Reorder Data in Log Files - Practice Coding Problems

Reorder Data in Log Files - Problem

You are given an array of logs. Each log is a space-delimited string of words, where the first word is the identifier.

There are two types of logs:

Letter-logs: All words (except the identifier) consist of lowercase English letters.
Digit-logs: All words (except the identifier) consist of digits.

Reorder these logs so that:

The letter-logs come before all digit-logs.
The letter-logs are sorted lexicographically by their contents. If their contents are the same, then sort them lexicographically by their identifiers.
The digit-logs maintain their relative ordering.

Return the final order of the logs.

Input & Output

Example 1 — Mixed Log Types

$ Input: logs = ["dig1 8 1 5 1","let1 art can","dig2 3 6","let2 own kit dig","let3 art zero"]

› Output: ["let1 art can","let3 art zero","let2 own kit dig","dig1 8 1 5 1","dig2 3 6"]

💡 Note: Letter-logs come first, sorted by content: "art can" < "art zero" < "own kit dig". Digit-logs maintain original order.

Example 2 — Same Content Different IDs

$ Input: logs = ["a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo"]

› Output: ["g1 act car","a8 act zoo","ab1 off key dog","a1 9 2 3 1","zo4 4 7"]

💡 Note: Letter-logs: "act car" < "act zoo" < "off key dog". For same content "act", sort by identifier: "g1" < "a8".

Example 3 — Only Digit Logs

$ Input: logs = ["dig1 8 1 5 1","dig2 3 6","dig3 1 9"]

› Output: ["dig1 8 1 5 1","dig2 3 6","dig3 1 9"]

💡 Note: All digit-logs maintain their original relative ordering.

Constraints

1 ≤ logs.length ≤ 100
3 ≤ logs[i].length ≤ 100
All tokens of logs[i] are separated by a single space
logs[i] is guaranteed to have an identifier and at least one word after the identifier

Visualization

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Understanding the Visualization

Input

Mixed array of letter-logs and digit-logs

Process

Sort letter-logs by content+ID, keep digit-logs in order

Output

Letter-logs first, then digit-logs

Key Takeaway

🎯 Key Insight: Custom comparator elegantly handles multiple sorting criteria in one pass

Asked in

a Amazon 45 G Google 12

The key insight is to use a custom comparator that handles different log types with specific sorting rules. Letter-logs are sorted lexicographically by content then identifier, while digit-logs maintain their relative order. Best approach uses custom comparator with built-in sort. Time: O(n log n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Separation	O(n²)	O(n)	Manually separate, sort letter logs, then combine
Custom Comparator	O(n log n)	O(1)	Use built-in sort with custom comparison function

Brute Force Separation — Algorithm Steps

Step 1: Separate logs into letter and digit categories
Step 2: Sort letter logs using bubble sort
Step 3: Combine letter logs + digit logs

Visualization

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Step-by-Step Walkthrough

Separate

Split into letter-logs and digit-logs

Bubble Sort

Sort letter-logs using nested loops

Combine

Merge letter-logs + digit-logs

Code -

solution.c — C

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

char** solution(char** logs, int logsSize, int* returnSize) {
    char** letterLogs = (char**)malloc(logsSize * sizeof(char*));
    char** digitLogs = (char**)malloc(logsSize * sizeof(char*));
    int letterCount = 0, digitCount = 0;
    
    // Separate logs
    for (int i = 0; i < logsSize; i++) {
        char* space = strchr(logs[i], ' ');
        if (isalpha(space[1])) {
            letterLogs[letterCount++] = logs[i];
        } else {
            digitLogs[digitCount++] = logs[i];
        }
    }
    
    // Bubble sort letter logs
    for (int i = 0; i < letterCount; i++) {
        for (int j = 0; j < letterCount - i - 1; j++) {
            char* space1 = strchr(letterLogs[j], ' ');
            char* space2 = strchr(letterLogs[j + 1], ' ');
            
            char content1[1000], content2[1000], id1[100], id2[100];
            strcpy(content1, space1 + 1);
            strcpy(content2, space2 + 1);
            
            int len1 = space1 - letterLogs[j];
            int len2 = space2 - letterLogs[j + 1];
            strncpy(id1, letterLogs[j], len1); id1[len1] = '\0';
            strncpy(id2, letterLogs[j + 1], len2); id2[len2] = '\0';
            
            int cmp = strcmp(content1, content2);
            if (cmp > 0 || (cmp == 0 && strcmp(id1, id2) > 0)) {
                char* temp = letterLogs[j];
                letterLogs[j] = letterLogs[j + 1];
                letterLogs[j + 1] = temp;
            }
        }
    }
    
    // Combine results
    char** result = (char**)malloc(logsSize * sizeof(char*));
    for (int i = 0; i < letterCount; i++) {
        result[i] = letterLogs[i];
    }
    for (int i = 0; i < digitCount; i++) {
        result[letterCount + i] = digitLogs[i];
    }
    
    *returnSize = logsSize;
    free(letterLogs);
    free(digitLogs);
    return result;
}

int main() {
    // Simplified parsing for demonstration
    char* logs[] = {"dig1 8 1 5 1", "let1 art can", "dig2 3 6", "let2 own kit dig", "let3 art zero"};
    int logsSize = 5;
    int returnSize;
    
    char** result = solution(logs, logsSize, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("\"%s\"", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Bubble sort on letter logs takes O(k²) where k is number of letter logs

⚠ Quadratic Growth

Space Complexity

O(n)

Two separate lists to store letter and digit logs

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Separation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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