};

C. vector<string> strs = ("flower", "flow", "flight");

D. std::vector strs = ("flower", "flow", "flight");

Solution

1. Step 1: Check correct vector initialization syntax

   In C++, to initialize a vector of strings with values, use curly braces {} and specify type with std::vector.

2. Step 2: Validate each option

   std::vector strs = {"flower", "flow", "flight"}; uses correct syntax with std::vector and braces {}. Versions missing std:: namespace, using parentheses instead of braces, or lacking template type <std::string> are invalid.

3. Final Answer:

   std::vector strs = {"flower", "flow", "flight"}; -> Option A

4. Quick Check:

   Correct vector initialization uses braces and type [OK]

Hint: Use curly braces {} and full type for vector initialization [OK]

Common Mistakes:

• Using parentheses instead of braces
• Omitting std:: namespace
• Missing template argument

3. What is the output of the following C++ code?

std::vector strs = {"interview", "internet", "internal"};
std::string prefix = strs[0];
for (int i = 1; i < strs.size(); i++) {
    int j = 0;
    while (j < prefix.size() && j < strs[i].size() && prefix[j] == strs[i][j]) {
        j++;
    }
    prefix = prefix.substr(0, j);
}
std::cout << prefix << std::endl;

medium

A. int

B. inte

C. intern

D. inter

Solution

1. Step 1: Initialize prefix and compare with each string

   Start prefix = "interview". Compare with "internet": common prefix is "inter". Update prefix = "inter".

2. Step 2: Compare updated prefix "inter" with "internal"

   Common prefix is "inter" again. So prefix remains "inter".

3. Final Answer:

   inter -> Option D

4. Quick Check:

   Common prefix of all = "inter" [OK]

Hint: Compare prefix with each string, shorten prefix on mismatch [OK]

Common Mistakes:

• Stopping after first comparison
• Not updating prefix correctly
• Confusing substring indices

4. Identify the bug in the following C++ code that tries to find the longest common prefix:

std::string longestCommonPrefix(std::vector& strs) {
    std::string prefix = strs[0];
    for (int i = 1; i < strs.size(); i++) {
        int j = 0;
        while (j < prefix.size() && j < strs[i].size() && prefix[j] == strs[i][j]) {
            j++;
        }
        prefix = prefix.substr(0, j);
        if (prefix.empty()) return "";
    }
    return prefix;
}

// Usage:
// std::vector words = {"dog", "racecar", "car"};
// std::cout << longestCommonPrefix(words) << std::endl;

medium

A. The function does not handle empty input vector

B. The function returns incorrect prefix for input {"dog", "racecar", "car"}

C. The while loop condition is incorrect and causes infinite loop

D. The function should return the longest common suffix instead

Solution

1. Step 1: Check empty input handling

   The code directly accesses strs[0] without checking if strs is empty, causing a runtime error (out-of-range access) for empty vectors.

2. Step 2: Verify other statements

   The loop logic correctly shortens the prefix, no infinite loop occurs, it returns prefix not suffix, and for {"dog", "racecar", "car"} it correctly returns "".

3. Final Answer:

   The function does not handle empty input vector -> Option A

4. Quick Check:

   Missing if(strs.empty()) return ""; [OK]

Hint: Test edge cases like empty lists first [OK]

Common Mistakes:

• Overlooking empty vector handling
• Misjudging the output for no common prefix cases
• Suspecting infinite loop in the while condition

5. Given a list of strings {"apple", "ape", "april", "apt"}, write a C++ function to find the longest common prefix using a vertical scanning approach. What will be the output of your function?

hard

A. "app"

B. "a"

C. "ap"

D. ""

Solution

1. Step 1: Understand vertical scanning approach

   Check characters column by column across all strings until mismatch or end of any string.

2. Step 2: Apply vertical scanning on given strings

   Index 0: all have 'a'. Index 1: all have 'p'. Index 2: "apple" has 'p', "ape" has 'e' (mismatch). Stop scanning.

3. Final Answer:

   "ap" -> Option C

4. Quick Check:

   Vertical scan stops at first mismatch, prefix = "ap" [OK]

Hint: Scan characters column-wise until mismatch found [OK]

Common Mistakes:

• Assuming prefix is longest word's start
• Stopping scan too early or too late
• Confusing vertical with horizontal scanning