Reorder Data in Log Files

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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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

✓ Brute Force Separation

⏱️ Time: O(n²) Space: O(n)

Split logs into letter-logs and digit-logs lists. Sort letter-logs using nested loops for comparison. Combine letter-logs first, then digit-logs.

Custom Comparator

⏱️ Time: O(n log n) Space: O(1)

Define a custom comparator that handles both letter-logs and digit-logs according to the rules. Letter-logs are sorted by content then identifier, digit-logs maintain relative order by using original index.

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];
    }

    // Stable bubble sort for 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 = space1 + 1;
            char* content2 = space2 + 1;

            int cmp = strcmp(content1, content2);

            if (cmp > 0 || (cmp == 0 &&
                strncmp(letterLogs[j], letterLogs[j + 1],
                        space1 - letterLogs[j]) > 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() {
    char input[10000];
    fgets(input, sizeof(input), stdin);

    char** logs = (char**)malloc(100 * sizeof(char*));
    int logsSize = 0;

    char* start = strchr(input, '[');
    if (start) {
        start++;  // skip [

        char* end = strrchr(start, ']');
        if (end) *end = '\0';

        char* token = strtok(start, ",");
        while (token && logsSize < 100) {

            // Trim leading spaces
            while (*token == ' ' || *token == '"')
                token++;

            // Trim trailing spaces and quotes
            char* tokenEnd = token + strlen(token) - 1;
            while (tokenEnd > token &&
                   (*tokenEnd == ' ' || *tokenEnd == '"')) {
                *tokenEnd = '\0';
                tokenEnd--;
            }

            logs[logsSize] = (char*)malloc(strlen(token) + 1);
            strcpy(logs[logsSize], token);
            logsSize++;

            token = strtok(NULL, ",");
        }
    }

    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");

    // Cleanup
    for (int i = 0; i < logsSize; i++)
        free(logs[i]);

    free(logs);
    free(result);

    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

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Related Problems

Common Approaches

Brute Force Separation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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