Longest Common Prefix in DSA C

Mental Model

Find the longest starting part that all words share exactly.

Analogy: Imagine several friends starting a race together and running the same path. The longest common prefix is like the longest stretch of road they all run together before some take different turns.

words: [flower, flow, flight]
indexes:  0123456789
prefix:  ↑↑↑

Dry Run Walkthrough

Input: words: [flower, flow, flight]

Goal: Find the longest prefix all words start with

Step 1: Start with the first word 'flower' as the prefix

prefix = 'flower'
words = [flower, flow, flight]

Why: We need a starting point to compare others against

Step 2: Compare prefix 'flower' with second word 'flow', shorten prefix to match

prefix = 'flow'
words = [flower, flow, flight]

Why: The prefix must be common to both words, so shorten to 'flow'

Step 3: Compare prefix 'flow' with third word 'flight', shorten prefix to match

prefix = 'fl'
words = [flower, flow, flight]

Why: Only 'fl' is common at start of 'flow' and 'flight'

Result:

prefix = 'fl'

Annotated Code

DSA C

#include <stdio.h>
#include <string.h>

// Function to find longest common prefix among array of strings
char* longestCommonPrefix(char** strs, int strsSize) {
    if (strsSize == 0) return "";
    char* prefix = strs[0];
    for (int i = 1; i < strsSize; i++) {
        int j = 0;
        while (prefix[j] && strs[i][j] && prefix[j] == strs[i][j]) {
            j++;
        }
        prefix[j] = '\0'; // shorten prefix to matched part
    }
    return prefix;
}

int main() {
    char str1[] = "flower";
    char str2[] = "flow";
    char str3[] = "flight";
    char* words[] = {str1, str2, str3};
    int size = 3;
    char* result = longestCommonPrefix(words, size);
    printf("%s\n", result);
    return 0;
}

char* prefix = strs[0];

initialize prefix with first word

for (int i = 1; i < strsSize; i++) {

iterate over remaining words

while (prefix[j] && strs[i][j] && prefix[j] == strs[i][j]) { j++; }

find common prefix length between prefix and current word

prefix[j] = '\0';

shorten prefix to matched length

Output

fl

Complexity Analysis

Time: O(S) because we compare characters of all words until mismatch, where S is total characters

Space: O(1) because we modify prefix in place without extra storage

vs Alternative: Compared to checking all pairs separately, this approach is efficient by progressively shortening prefix

Edge Cases

empty array of words

returns empty string immediately

DSA C

if (strsSize == 0) return "";

array with one word

returns that word as prefix

DSA C

char* prefix = strs[0];

no common prefix at all

prefix becomes empty string

DSA C

prefix[j] = '\0';

When to Use This Pattern

When you need to find the longest starting substring common to many strings, use the longest common prefix pattern to compare and shorten progressively.

Common Mistakes

Mistake: Not shortening the prefix string properly, causing incorrect longer prefix
Fix: Set prefix[j] = '\0' to cut prefix to matched length after each comparison

Summary

Finds the longest starting substring common to all strings in a list.

Use when you want to find shared beginnings among multiple words or strings.

Start with the first word as prefix and shorten it step-by-step by comparing with others.

Practice

1. What does the Longest Common Prefix (LCP) problem ask for when given an array of strings?

easy

A. The longest ending substring common to all strings

B. The longest substring common to any two strings

C. The longest starting substring common to all strings

D. The shortest string in the array

Solution

1. Step 1: Understand the problem definition

   The Longest Common Prefix means the longest substring starting from the beginning that is shared by all strings in the array.

2. Step 2: Compare options with definition

   Only The longest starting substring common to all strings matches this definition exactly. Options A and C talk about suffix or any substring, and D is unrelated.

3. Final Answer:

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

4. Quick Check:

   Longest Common Prefix = longest starting substring [OK]

Hint: Focus on the start of strings for common parts [OK]

Common Mistakes:

• Confusing prefix with suffix
• Thinking about any substring instead of prefix
• Choosing shortest string instead of prefix

2. Which of the following C code snippets correctly initializes a variable to store the longest common prefix length?

easy

A. int lcp_length = 0;

B. char lcp_length = '';

C. int lcp_length = NULL;

D. string lcp_length = 0;

Solution

1. Step 1: Identify correct data type for prefix length

   The length of a prefix is a number, so it should be stored in an integer variable.

2. Step 2: Check initialization syntax

   int lcp_length = 0; correctly declares an int and initializes it to 0. char lcp_length = ''; uses char with invalid empty quotes, C uses NULL which is for pointers, and D uses string which is not a C type.

