Longest Common Prefix

Longest Common Prefix - Problem

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

Note: All given inputs are in lowercase letters a-z.

Input & Output

Example 1 — Basic Case

$ Input: strs = ["flower","flow","flight"]

› Output: "fl"

💡 Note: The common prefix is "fl". All three strings start with "fl", but at position 2, "flower" and "flow" have 'o' while "flight" has 'i', so the common prefix stops there.

Example 2 — No Common Prefix

$ Input: strs = ["dog","racecar","car"]

› Output: ""

💡 Note: There is no common prefix among the input strings. The first characters are 'd', 'r', and 'c' respectively, which are all different.

Example 3 — Complete Match

$ Input: strs = ["test","test","test"]

› Output: "test"

💡 Note: All strings are identical, so the entire string "test" is the common prefix.

Constraints

1 ≤ strs.length ≤ 200
0 ≤ strs[i].length ≤ 200
strs[i] consists of only lowercase English letters

Visualization

Tap to expand

Asked in

G Google 45 a Amazon 38 f Facebook 32 M Microsoft 28

The key insight is to scan vertically character by character across all strings until a mismatch is found. The optimal approach uses vertical scanning with early termination. Time: O(S), Space: O(1) where S is the sum of all characters.

Common Approaches

✓ Brute Force - Character by Character

⏱️ Time: O(S) Space: O(1)

Check each character position from left to right across all strings. Stop when any string differs or ends.

Vertical Scanning

⏱️ Time: O(S) Space: O(1)

Instead of comparing strings horizontally, scan vertically by checking the same character position across all strings simultaneously.

Brute Force - Character by Character — Algorithm Steps

Start from position 0
Check if all strings have character at current position
If all match, add to result and move to next position
If any differ or string ends, return current result

Visualization

Tap to expand

Step-by-Step Walkthrough

Position 0

Check first character in all strings

Position 1

Check second character in all strings

Mismatch

Stop when characters don't match

Code -

solution.c — C

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

char* solution(char** strs, int strsSize) {
    if (strsSize == 0) {
        char* result = malloc(1);
        result[0] = '\0';
        return result;
    }
    
    // Find minimum length to avoid index out of bounds
    int minLen = strlen(strs[0]);
    for (int i = 1; i < strsSize; i++) {
        int len = strlen(strs[i]);
        if (len < minLen) {
            minLen = len;
        }
    }
    
    char* prefix = malloc(minLen + 1);
    int prefixLen = 0;
    
    for (int i = 0; i < minLen; i++) {
        // Get character at position i from first string
        char c = strs[0][i];
        
        // Check if all strings have same character at position i
        for (int j = 1; j < strsSize; j++) {
            if (strs[j][i] != c) {
                prefix[prefixLen] = '\0';
                return prefix;
            }
        }
        
        // All strings match at position i, add to prefix
        prefix[prefixLen++] = c;
    }
    
    prefix[prefixLen] = '\0';
    return prefix;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Simple parsing for array of strings
    char** strs = malloc(1000 * sizeof(char*));
    int strsSize = 0;
    
    if (strlen(line) > 2) {
        char* token = strtok(line + 1, ",]"); // Skip [ and split by comma
        while (token != NULL) {
            // Remove quotes and spaces
            while (*token == ' ' || *token == '"') token++;
            int len = strlen(token);
            while (len > 0 && (token[len-1] == ' ' || token[len-1] == '"')) {
                token[len-1] = '\0';
                len--;
            }
            strs[strsSize] = malloc(len + 1);
            strcpy(strs[strsSize], token);
            strsSize++;
            token = strtok(NULL, ",]");
        }
    }
    
    char* result = solution(strs, strsSize);
    printf("%s\n", result);
    
    // Cleanup
    for (int i = 0; i < strsSize; i++) {
        free(strs[i]);
    }
    free(strs);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(S)

S is sum of all characters in all strings - worst case we scan all characters

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

52.0K Views

High Frequency

~15 min Avg. Time

1.9K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Character by Character — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler