Today I have done leetcode's March LeetCoding Challenge with cpp.

March LeetCoding Challenge 30

Description

Partition Labels

You are given a string s. We want to partition the string into as many parts as possible so that each letter appears in at most one part. For example, the string "ababcc" can be partitioned into ["abab", "cc"], but partitions such as ["aba", "bcc"] or ["ab", "ab", "cc"] are invalid.

Note that the partition is done so that after concatenating all the parts in order, the resultant string should be s.

Return a list of integers representing the size of these parts.

Example 1:

Input: s = "ababcbacadefegdehijhklij"
Output: [9,7,8]
Explanation:
The partition is "ababcbaca", "defegde", "hijhklij".
This is a partition so that each letter appears in at most one part.
A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits s into less parts.

Example 2:

Input: s = "eccbbbbdec"
Output: [10]

Constraints:

• 1 <= s.length <= 500
• s consists of lowercase English letters.

Solution

class Solution {
public:
  vector<int> partitionLabels(string s) {
    vector<pair<int, int>> pos(26, {-1, -1});
    for(int i = 0; i < s.length(); ++i) {
      if(pos[s[i] - 'a'].first == -1) {
        pos[s[i] - 'a'].first = i;
      }
      pos[s[i] - 'a'].second = i;
    }
    sort(pos.begin(), pos.end());
    vector<int> answer;
    int begin = pos.front().first;
    int end = pos.front().second;
    for(auto [b, e] : pos) {
      if(b > end) {
        if(begin != -1) {
          answer.push_back(end - begin + 1);
        }
        begin = b;
        end = e;
      } else {
        end = max(end, e);
      }
    }
    answer.push_back(end - begin + 1);
    return answer;
  }
};

// Accepted
// 118/118 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 71.56 % of cpp submissions (8.8 MB)