[leetcode-763] Partition Labels

Description

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.

Solution

1. Select a partition with the first and last occurrence of a letter, mark them as start and end of a list respectively.
2. Iterate all letters within the list.
3. If found letter have last occurrence bigger than end, update the end accordingly.
4. If end was reached, update count with end-start+1
5. Start new round with end + 1

code:

class Solution {
     
public:
    vector<int> partitionLabels(string S) {
     
        vector<int> result;
        int lastOccurence[26];
        
        for(int i = 0; i < S.length(); i++) {
     
            lastOccurence[S[i] - 'a'] = i;
        }
        
        int i = 0;
        while(i < S.length()) {
     
            int end = lastOccurence[S[i] - 'a'];
            
            int j = i;
            
            while(j<end) {
     
                end = max(end, lastOccurence[S[j] - 'a']);
                j++;
            }
            
            int count = j - i + 1;
            result.push_back(count);
            
            i = j + 1;
        }
        
        return result;
        
    }
};