17. Partition Labels

A string S of lowercase English letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and 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.

Solution:

Approach:

We keep track of max last occurrence of a new character from start. When we reach the end of occurrence of all characters we push and increase start

class Solution
{
public:
    vector<int> partitionLabels(string s)
    {

        vector<int> res;
        vector<int> v(26, 0);
        
        for (int i = 0; i < s.length(); i++)
        {
            v[s[i] - 97] = i;
        }

        int start = 0, end = 0;

        for (int i = 0; i < s.length(); i++)
        {
            end = max(end, v[s[i] - 97]);

            if (end == i)
            {
                int x = (end - start) + 1;
                res.push_back(x);
                start = i + 1;
                end = i + 1;
            }
        }

        return res;
    }
};

Time Complexity: O(n)