3. Final Answer:

   int lcp_length = 0; -> Option A

4. Quick Check:

   Prefix length is an integer initialized to zero [OK]

Hint: Use int for lengths, not char or string [OK]

Common Mistakes:

• Using char for numeric length
• Assigning NULL to int variable
• Using non-C types like string

3. What is the output of the following C code snippet that finds the longest common prefix length?

char *strs[] = {"flower", "flow", "flight"};
int i = 0;
while (strs[0][i] && strs[1][i] && strs[2][i] && strs[0][i] == strs[1][i] && strs[1][i] == strs[2][i]) {
    i++;
}
printf("%d", i);

medium

A. 4

B. 2

C. 3

D. 0

Solution

1. Step 1: Compare characters at each index

   Index 0: 'f' == 'f' == 'f' (true), index 1: 'l' == 'l' == 'l' (true), index 2: 'o' == 'o' == 'i' (false)

2. Step 2: Count matching prefix length

   Matching characters are at indices 0 and 1 only, so length is 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Matching prefix length = 2 [OK]

Hint: Check characters index by index until mismatch [OK]

Common Mistakes:

• Counting beyond first mismatch
• Confusing index with length
• Ignoring all strings in comparison

4. Identify the bug in this C code snippet that tries to find the longest common prefix of an array of strings:

int longestCommonPrefix(char **strs, int strsSize) {
    int i = 0;
    while (strs[0][i]) {
        for (int j = 1; j < strsSize; j++) {
            if (strs[j][i] != strs[0][i])
                return i;
        }
        i++;
    }
    return i;
}

medium

A. Does not initialize i before use

B. Returns i too early inside the loop

C. Uses wrong loop variable in inner loop

D. Does not check if strs[j][i] is null before comparison

Solution

1. Step 1: Analyze character comparison in inner loop

   The code compares strs[j][i] with strs[0][i] but does not check if strs[j][i] is '\0' (string end).

2. Step 2: Understand consequence of missing null check

   If strs[j] is shorter than strs[0], accessing strs[j][i] beyond its length causes undefined behavior or wrong result.

3. Final Answer:

   Does not check if strs[j][i] is null before comparison -> Option D

4. Quick Check:

   Check string length before comparing characters [OK]

Hint: Always check string end before character compare [OK]

Common Mistakes:

• Ignoring string length in loops
• Returning too early without full check
• Using wrong loop indices

5. Given an array of strings ["interview", "interval", "internal", "internet"], which of the following C code snippets correctly returns the longest common prefix string?

hard

A. int i = 0;
while (1) {
char c = strs[0][i];
if (c == '\0') break;
for (int j = 1; j < strsSize; j++) {
if (strs[j][i] != c || strs[j][i] == '\0')
goto end;
}
i++;
}
end:
char *prefix = malloc(i + 1);
strncpy(prefix, strs[0], i);
prefix[i] = '\0';
return prefix;

B. int i = 0;
while (strs[0][i]) {
for (int j = 1; j < strsSize; j++) {
if (strs[j][i] != strs[0][i])
break;
}
i++;
}
char *prefix = malloc(i + 1);
strncpy(prefix, strs[0], i);
prefix[i] = '\0';
return prefix;

C. int i = 0;
while (strs[0][i]) {
for (int j = 1; j < strsSize; j++) {
if (strs[j][i] != strs[0][i])
return NULL;
}
i++;
}
return strs[0];

D. int i = 0;
while (strs[0][i]) {
for (int j = 1; j < strsSize; j++) {
if (strs[j][i] != strs[0][i])
return "";
}
i++;
}
return strs[0];

Solution

1. Step 1: Check loop and character comparison logic

   The correct code uses while(1) with break if strs[0][i] == '\0', caches c = strs[0][i], and for each j checks strs[j][i] != c || strs[j][i] == '\0' then goto end before i++. This stops exactly at the first mismatch or shorter string.

2. Step 2: Verify prefix extraction and return

   After loop, malloc(i + 1), strncpy(prefix, strs[0], i), prefix[i] = '\0', return prefix. This creates and returns the new prefix string correctly.

3. Final Answer:

   Uses while(1), goto on mismatch/null, malloc/strncpy new prefix -> Option A

4. Quick Check:

   Check all strings char by char, then copy prefix [OK]

Hint: Use memory copy after finding prefix length [OK]

Common Mistakes:

• Not handling string ends properly
• Returning original string instead of prefix
• Not allocating memory for prefix