Reorder Data In Log Files Problem
Description
LeetCode Problem 937.
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.
Example 1:
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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"]
Explanation:
The letter-log contents are all different, so their ordering is "art can", "art zero", "own kit dig".
The digit-logs have a relative order of "dig1 8 1 5 1", "dig2 3 6".
Example 2:
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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"]
Constraints:
- 1 <= logs.length <= 100
- 3 <= logs[i].length <= 100
- All the 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.
Sample C++ Code
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class Solution {
public:
vector<string> reorderLogFiles(vector<string>& logs) {
vector<string> digitLogs, ans;
vector<pair<string, string>> letterLogs;
for (string &s : logs) {
int i = 0;
while (s[i] != ' ')
++i;
if (isalpha(s[i + 1]))
letterLogs.emplace_back(s.substr(0, i), s.substr(i + 1));
else
digitLogs.push_back(s);
}
sort(letterLogs.begin(), letterLogs.end(), [&](auto& a, auto& b) {
return a.second == b.second ? a.first < b.first : a.second < b.second;
});
for (auto &p : letterLogs)
ans.push_back(p.first + " " + p.second);
for (string &s : digitLogs)
ans.push_back(s);
return ans;
}
};