Skip to content

Latest commit

 

History

History
101 lines (84 loc) · 5.46 KB

README.md

File metadata and controls

101 lines (84 loc) · 5.46 KB

You are given a string, message, and a positive integer, limit.

You must split message into one or more parts based on limit. Each resulting part should have the suffix "<a/b>", where "b" is to be replaced with the total number of parts and "a" is to be replaced with the index of the part, starting from 1 and going up to b. Additionally, the length of each resulting part (including its suffix) should be equal to limit, except for the last part whose length can be at most limit.

The resulting parts should be formed such that when their suffixes are removed and they are all concatenated in order, they should be equal to message. Also, the result should contain as few parts as possible.

Return the parts message would be split into as an array of strings. If it is impossible to split message as required, return an empty array.

 

Example 1:

Input: message = "this is really a very awesome message", limit = 9
Output: ["thi<1/14>","s i<2/14>","s r<3/14>","eal<4/14>","ly <5/14>","a v<6/14>","ery<7/14>"," aw<8/14>","eso<9/14>","me<10/14>"," m<11/14>","es<12/14>","sa<13/14>","ge<14/14>"]
Explanation:
The first 9 parts take 3 characters each from the beginning of message.
The next 5 parts take 2 characters each to finish splitting message. 
In this example, each part, including the last, has length 9. 
It can be shown it is not possible to split message into less than 14 parts.

Example 2:

Input: message = "short message", limit = 15
Output: ["short mess<1/2>","age<2/2>"]
Explanation:
Under the given constraints, the string can be split into two parts: 
- The first part comprises of the first 10 characters, and has a length 15.
- The next part comprises of the last 3 characters, and has a length 8.

 

Constraints:

  • 1 <= message.length <= 104
  • message consists only of lowercase English letters and ' '.
  • 1 <= limit <= 104

Companies: Uber

Related Topics:
String, Binary Search

Similar Questions:

Hints:

  • Could you solve the problem if you knew how many digits the total number of parts has?
  • Try all possible lengths of the total number of parts, and see if the string can be split such that the total number of parts has that length.
  • Binary search can be used for each part length to find the precise number of parts needed.

Solution 1.

// OJ: https://leetcode.com/problems/split-message-based-on-limit
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<string> splitMessage(string s, int limit) {
        int N = s.size();
        vector<int> cnt{9};
        vector<string> ans;
        for (int d = 1; true; ++d) { // split the message into `d`-digit parts
            int total = N, totalParts = 0;
            for (int i = 1; i <= d && total > 0; ++i) { // fill the `i`-digit parts
                int len = 3 + i + d, diff = limit - len; // the tag part takes `len` space. We fill rest `diff` characters with characters from `s`
                if (diff <= 0) break; // can't consume any characters from `s`. It's meaningless to add more parts. Break
                int parts = min(total / diff + (total % diff > 0), cnt[i - 1]); // we can take this many `i`-digit parts.
                totalParts += parts;
                total -= parts * diff;
            }
            if (totalParts == 0) break; // We can't fill any parts. It's impossible to split.
            if (total <= 0) { // We've found a split
                total = N;
                for (int i = 1, id = 1, j = 0; i <= d && total > 0; ++i) {
                    int len = 3 + i + d, diff = limit - len, parts = min(total / diff + (total % diff > 0), cnt[i - 1]);
                    for (int k = 0; k < parts; ++k) {
                        ans.push_back(s.substr(j, diff) + "<" + to_string(id++) + "/" + to_string(totalParts) + ">");
                        j += diff;
                    }
                    total -= parts * diff;
                }
                break;
            }
            cnt.push_back(cnt.back() * 10);
        }
        return ans;
    }
};